Aimathic
Login | English | Deutsch

Free Math Worksheets

Build your own math worksheets from 21,000 problems for grades 3 to 12, from fractions to calculus. Every problem includes step-by-step solutions.

Multi-step word problems

Click problems to add them to your worksheet.

5161464
Four children watch a \(90\)-minute movie together. Leo calculates \(90 \div 4 = 22.5\) and says, “Because there are four of us, each person watched only \(22.5\) minutes of the movie.” Explain why Leo’s calculation does not make sense in this situation. How long did each child actually watch the movie?

Hints

- Does the screen show the movie for only one fourth of the time to each child? - Are the children watching one after another or at the same time? - In what kind of situation would dividing a time by \(4\) make sense?

Solution

1. The children watch the movie at the same time, so the movie’s duration is not divided among the viewers. 2. Dividing \(90\) by \(4\) would make sense only if the \(90\) minutes were being split into four separate time periods. 3. Each child watched the entire \(90\)-minute movie.

Answer

Leo’s calculation does not make sense because all four children watched the movie at the same time. Each child watched \(90\) minutes.
5161474
A family of five drives \(420\) miles from home to a vacation cabin. A daughter says, “Altogether, our family traveled \(2100\) miles today.” 1. How did she get \(2100\)? 2. Does \(2100\) miles tell how far the cabin is from home? Explain.

Hints

- Which operation could combine the distance and the number of family members to produce \(2100\)? - Does the car’s odometer change differently when another passenger gets in? - Focus on the distance between the two locations.

Solution

1. She multiplied the driving distance by the number of family members: \(420 \times 5 = 2100\). 2. The car traveled the route once. The number of passengers does not change the distance between the two places. 3. The cabin is \(420\) miles from home, not \(2100\) miles.

Answer

1. She calculated \(420 \times 5 = 2100\). 2. No. The cabin is \(420\) miles from home because the distance does not change with the number of passengers.
5162964
At a grocery store, \(3\) boxes of pasta cost \(\$6.00\). Leon wants to find the total cost of \(5\) boxes of the pasta and one jar of tomato sauce. Explain why he cannot find the total from the information given.

Hints

- List all the items Leon plans to buy. - Can you find the cost of the pasta? - Is the price known for every item? - What information is needed to add the complete total?

Solution

1. Find the price of one box of pasta: \(\$6.00 \div 3 = \$2.00\). 2. Find the price of five boxes: \(5 \times \$2.00 = \$10.00\). 3. The price of the jar of tomato sauce is not given. 4. Therefore, the total cost cannot be determined.

Answer

Leon cannot find the total because the price of the tomato sauce is missing.
5165954
Park Elementary offers several after-school sports clubs. The table shows how many students are enrolled in each club. <table> <tr><th>Club</th><th>Boys</th><th>Girls</th></tr> <tr><td>Basketball</td><td>\(24\)</td><td>\(18\)</td></tr> <tr><td>Gymnastics</td><td>\(12\)</td><td>\(35\)</td></tr> <tr><td>Swimming</td><td>\(22\)</td><td>\(26\)</td></tr> </table> a) How many boys are enrolled in the gymnastics club? b) How many students are enrolled in the basketball club altogether? c) How many girls are enrolled in the three clubs altogether?

Hints

- Use the row labels to find a club and the column labels to find a group of students. - For part a), you only need to read one entry from the table. - When a question asks for a total, add the relevant entries.

Solution

1. Read the entry in the Boys column for Gymnastics: \(12\). 2. Add the numbers of boys and girls in Basketball: \(24 + 18 = 42\). 3. Add the entries in the Girls column: \(18 + 35 + 26 = 79\).

Answer

a) \(12\) boys b) \(42\) students c) \(79\) girls
5165964
A snack stand by a lake sells popsicles on weekends. The table shows the number of each flavor sold on Saturday and Sunday. <table> <tr><th>Flavor</th><th>Saturday</th><th>Sunday</th></tr> <tr><td>Chocolate</td><td>\(156\)</td><td>\(213\)</td></tr> <tr><td>Strawberry</td><td>\(98\)</td><td>\(145\)</td></tr> <tr><td>Vanilla</td><td>\(112\)</td><td>\(128\)</td></tr> </table> a) How many popsicles were sold on Saturday altogether? b) Which flavor sold the most on Sunday? c) How many strawberry popsicles were sold over the entire weekend?

Hints

- Which column lists the sales for Saturday? - For part b), compare all the numbers in the Sunday column. - For part c), combine the two entries in one row.

Solution

1. Add the three Saturday sales: \(156 + 98 + 112 = 366\). 2. Compare the Sunday sales: \(213\) chocolate, \(145\) strawberry, and \(128\) vanilla. The greatest value is \(213\), so chocolate sold the most. 3. Add the strawberry sales from both days: \(98 + 145 = 243\).

Answer

a) \(366\) popsicles b) Chocolate c) \(243\) strawberry popsicles
5166314
The table shows the populations of three towns in 2010 and 2020, along with projections for 2030. <table> <thead> <tr> <th>Town</th> <th>2010</th> <th>2020</th> <th>2030 projection</th> </tr> </thead> <tbody> <tr> <td>Pineville</td> <td>\(45{,}600\)</td> <td>\(48{,}200\)</td> <td>\(51{,}000\)</td> </tr> <tr> <td>Oakton</td> <td>\(32{,}100\)</td> <td>\(30{,}900\)</td> <td>\(29{,}500\)</td> </tr> <tr> <td>Lakeview</td> <td>\(54{,}300\)</td> <td>\(56{,}700\)</td> <td>\(55{,}200\)</td> </tr> </tbody> </table> a) In which town does the population decrease from each listed year to the next? b) By how many people is Pineville’s population projected to increase from 2010 to 2030?

Hints

- For part a), read each row from left to right and check whether both changes are decreases. - For part b), identify the entries for the two requested years. - Use subtraction to find the difference between two populations.

Solution

1. Compare each row from left to right. Oakton decreases at both steps because \(32{,}100 > 30{,}900 > 29{,}500\). Pineville increases at both steps, and Lakeview increases and then decreases. 2. Subtract Pineville’s 2010 population from its projected 2030 population: \(51{,}000 - 45{,}600 = 5400\).

Answer

a) Oakton b) \(5400\) people
5173914
A tower is \(68\,\text{m}\) tall. One story of an apartment building is about \(3\,\text{m}\) tall. 1. Estimate how many stories tall the tower is. 2. Round the tower’s height to the nearest ten meters.

Hints

- Find a nearby multiple of \(3\) that is easy to divide. - Estimate how many groups of \(3\) meters fit into \(68\) meters. - Use the ones digit to round to the nearest ten.

Solution

1. Since \(66\div3=22\) and \(69\div3=23\), \(68\div3\) is about \(23\). The tower is about \(23\) stories tall. 2. The ones digit of \(68\) is \(8\), so \(68\,\text{m}\) rounds to \(70\,\text{m}\).

Answer

1. About \(23\) stories 2. \(70\,\text{m}\)
5195504
A beekeeper has \(360\) jars of honey and packs them into cartons that hold \(6\) jars each. a) How many cartons can the beekeeper fill? b) If the beekeeper uses larger cartons that hold \(9\) jars each, will more or fewer cartons be needed? Explain without calculating the exact number of larger cartons.

Hints

- Divide \(360\) by \(6\) for part a. - For part b, compare the two carton capacities. - Think about how group size affects the number of groups when the total stays the same.

Solution

1. For part a, divide the number of jars by the number in each carton: \(360 \div 6 = 60\). 2. For part b, each larger carton holds more jars. Packing the same total into larger groups produces fewer groups, so fewer cartons are needed.

Answer

a) The beekeeper can fill \(60\) cartons. b) Fewer cartons are needed because each carton holds more jars.
5205714
A bakery has \(1500\) fresh rolls in the morning. It sells \(900\) rolls before noon and then bakes \(400\) more. A student writes: Step 1: \(1500-900=600\) Step 2: \(600+400=1000\) Explain what the \(600\) represents. How many rolls are in the bakery after the additional baking?

Hints

- Identify the two quantities used in the first subtraction. - Sold rolls are no longer in the bakery. - The second step adds the rolls baked later.

Solution

1. The calculation \(1500-900=600\) subtracts the rolls sold from the starting amount. 2. Therefore, \(600\) represents the rolls remaining after the morning sales. 3. Adding the newly baked rolls gives \(600+400=1000\) rolls.

Answer

The \(600\) represents the rolls left after the morning sales. After baking more, the bakery has \(1000\) rolls.
5209084
An old school building was constructed in \(1894\). A large celebration was held for its \(125\)th anniversary. a) In what year was the celebration held? b) In \(2025\), how many years had passed since the building was constructed?

Hints

- An anniversary year is found by adding the anniversary number to the starting year. - To find elapsed years, subtract the earlier year from the later year. - Use \(1894\) as the starting year for both parts.

Solution

1. Find the anniversary year: \(1894 + 125 = 2019\). 2. Find the elapsed years by \(2025\): \(2025 - 1894 = 131\) years.

Answer

a) The celebration was held in \(2019\). b) In \(2025\), \(131\) years had passed since construction.
5362864
At one vertex of a rectangular prism, three edges have lengths \(5\,\text{cm}\), \(4\,\text{cm}\), and \(3\,\text{cm}\). a) What is the sum of these three edge lengths? b) A rectangular prism has \(12\) edges. Find the total length of all its edges.
Figure for problem 536286

Hints

- A rectangular prism has four edges of each dimension. - Use your result from part a) to find the total for all four sets.

Solution

1. The three edge lengths at one vertex add to \(5\,\text{cm}+4\,\text{cm}+3\,\text{cm}=12\,\text{cm}\). 2. A rectangular prism has four edges of each length. 3. The total edge length is \(4 \times (5\,\text{cm}+4\,\text{cm}+3\,\text{cm})=4 \times 12\,\text{cm}=48\,\text{cm}\).

Answer

a) \(12\,\text{cm}\) b) \(48\,\text{cm}\)
5374364
A five-week calendar diagram shows \(35\) days. Ten weekend days are marked in gray, and \(3\) additional vacation days are marked in red. None of the vacation days falls on a weekend. How many school days are shown?
Figure for problem 537436

Hints

- The vacation days do not overlap the weekend days. - Subtract both types of days off from \(35\).

Solution

1. Start with all \(35\) days. 2. Remove the \(10\) weekend days and the \(3\) separate vacation days: \(35-10-3=22\).

Answer

The calendar shows \(22\) school days.
5156534
Mr. Weber buys \(15\) packs of colored pencils for \(\$4\) each and \(20\) sketch pads for \(\$3\) each for the school art club. How much does he spend altogether?

Hints

- Find the cost of each group of supplies. - Then add the two costs. - Break the problem into smaller steps.

Solution

1. Find the cost of the colored pencils: \(15 \times 4 = 60\), so they cost \(\$60\). 2. Find the cost of the sketch pads: \(20 \times 3 = 60\), so they cost \(\$60\). 3. Add the two costs: \(\$60 + \$60 = \$120\).

Answer

Mr. Weber spends \(\$120\) altogether.
5156544
A third-grade class visits a museum. There are \(25\) students and \(4\) adult chaperones. Admission costs \(\$6\) per student and \(\$9\) per chaperone. What is the total admission cost for the group?

Hints

- Find the cost for each group separately. - Use the different admission price for each group. - Add the two group costs.

Solution

1. Find the admission cost for the students: \(25 \times 6 = 150\), so the students cost \(\$150\). 2. Find the admission cost for the chaperones: \(4 \times 9 = 36\), so the chaperones cost \(\$36\). 3. Add the two costs: \(\$150 + \$36 = \$186\).

Answer

The total admission cost is \(\$186\).
5156554
A gardener delivers \(8\) boxes with \(25\) flower bulbs in each box. Twenty bulbs are damaged and cannot be planted. The remaining bulbs are divided equally among \(6\) flower beds. How many bulbs are planted in each bed?

Hints

- First find the total number of bulbs. - Subtract the bulbs that cannot be used. - Equal sharing indicates division.

Solution

1. Find the total number delivered: \(8 \times 25 = 200\). 2. Remove the damaged bulbs: \(200 - 20 = 180\). 3. Divide the remaining bulbs equally: \(180 \div 6 = 30\).

Answer

\(30\) flower bulbs are planted in each bed.
5156564
Paul has \(160\) marbles. He gives half of them to his younger brother. His brother then gives \(15\) marbles to a friend. How many marbles does Paul’s brother have left?

Hints

- First find how many marbles the brother receives. - Giving marbles away decreases his amount. - Solve the problem in two steps.

Solution

1. Find half of the marbles: \(160 \div 2 = 80\). 2. Subtract the marbles given to the friend: \(80 - 15 = 65\).

Answer

Paul’s brother has \(65\) marbles left.
5156574
A movie theater has \(720\) seats. Half of the seats are reserved for an afternoon show. Just before the show, the theater sells another \(85\) tickets. How many seats are now reserved or sold?

Hints

- First find half of the total seats. - The tickets sold add to the occupied seats. - Combine the reserved and sold seats.

Solution

1. Find the number reserved: \(720 \div 2 = 360\). 2. Add the tickets sold: \(360 + 85 = 445\).

Answer

\(445\) seats are reserved or sold.
5156584
A gardener has \(840\) flower bulbs. On Monday, the gardener plants exactly half of them. On Tuesday, the gardener plants another \(155\) bulbs. How many bulbs remain unplanted?

Hints

- Find the number left after Monday. - Planting more bulbs decreases the amount remaining. - The question asks for the bulbs not yet planted.

Solution

1. After half are planted, \(840 \div 2 = 420\) bulbs remain. 2. Subtract the bulbs planted Tuesday: \(420 - 155 = 265\).

Answer

\(265\) flower bulbs remain unplanted.
5156594
A school library installs \(3\) new bookcases. Each bookcase holds exactly \(85\) books. Students have already placed \(140\) books on the new bookcases. How many more books can the bookcases hold altogether?

Hints

- First find how many books all three bookcases can hold. - Then subtract the books already on the shelves. - Use the capacity of one bookcase to find the total capacity.

Solution

1. Find the total capacity of the three bookcases: \(3 \times 85 = 255\). 2. Subtract the books already placed on the shelves: \(255 - 140 = 115\).

Answer

The bookcases can hold \(115\) more books.
5156604
For a school fair, Ms. Miller buys \(5\) cases with \(12\) bottles of apple juice in each case and \(4\) cases with \(20\) bottles of water in each case. Guests drink \(85\) bottles during the fair. How many bottles are left?

Hints

- Find the number of juice bottles and water bottles separately. - Add to find the total number purchased. - Subtract the number that guests drank.

Solution

1. Find the number of juice bottles: \(5 \times 12 = 60\). 2. Find the number of water bottles: \(4 \times 20 = 80\). 3. Find the total number of bottles: \(60 + 80 = 140\). 4. Subtract the bottles that were consumed: \(140 - 85 = 55\).

Answer

There are \(55\) bottles left.
5156614
A gardener has \(400\) tulip bulbs. The gardener plants \(6\) small beds with \(45\) bulbs in each bed. The remaining bulbs are divided equally between \(2\) large planters. How many bulbs go in each planter?

Hints

- First find the number planted in the small beds. - Subtract to find the number left. - Divide the remaining bulbs equally between two planters.

Solution

1. Find the number planted in the beds: \(6 \times 45 = 270\). 2. Find the number remaining: \(400 - 270 = 130\). 3. Divide equally between the planters: \(130 \div 2 = 65\).

Answer

Each planter receives \(65\) tulip bulbs.
5156634
Two classes bake muffins for a school fair. One class fills \(4\) trays with \(12\) muffins on each tray. The other class bakes \(55\) muffins. Which class bakes more muffins, and how many more does it bake?

Hints

- Find the total number of muffins on the four trays. - Compare that total with \(55\). - Subtract to find the difference.

Solution

1. Find the number baked by the first class: \(4 \times 12 = 48\). 2. Compare the amounts: \(55 > 48\), so the second class bakes more. 3. Find the difference: \(55 - 48 = 7\).

Answer

The second class bakes more muffins. It bakes \(7\) more muffins.
5156644
A gardener plants \(6\) flower beds with \(15\) red tulips in each bed and \(4\) flower beds with \(20\) yellow tulips in each bed. How many tulips does the gardener plant altogether?

Hints

- Find the total number of red tulips. - Find the total number of yellow tulips. - Add the two totals.

Solution

1. Find the number of red tulips: \(6 \times 15 = 90\). 2. Find the number of yellow tulips: \(4 \times 20 = 80\). 3. Add the two amounts: \(90 + 80 = 170\).

Answer

The gardener plants \(170\) tulips altogether.
5157404
Students sell used toys at a school fundraiser. The school receives \(\$3\) for each toy sold. The table shows how many toys six classes sell. <table> <tr> <td>Class</td> <td>A</td> <td>B</td> <td>C</td> <td>D</td> <td>E</td> <td>F</td> </tr> <tr> <td>Toys sold</td> <td>34</td> <td>41</td> <td>38</td> <td>45</td> <td>42</td> <td>50</td> </tr> </table> a) How many toys do the six classes sell altogether? b) How much money does the school receive?

Hints

- First add the numbers in the table. - The school receives the same amount for each toy. - Multiply the total number of toys by \(3\).

Solution

1. Add the numbers of toys sold: \(34 + 41 + 38 + 45 + 42 + 50 = 250\). 2. Multiply by the amount received for each toy: \(250 \times 3 = 750\).

Answer

a) The classes sell \(250\) toys altogether. b) The school receives \(\$750\).
5157424
Four teams collect boxes of recyclable materials for a school garden project. The school receives a \(\$2\) recycling credit for each box. <table> <tr> <td>Team</td> <td>Blue</td> <td>Yellow</td> <td>Green</td> <td>Red</td> </tr> <tr> <td>Boxes</td> <td>115</td> <td>122</td> <td>108</td> <td>135</td> </tr> </table> a) How many boxes do the four teams collect altogether? b) What is the total recycling credit?

Hints

- First add the numbers of boxes collected by all four teams. - Each box is worth \(\$2\). - Double the total number of boxes to find the credit.

Solution

1. Add the numbers of boxes: \(115 + 122 + 108 + 135 = 480\). 2. Multiply by the credit for each box: \(480 \times 2 = 960\).

Answer

a) The teams collect \(480\) boxes altogether. b) The school receives a \(\$960\) recycling credit.
5158294
Lucas has \(82\) stickers. He puts \(9\) stickers on each full page of his album. a) How many pages can he fill completely? b) How many stickers will be on the next page? c) How many more stickers does Lucas need to fill that page?

Hints

- Divide the total number of stickers by the number on each full page. - What does the remainder tell you about the next page? - Subtract the number already on that page from \(9\).

Solution

1. Divide to find the quotient and remainder: \(82 \div 9 = 9\) remainder \(1\). 2. The quotient shows that Lucas can fill \(9\) pages completely. 3. The remainder shows that \(1\) sticker will be on the next page. 4. Subtract to find how many more stickers are needed: \(9 - 1 = 8\).

Answer

a) Lucas can fill \(9\) pages completely. b) \(1\) sticker will be on the next page. c) Lucas needs \(8\) more stickers to fill that page.
5158404
A trampoline park sells jump passes at these prices: <table> <tr><th>Number of passes</th><th>Price</th></tr> <tr><td>\(1\) pass</td><td>\(\$4\)</td></tr> <tr><td>\(3\) passes</td><td>\(\$10\)</td></tr> <tr><td>\(5\) passes</td><td>\(\$16\)</td></tr> </table> a) What is the least expensive way to buy exactly \(4\) passes, and how much does it cost? b) The Miller family needs exactly \(8\) passes. Find two different ways to buy them and calculate the price of each way.

Hints

- Combine the package sizes to make exactly the required number of passes. - Try more than one combination. - Add the prices and compare the totals.

Solution

1. For \(4\) passes, one \(3\)-pass package plus one single pass costs \(10 + 4 = 14\) dollars. This costs less than four single passes. 2. For \(8\) passes, one \(5\)-pass package plus one \(3\)-pass package costs \(16 + 10 = 26\) dollars. 3. Another way is two \(3\)-pass packages plus two single passes, which costs \(10 + 10 + 4 + 4 = 28\) dollars.

Answer

a) One \(3\)-pass package and one single pass cost \(\$14\). b) One \(5\)-pass package plus one \(3\)-pass package costs \(\$26\). Two \(3\)-pass packages plus two single passes cost \(\$28\).
5158424
Follow the steps in each number riddle. a) Double \(300\), then add \(150\). b) Find half of \(800\), then subtract \(60\).

Hints

- Complete the operations in the order given. - Decide whether the first step asks you to double or halve. - Use the result of the first step in the second step.

Solution

1. a) \(300 \times 2 = 600\), and \(600 + 150 = 750\). 2. b) \(800 \div 2 = 400\), and \(400 - 60 = 340\).

Answer

a) \(750\) b) \(340\)
5158514
Suppose you want to save exactly \(1000\) cents. a) How many dollars is that? b) You save one dime each day. How many days will it take to save \(1000\) cents? c) Is that more or less than half a year? Use \(182\) days for half a year.

Hints

- Use \(100\) cents in one dollar and \(10\) cents in one dime. - Divide the total number of cents by the amount saved each day. - Compare your number of days with \(182\).

Solution

1. Since \(100\) cents equals \(1\) dollar, \(1000 \div 100 = 10\). Therefore, \(1000\) cents is \(\$10\). 2. One dime is \(10\) cents. Divide the goal by the daily amount: \(1000 \div 10 = 100\) days. 3. Compare \(100\) and \(182\). Since \(100 < 182\), the saving time is less than half a year.

Answer

a) \(\$10\) b) \(100\) days c) Less than half a year
5158524
A robotics club builds robots from blocks. Each package contains \(20\) blocks. Complete the table and determine how many packages contain \(1000\) blocks. | Packages | Blocks | | :--- | :--- | | \(1\) | \(20\) | | \(2\) | \(...\) | | \(10\) | \(...\) | | \(...\) | \(1000\) |

Hints

- Use the number of blocks in one package. - Multiply to find the number in \(2\) and \(10\) packages. - Divide \(1000\) by \(20\) to find the number of packages.

Solution

1. Two packages contain \(2 \times 20 = 40\) blocks. 2. Ten packages contain \(10 \times 20 = 200\) blocks. 3. Since \(1000 \div 20 = 50\), \(50\) packages contain \(1000\) blocks.

Answer

| Packages | Blocks | | :--- | :--- | | \(1\) | \(20\) | | \(2\) | \(40\) | | \(10\) | \(200\) | | \(50\) | \(1000\) |
5159444
Ms. Miller has a \(\$150\) classroom-supply budget. She buys \(7\) packages of construction paper for \(\$12\) each and \(6\) watercolor sets for \(\$9\) each. She uses the remaining money to buy glue sticks costing \(\$2\) each. How many glue sticks can she buy?

Hints

- Find the cost of the first two types of supplies. - Subtract that total from the budget. - Determine how many \(\$2\) glue sticks fit within the remaining amount.

Solution

1. Find the cost of the construction paper: \(7 \times \$12 = \$84\). 2. Find the cost of the watercolor sets: \(6 \times \$9 = \$54\). 3. Find the total spent: \(\$84 + \$54 = \$138\). 4. Find the money remaining: \(\$150 - \$138 = \$12\). 5. Find the number of glue sticks: \(\$12 \div \$2 = 6\).

Answer

Ms. Miller can buy \(6\) glue sticks.
5161274
The Sunny Scoop ice cream shop has this menu: <table> <tr><th>Ice cream</th><th>Price</th><th>Drinks</th><th>Price</th></tr> <tr><td>Fruit sundae</td><td>\(\$5.80\)</td><td>Apple juice</td><td>\(\$2.40\)</td></tr> <tr><td>Chocolate sundae</td><td>\(\$6.20\)</td><td>Bottled water</td><td>\(\$1.90\)</td></tr> <tr><td>Vanilla sundae</td><td>\(\$5.50\)</td><td>Iced tea</td><td>\(\$2.70\)</td></tr> <tr><td>Strawberry sundae</td><td>\(\$6.50\)</td><td>Lemonade</td><td>\(\$2.30\)</td></tr> </table> a) Leonie buys a vanilla sundae and an iced tea. How much does she pay? b) What is the price difference between the most expensive and least expensive sundaes? c) Jonas has \(\$10.00\). He wants a chocolate sundae and one drink. Which drinks can he afford?

Hints

- Use the table to find each price. - For part b), identify the greatest and least sundae prices. - For part c), find how much money remains after buying the sundae.

Solution

1. Leonie pays \(\$5.50 + \$2.70 = \$8.20\). 2. The strawberry sundae costs \(\$6.50\), and the vanilla sundae costs \(\$5.50\). The difference is \(\$6.50 - \$5.50 = \$1.00\). 3. After buying the chocolate sundae, Jonas has \(\$10.00 - \$6.20 = \$3.80\) left. Every drink costs less than \(\$3.80\), so he can afford any drink.

Answer

a) \(\$8.20\) b) \(\$1.00\) c) Apple juice, bottled water, iced tea, or lemonade
5161294
The Fisher family goes out for lunch. The menu lists these prices: <table> <tr><th>Item</th><th>Price</th></tr> <tr><td>Kids' meal</td><td>\(\$4.50\)</td></tr> <tr><td>Baked ziti</td><td>\(\$6.80\)</td></tr> <tr><td>Turkey entrée</td><td>\(\$9.20\)</td></tr> <tr><td>Soup of the day</td><td>\(\$3.90\)</td></tr> <tr><td>Large salad</td><td>\(\$7.50\)</td></tr> </table> The family has two adults and two children. Each child orders a kids' meal. The adults order one turkey entrée and one baked ziti. All four people also order soup. Is \(\$50\) enough to pay the bill? Justify your answer.

Hints

- Find the cost of each group of items. - Add the subtotals. - Compare the total with \(\$50\).

Solution

1. The two kids' meals cost \(2 \times \$4.50 = \$9.00\). 2. The adults' entrées cost \(\$9.20 + \$6.80 = \$16.00\). 3. Four soups cost \(4 \times \$3.90 = \$15.60\). 4. The total is \(\$9.00 + \$16.00 + \$15.60 = \$40.60\). 5. Since \(\$40.60 < \$50.00\), \(\$50\) is enough. The change is \(\$50.00 - \$40.60 = \$9.40\).

Answer

Yes. The bill is \(\$40.60\), so \(\$50\) is enough and \(\$9.40\) remains.
5161424
Two signs along the same bike route show distances to Salem and Albany. Sign 1: Salem \(54\,\text{miles}\), Albany \(92\,\text{miles}\) Sign 2: Salem \(38\,\text{miles}\), Albany \(76\,\text{miles}\) a) Use each sign to find the distance from Salem to Albany. b) Compare your results. What do you notice?

Hints

- On each sign, subtract the smaller distance from the larger distance. - Does the distance between two fixed cities change as you move closer to both of them? - Compare the two results from part a).

Solution

1. Using Sign 1, subtract the distance to Salem from the distance to Albany: \(92 - 54 = 38\,\text{miles}\). 2. Using Sign 2, subtract again: \(76 - 38 = 38\,\text{miles}\). 3. Both signs give the same distance because the distance between two fixed places does not change as the rider moves along the route.

Answer

a) Sign 1 gives \(38\,\text{miles}\), and Sign 2 gives \(38\,\text{miles}\). b) Both results are the same. Salem and Albany are \(38\,\text{miles}\) apart.
5161434
A highway sign lists the distances to three towns ahead: Pine City: \(155\,\text{miles}\) Lakeview: \(82\,\text{miles}\) Riverdale: \(15\,\text{miles}\) a) What is the distance from Riverdale to Lakeview? b) What is the distance from Lakeview to Pine City? c) Which of those two sections is longer, and by how many miles?

Hints

- Imagine the three towns in order along one line. - Subtract the smaller sign distance from the larger sign distance to find each section. - Compare the two section lengths and subtract to find the difference.

Solution

1. Find the distance from Riverdale to Lakeview: \(82 - 15 = 67\,\text{miles}\). 2. Find the distance from Lakeview to Pine City: \(155 - 82 = 73\,\text{miles}\). 3. Since \(73 > 67\), the Lakeview-to-Pine City section is longer. 4. Find the difference: \(73 - 67 = 6\,\text{miles}\).

Answer

a) The distance is \(67\,\text{miles}\). b) The distance is \(73\,\text{miles}\). c) The Lakeview-to-Pine City section is longer by \(6\,\text{miles}\).
5161484
A sporting goods store lists these prices: <table> <tr><td>Inline skates</td><td>\(\$59\)</td></tr> <tr><td>Helmet</td><td>\(\$24\)</td></tr> <tr><td>Knee pads</td><td>\(\$12\)</td></tr> <tr><td>Elbow pads</td><td>\(\$9\)</td></tr> </table> Sarah receives \(\$100\) for her birthday and buys the inline skates and helmet. a) How much does she pay for those two items? b) How much money remains? c) Is the remaining money enough to buy both the knee pads and elbow pads? Explain.

Hints

- Add the prices of the skates and helmet. - Subtract that total from \(\$100\). - Add the prices of the knee pads and elbow pads. - Compare the pad cost with the money remaining.

Solution

1. The skates and helmet cost \(\$59 + \$24 = \$83\). 2. The money remaining is \(\$100 - \$83 = \$17\). 3. The knee pads and elbow pads cost \(\$12 + \$9 = \$21\). 4. Since \(\$21 > \$17\), the remaining money is not enough.

Answer

a) \(\$83\) b) \(\$17\) c) No. The pads cost \(\$21\), but Sarah has only \(\$17\) left.
5161494
Tim and Anna visit a zoo gift shop. Each has saved \(\$25\). <table> <tr><td>Stuffed tiger</td><td>\(\$18\)</td></tr> <tr><td>Elephant figure</td><td>\(\$7\)</td></tr> <tr><td>Zoo puzzle</td><td>\(\$12\)</td></tr> <tr><td>Postcard set</td><td>\(\$4\)</td></tr> </table> Tim buys a stuffed tiger and a postcard set. Anna buys two zoo puzzles. Who spends more money? How much money does each person have left?

Hints

- Find each person's total spending separately. - Remember that Anna buys two identical items. - Subtract each total from \(\$25\) to find the money left. - Compare the two spending totals.

Solution

1. Tim spends \(\$18 + \$4 = \$22\) and has \(\$25 - \$22 = \$3\) left. 2. Anna spends \(2 \times \$12 = \$24\) and has \(\$25 - \$24 = \$1\) left. 3. Since \(\$24 > \$22\), Anna spends more.

Answer

Anna spends more. Tim has \(\$3\) left, and Anna has \(\$1\) left.
5161504
Mr. Miller has \(\$70\) to buy books for his classroom reading corner. <table> <tr><td>Mystery book</td><td>\(\$9\)</td></tr> <tr><td>Horse book</td><td>\(\$12\)</td></tr> <tr><td>Comic book</td><td>\(\$6\)</td></tr> <tr><td>Science book</td><td>\(\$15\)</td></tr> </table> He chooses two science books and three mystery books. How much has he spent? How many comic books can he buy with the money left?

Hints

- Find the cost of each group of books. - Subtract the total from \(\$70\). - Determine how many \(\$6\) comic books fit within the remaining budget.

Solution

1. Two science books cost \(2 \times \$15 = \$30\). 2. Three mystery books cost \(3 \times \$9 = \$27\). 3. The books cost \(\$30 + \$27 = \$57\) altogether. 4. He has \(\$70 - \$57 = \$13\) left. 5. Since two comic books cost \(2 \times \$6 = \$12\) and three would cost \(\$18\), he can buy \(2\) comic books.

Answer

He has spent \(\$57\) and can buy \(2\) comic books with the \(\$13\) left.
5162424
A school bus trip costs \(\$13\) per student for \(24\) students. The school’s parent organization contributes \(\$100\) toward the total cost. How much must the students pay altogether?

Hints

- First find the bus cost for the entire class without the contribution. - Subtract the amount paid by the parent organization.

Solution

1. Find the full cost for \(24\) students: \(24 \times \$13 = \$312\). 2. Subtract the contribution: \(\$312 - \$100 = \$212\).

Answer

\(\$212\)
5162434
Four siblings are saving for a trampoline that costs \(\$320\). Each child puts \(\$8\) into a shared savings jar every month. After how many months will they have enough money?

Hints

- How much do all four children save in one month? - How many monthly deposits of that amount are needed to reach \(\$320\)?

Solution

1. Find the amount the four children save each month: \(4 \times \$8 = \$32\). 2. Let \(m\) be the number of months. The total saved is \(\$32 \times m\). Since \(\$32 \times 10 = \$320\), \(m=10\). 3. They will reach the goal after \(10\) months.

Answer

They will have enough money after \(10\) months.
5162534
A movie theater has \(15\) rows with \(20\) seats in each row. All seats are occupied, and \(180\) of the moviegoers are adults. A child’s ticket costs \(\$5.50\), and an adult’s ticket costs \(\$8.50\). The movie starts at \(3{:}30\) p.m. and lasts \(90\) minutes. How many children are in the theater?

Hints

- Find the total number of seats from the rows and seats per row. - What does it mean that all seats are occupied? - Subtract the number of adults from the total number of moviegoers. - Decide which details are not needed.

Solution

1. Find the total number of seats: \(15 \times 20 = 300\). 2. Because all seats are occupied, there are \(300\) moviegoers. 3. Subtract the number of adults: \(300-180=120\). 4. The ticket prices, start time, and movie duration are not needed.

Answer

There are \(120\) children in the theater.
5162854
A truck driver travels \(840\) miles from Monday through Thursday. The driver travels \(195\) miles on Monday, \(210\) miles on Tuesday, and \(185\) miles on Wednesday. The remaining distance is traveled on Thursday. On which day does the driver travel the farthest? How many more miles is that than the distance traveled on Wednesday?

Hints

- First find the distance traveled on Thursday. - Compare all four daily distances. - Subtract Wednesday's distance from the greatest distance.

Solution

1. Add the distances from Monday through Wednesday: \(195 + 210 + 185 = 590\) miles. 2. Find Thursday's distance: \(840 - 590 = 250\) miles. 3. Compare \(195\), \(210\), \(185\), and \(250\). Thursday has the greatest distance. 4. Find the difference between Thursday and Wednesday: \(250 - 185 = 65\) miles.

Answer

The driver travels farthest on Thursday, at \(250\) miles. That is \(65\) miles farther than Wednesday.
5164504
A \(\$1{,}000{,}000\) grant is divided among three programs in each plan. Complete the table so that each row totals exactly \(\$1{,}000{,}000\). <table> <tr><th>Program 1</th><th>Program 2</th><th>Program 3</th></tr> <tr><td>\(\$450{,}000\)</td><td>\(\$250{,}000\)</td><td>?</td></tr> <tr><td>\(\$125{,}000\)</td><td>?</td><td>\(\$375{,}000\)</td></tr> <tr><td>?</td><td>\(\$660{,}000\)</td><td>\(\$140{,}000\)</td></tr> </table>

Hints

- Add the two known amounts in each row. - Subtract that sum from \(\$1{,}000{,}000\). - Check that all three amounts in each row total \(\$1{,}000{,}000\).

Solution

1. In row 1, the known amounts total \(450{,}000 + 250{,}000 = 700{,}000\). The missing amount is \(1{,}000{,}000 - 700{,}000 = 300{,}000\). 2. In row 2, the known amounts total \(125{,}000 + 375{,}000 = 500{,}000\). The missing amount is \(1{,}000{,}000 - 500{,}000 = 500{,}000\). 3. In row 3, the known amounts total \(660{,}000 + 140{,}000 = 800{,}000\). The missing amount is \(1{,}000{,}000 - 800{,}000 = 200{,}000\).

Answer

Row 1: \(\$300{,}000\) Row 2: \(\$500{,}000\) Row 3: \(\$200{,}000\)
5164834
A warehouse receives \(345\) packages on Monday and \(287\) packages on Tuesday. a) How many packages have arrived altogether? b) A large order requires \(900\) packages. How many more packages are needed after Tuesday’s delivery?

Hints

- First decide which operation finds the total delivered. - Then find the difference between the required number and the number already delivered.

Solution

1. Add the two deliveries: \(345+287=632\). A total of \(632\) packages have arrived. 2. Subtract the amount received from the amount needed: \(900-632=268\). 3. The warehouse needs \(268\) more packages.

Answer

a) \(632\) packages b) \(268\) more packages
5165174
The Miller family is taking a ferry to an island. The family has \(2\) adults and \(3\) children. <table> <tr><th>Passenger</th><th>Price per person</th></tr> <tr><td>Adult</td><td>\(\$14\)</td></tr> <tr><td>Child</td><td>\(\$9\)</td></tr> </table> How much does the family pay altogether?

Hints

- Find the adult and child costs separately. - Multiply each price by the number of people. - Add the two subtotals.

Solution

1. The adult tickets cost \(2 \times \$14 = \$28\). 2. The child tickets cost \(3 \times \$9 = \$27\). 3. The total is \(\$28 + \$27 = \$55\).

Answer

The family pays \(\$55\) altogether.
5165194
The Smith family of \(4\) compares two ways to travel to a vacation destination. <table> <tr><th>Transportation</th><th>Cost</th><th>Travel time</th></tr> <tr><td>Train</td><td>\(\$21\) per person</td><td>\(2\) hours</td></tr> <tr><td>Car</td><td>\(\$60\) total</td><td>\(4\) hours</td></tr> </table> How much money would the family save by driving instead of taking the train? How many hours of travel time would the family save by taking the train instead of driving?

Hints

- Find the total train cost for all \(4\) people. - Compare the train cost with the total car cost. - Subtract the travel times to find the time savings.

Solution

1. The train costs \(4 \times \$21 = \$84\) for the family. 2. Driving saves \(\$84 - \$60 = \$24\). 3. Taking the train saves \(4 - 2 = 2\) hours.

Answer

Driving saves \(\$24\). Taking the train saves \(2\) hours of travel time.
5165974
A public library added books to its collection. The table groups the new books by age level, type, and how the library received them. <table> <tr><th></th><th colspan="2">Children (ages 12 and under)</th><th colspan="2">Teens (ages 13 and older)</th></tr> <tr><td></td><td>Nonfiction</td><td>Fiction</td><td>Nonfiction</td><td>Fiction</td></tr> <tr><td>Purchased</td><td>\(1450\)</td><td>\(2380\)</td><td>\(870\)</td><td>\(1560\)</td></tr> <tr><td>Donated</td><td>\(320\)</td><td>\(540\)</td><td>\(150\)</td><td>\(420\)</td></tr> </table> a) How many children’s fiction books were added altogether? b) How many teen nonfiction books were added altogether? c) Which age level received more books altogether?

Hints

- Notice that each age level has two columns. - Identify which columns belong to children and which belong to teens. - For part c), add all four entries for each age level, and then compare the totals.

Solution

1. Add the purchased and donated children’s fiction books: \(2380 + 540 = 2920\). 2. Add the purchased and donated teen nonfiction books: \(870 + 150 = 1020\). 3. Find the children’s total: \(1450 + 2380 + 320 + 540 = 4690\). 4. Find the teen total: \(870 + 1560 + 150 + 420 = 3000\). 5. Since \(4690 > 3000\), the children’s collection received more books.

Answer

a) \(2920\) children’s fiction books b) \(1020\) teen nonfiction books c) The children’s collection
5166264
A technology museum celebrates its \(125\)th anniversary on September 20, 2025. On that date, the museum director is exactly \(62\) years old. a) In what year did the museum open? b) How old was the director when the museum celebrated its \(100\)th anniversary?

Hints

- First find the year the museum opened. - In what year did the museum celebrate its \(100\)th anniversary? - How many years passed between the \(100\)th and \(125\)th anniversaries?

Solution

1. Find the opening year: \(2025 - 125 = 1900\). 2. Find the year of the \(100\)th anniversary: \(1900 + 100 = 2000\). 3. Find the number of years from the \(100\)th anniversary to the \(125\)th anniversary: \(2025 - 2000 = 25\) years. 4. Find the director’s age in \(2000\): \(62 - 25 = 37\).

Answer

a) The museum opened in \(1900\). b) The director was \(37\) years old.
5166274
Grandpa Henry celebrated his \(75\)th birthday in \(2023\). That same year, he also celebrated \(50\) years of working at the same cabinet shop. a) In what year was Grandpa Henry born? b) In what year did he begin working at the cabinet shop?

Hints

- Which year is the starting point for both calculations? - To find a birth year from an age, count backward from the current year. - What does a \(50\)-year work anniversary tell you about the starting year?

Solution

1. Find his birth year: \(2023 - 75 = 1948\). 2. Find the year he began working at the cabinet shop: \(2023 - 50 = 1973\).

Answer

a) Grandpa Henry was born in \(1948\). b) He began working at the cabinet shop in \(1973\).
5166324
A state forestry department tracks newly planted trees in three forest districts. <table> <thead> <tr> <th>District</th> <th>Evergreen trees</th> <th>Deciduous trees</th> </tr> </thead> <tbody> <tr> <td>Oak Forest</td> <td>\(125{,}000\)</td> <td>\(88{,}500\)</td> </tr> <tr> <td>Pine Ridge</td> <td>\(94{,}200\)</td> <td>\(112{,}000\)</td> </tr> <tr> <td>Beech Grove</td> <td>\(108{,}600\)</td> <td>\(95{,}400\)</td> </tr> </tbody> </table> a) How many trees were planted in Pine Ridge altogether? b) How many evergreen trees were planted in the three districts altogether? c) In which district is the difference between the numbers of evergreen and deciduous trees greatest?

Hints

- For part a), add only the two entries in one row. - For part b), add the entries in one column. - For part c), find the difference in each row and compare the results.

Solution

1. Add the two entries for Pine Ridge: \(94{,}200 + 112{,}000 = 206{,}200\). 2. Add the evergreen entries: \(125{,}000 + 94{,}200 + 108{,}600 = 327{,}800\). 3. Find the difference in each district. Oak Forest: \(125{,}000 - 88{,}500 = 36{,}500\). Pine Ridge: \(112{,}000 - 94{,}200 = 17{,}800\). Beech Grove: \(108{,}600 - 95{,}400 = 13{,}200\). The greatest difference is in Oak Forest.

Answer

a) \(206{,}200\) trees b) \(327{,}800\) evergreen trees c) Oak Forest
5166334
The table shows the number of visitors to four large amusement parks in 2022 and 2023. <table> <thead> <tr> <th>Amusement park</th> <th>2022</th> <th>2023</th> </tr> </thead> <tbody> <tr> <td>Sunshine Park</td> <td>\(245{,}000\)</td> <td>\(261{,}000\)</td> </tr> <tr> <td>Adventure World</td> <td>\(198{,}500\)</td> <td>\(192{,}000\)</td> </tr> <tr> <td>Wildlife West</td> <td>\(312{,}400\)</td> <td>\(315{,}000\)</td> </tr> <tr> <td>Water Paradise</td> <td>\(156{,}700\)</td> <td>\(168{,}200\)</td> </tr> </tbody> </table> a) Which park had the most visitors in 2023? b) By how many visitors did attendance at Water Paradise increase from 2022 to 2023? c) Is the combined 2023 attendance at Sunshine Park and Adventure World greater than or less than the 2023 attendance at Wildlife West?

Hints

- For part a), find the greatest entry in the correct column. - Use subtraction to find the increase in part b). - For part c), add two entries first, and then compare the sum with a third entry.

Solution

1. Compare the entries in the 2023 column. The greatest value is \(315{,}000\), for Wildlife West. 2. Subtract the 2022 attendance from the 2023 attendance for Water Paradise: \(168{,}200 - 156{,}700 = 11{,}500\). 3. Add the 2023 attendance at Sunshine Park and Adventure World: \(261{,}000 + 192{,}000 = 453{,}000\). 4. Compare the result with Wildlife West: \(453{,}000 > 315{,}000\), so the combined attendance is greater.

Answer

a) Wildlife West b) \(11{,}500\) visitors c) Greater than
5166404
A state had \(435{,}600\) elementary school students in 2021. In 2022, enrollment increased by \(12{,}850\) students. For 2023, enrollment was projected to decrease by \(9420\) students from the 2022 level. a) What was the projected enrollment for 2023? b) How many more students were projected for 2023 than were enrolled in 2021?

Hints

- Decide whether enrollment increases or decreases at each step. - Work from one year to the next in order. - For part b, compare the 2023 result with the original 2021 value.

Solution

1. Find the 2022 enrollment: \(435{,}600+12{,}850=448{,}450\). 2. Subtract the projected decrease: \(448{,}450-9420=439{,}030\). 3. Compare the 2023 projection with 2021: \(439{,}030-435{,}600=3430\).

Answer

a) The projected 2023 enrollment was \(439{,}030\) students. b) That is \(3430\) more students than in 2021.
5166414
Three large school districts report their elementary school enrollment: - District A has \(84{,}320\) students. - District B has \(15{,}650\) fewer students than District A. - District C has \(7480\) more students than District B. How many elementary school students are enrolled in the three districts altogether?

Hints

- Find the enrollments of Districts B and C before finding the total. - Pay attention to which district each “fewer than” or “more than” statement compares. - Add all three district enrollments at the end.

Solution

1. Find District B’s enrollment: \(84{,}320-15{,}650=68{,}670\). 2. Find District C’s enrollment: \(68{,}670+7480=76{,}150\). 3. Add all three enrollments: \(84{,}320+68{,}670+76{,}150=229{,}140\).

Answer

The three districts have \(229{,}140\) elementary school students altogether.
5166424
A state education department plans for \(950{,}000\) elementary school seats in Grades 1 through 4. Current enrollment is: - Grade 1: \(234{,}500\) students - Grade 2: \(241{,}200\) students - Grade 3: \(238{,}900\) students - Grade 4: \(232{,}150\) students How many seats will remain open if every enrolled student receives a seat?

Hints

- First find the total enrollment across all four grades. - Compare the total enrollment with the total number of seats. - The word “altogether” signals that addition is needed first.

Solution

1. Find the total enrollment: \(234{,}500+241{,}200+238{,}900+232{,}150=946{,}750\). 2. Subtract the enrollment from the available seats: \(950{,}000-946{,}750=3250\).

Answer

\(3250\) seats will remain open.
5166624
Two fruit crates are weighed at a wholesale market. Crate A, filled with apples, has a total mass of \(12{,}450\,\text{g}\). The empty crate has a mass of \(1120\,\text{g}\). Crate B, filled with pears, has a total mass of \(15{,}200\,\text{g}\). The empty crate has a mass of \(1450\,\text{g}\). Which fruit has the greater mass, and what is the difference in grams?

Hints

- Find the mass of the fruit in each crate by subtracting the empty crate's mass. - Compare the two fruit masses. - Subtract to find the difference.

Solution

1. Start with the filled mass of Crate A, \(12{,}450\,\text{g}\). Subtract the empty crate's mass, \(1120\,\text{g}\). The apples have a mass of \(11{,}330\,\text{g}\). 2. Start with the filled mass of Crate B, \(15{,}200\,\text{g}\). Subtract the empty crate's mass, \(1450\,\text{g}\). The pears have a mass of \(13{,}750\,\text{g}\). 3. The pears have the greater mass because \(13{,}750\,\text{g} > 11{,}330\,\text{g}\). 4. Subtract \(11{,}330\,\text{g}\) from \(13{,}750\,\text{g}\). The difference is \(2420\,\text{g}\).

Answer

The pears have a greater mass by \(2420\,\text{g}\).
5166634
A small truck may have a maximum total mass of \(14{,}250\,\text{kg}\). The empty truck has a mass of \(10{,}150\,\text{kg}\). Two machines with a combined mass of \(1240\,\text{kg}\) and a tool pallet with a mass of \(980\,\text{kg}\) are loaded. How many more kilograms can be added without exceeding the limit?

Hints

- Add the empty truck and all items already loaded. - Compare the current mass with the maximum allowed mass. - Subtract to find the remaining capacity.

Solution

1. Find the combined mass of the machines and tool pallet: \(1240\,\text{kg} + 980\,\text{kg} = 2220\,\text{kg}\). 2. Add \(2220\,\text{kg}\) to the empty truck's mass of \(10{,}150\,\text{kg}\). The truck's current mass is \(12{,}370\,\text{kg}\). 3. Subtract the current mass from the limit. Start with \(14{,}250\,\text{kg}\) and subtract \(12{,}370\,\text{kg}\). The remaining capacity is \(1880\,\text{kg}\).

Answer

The truck can carry \(1880\,\text{kg}\) more.
5166784
A horse trailer may have a maximum total mass of \(2000\,\text{kg}\). The empty trailer has a mass of \(840\,\text{kg}\). One horse with a mass of \(565\,\text{kg}\) is already inside. What is the greatest combined mass the second horse and the remaining equipment can have?

Hints

- First find how much cargo the empty trailer can carry. - The first horse uses part of that capacity. - Subtract the first horse's mass from the total payload capacity.

Solution

1. Find the trailer's total payload capacity: \(2000\,\text{kg} - 840\,\text{kg} = 1160\,\text{kg}\). 2. Subtract the first horse's mass: \(1160\,\text{kg} - 565\,\text{kg} = 595\,\text{kg}\).

Answer

The second horse and equipment can have a combined mass of at most \(595\,\text{kg}\).
5167004
A soccer club buys \(6\) new soccer balls at \(\$24\) each and one large goal net for \(\$155\). What is the total cost?

Hints

- Which cost occurs once, and which cost occurs for every ball? - First find the total cost of all the balls. - Then add the net’s price.

Solution

1. Find the cost of the soccer balls: \(6 \times \$24 = \$144\). 2. Add the cost of the goal net: \(\$144 + \$155 = \$299\).

Answer

\(\$299\)
5167014
The Huber family buys \(4\) young apple trees at \(\$142\) each. The nursery charges a single \(\$75\) fee to deliver and plant all the trees. What is the total cost?

Hints

- First find the cost of the \(4\) trees. - The delivery and planting fee is charged only once. - Use multiplication, then addition.

Solution

1. Find the cost of the trees: \(4 \times \$142 = \$568\). 2. Add the delivery and planting fee: \(\$568 + \$75 = \$643\).

Answer

\(\$643\)
5167024
For a school trip, \(26\) students each pay \(\$14\) for admission to an adventure park. The class also pays a single \(\$85\) bus fee. What is the total cost of the trip?

Hints

- Find the admission cost for all the students. - Add the one-time bus fee. - Written multiplication may help with the first step.

Solution

1. Find the total admission cost: \(26 \times \$14 = \$364\). 2. Add the bus fee: \(\$364 + \$85 = \$449\).

Answer

\(\$449\)
5167424
A mouse takes \(163\) breaths per minute, and a guinea pig takes \(90\) breaths per minute. How many breaths do the two animals take altogether in \(10\) minutes?

Hints

- First find how many breaths the two animals take altogether in one minute. - How does that amount change when the time is \(10\) times as long?

Solution

1. Find the combined number of breaths per minute: \(163 + 90 = 253\). 2. Multiply by \(10\) minutes: \(253 \times 10 = 2530\) breaths.

Answer

Together, the two animals take \(2530\) breaths in \(10\) minutes.
5167434
An elephant’s heart beats \(24\) times per minute, a horse’s heart beats \(36\) times per minute, and an adult human’s heart beats \(65\) times per minute. In one hour, how many more times does the human heart beat than the elephant and horse hearts combined?

Hints

- How many minutes are in one hour? - First combine the two animal rates, then scale that rate to one hour. - Compare the animals’ total with the human total.

Solution

1. Find the combined animal heartbeats per minute: \(24 + 36 = 60\). 2. Find the combined animal heartbeats in \(60\) minutes: \(60 \times 60 = 3600\). 3. Find the human heartbeats in \(60\) minutes: \(65 \times 60 = 3900\). 4. Find the difference: \(3900 - 3600 = 300\).

Answer

The human heart beats \(300\) more times in one hour.
5167444
A hedgehog’s heart beats \(280\) times per minute while it is awake and only \(18\) times per minute during hibernation. Over \(20\) minutes, how many fewer heartbeats occur during hibernation than while the hedgehog is awake?

Hints

- How many fewer heartbeats occur in one minute? - Once you know the one-minute difference, how can you find the difference over \(20\) minutes?

Solution

1. Find the difference in heartbeats per minute: \(280 - 18 = 262\). 2. Multiply the difference by \(20\) minutes: \(262 \times 20 = 5240\) heartbeats.

Answer

During hibernation, the hedgehog’s heart beats \(5240\) fewer times over \(20\) minutes.
5167494
A mail carrier drives \(36\,\text{miles}\) each day from Monday through Friday. Her Saturday route is only \(15\,\text{miles}\). How many miles does she drive in \(4\) full weeks?

Hints

- Find the total distance for one week first. - Count how many days she drives the longer route and how many days she drives the shorter route. - Then multiply the weekly distance by \(4\).

Solution

1. Find the distance from Monday through Friday: \(5 \times 36 = 180\,\text{miles}\). 2. Add Saturday's route: \(180 + 15 = 195\,\text{miles}\) per week. 3. Find the distance for four weeks: \(195 \times 4 = 780\,\text{miles}\).

Answer

The mail carrier drives \(780\,\text{miles}\) in \(4\) weeks.
5167504
An airport shuttle travels \(12\,\text{miles}\) one way between a hotel and the airport. Each day, it makes \(8\) complete round trips. How many miles does the shuttle travel in a \(30\)-day month?

Hints

- How far is one complete round trip? - Find the distance traveled in one day. - Then multiply the daily distance by \(30\).

Solution

1. Find the distance of one round trip: \(12 \times 2 = 24\,\text{miles}\). 2. Find the daily distance: \(24 \times 8 = 192\,\text{miles}\). 3. Find the distance for \(30\) days: \(192 \times 30 = 5760\,\text{miles}\).

Answer

The shuttle travels \(5760\,\text{miles}\) in the month.
5167634
A warehouse has \(4\) large shipping crates. Each shipping crate contains \(4\) wooden boxes. Each wooden box contains \(4\) metal cases. Each metal case holds \(4\) watches. a) How many watches are in one shipping crate? b) How many watches are in the warehouse altogether? c) How many containers—shipping crates, wooden boxes, and metal cases—are used altogether?

Hints

- Work from the smallest container up to one shipping crate. - Multiply the watches in one shipping crate by the number of shipping crates. - Count each type of container separately before adding them.

Solution

1. One shipping crate contains \(4\times 4\times 4=64\) watches. 2. With \(4\) shipping crates, the warehouse contains \(64\times 4=256\) watches. 3. There are \(4\) shipping crates, \(4\times 4=16\) wooden boxes, and \(16\times 4=64\) metal cases. 4. The total number of containers is \(4+16+64=84\).

Answer

a) \(64\) watches b) \(256\) watches c) \(84\) containers
5167664
A farm has \(5\) barns. Each barn contains \(8\) haystacks. Each haystack is home to \(12\) mouse families, and each family has \(6\) baby mice. How many baby mice live in the haystacks altogether?

Hints

- Break the problem into smaller steps. - First find the total number of haystacks. - After finding the number of families, multiply by the number of baby mice in each family. - Work from the largest group, barns, down to the smallest group, baby mice.

Solution

1. Find the total number of haystacks: \(5\times 8=40\). 2. Find the total number of mouse families: \(40\times 12=480\). 3. Find the total number of baby mice: \(480\times 6=2880\).

Answer

There are \(2880\) baby mice altogether.
5167694
A television costs \(\$565\) when paid for in full. A payment plan charges \(\$52\) per month for \(12\) months. How much more does the payment plan cost than paying in full?

Hints

- Find the total amount paid over all \(12\) months. - Use multiplication to combine the monthly payments. - Subtract the full-payment price from the payment-plan total.

Solution

1. Find the total cost of the payment plan: \(12 \times \$52 = \$624\). 2. Find the difference from the full-payment price: \(\$624 - \$565 = \$59\).

Answer

The payment plan costs \(\$59\) more.
5167724
A museum sells \(312\) child tickets at \(\$6\) each and \(145\) adult tickets at \(\$9\) each during one morning. What is the museum’s total ticket revenue?

Hints

- Find how much each group of visitors pays in total. - Multiplication gives the revenue when the same price is paid many times. - Add the two revenue amounts.

Solution

1. Find the revenue from child tickets: \(312 \times \$6 = \$1872\). 2. Find the revenue from adult tickets: \(145 \times \$9 = \$1305\). 3. Add the two amounts: \(\$1872 + \$1305 = \$3177\).

Answer

The museum’s total ticket revenue is \(\$3177\).
5167754
Ms. Meyer wants to borrow \(\$2000\). She compares two payment plans: - Plan A: \(12\) monthly payments of \(\$178\) - Plan B: \(24\) monthly payments of \(\$95\) What is the total paid under each plan? Which plan costs less overall, and by how much?

Hints

- How many times is each monthly payment made? - Find the total for each plan separately. - “Costs less” means the smaller total. - Subtract the two totals to find the difference.

Solution

1. Find the total for Plan A: \(12 \times \$178 = \$2136\). 2. Find the total for Plan B: \(24 \times \$95 = \$2280\). 3. Since \(\$2136 < \$2280\), Plan A costs less. 4. Find the difference: \(\$2280 - \$2136 = \$144\).

Answer

Plan A totals \(\$2136\), and Plan B totals \(\$2280\). Plan A costs \(\$144\) less.
5167784
Lucas wants to buy a tablet for \(\$240\). He has saved \(\$60\). He plans to repay the remaining amount to his parents in monthly payments of \(\$20\). a) How much does Lucas still need to repay? b) How many months will it take him to repay the full amount? c) Lucas receives a \(\$25\) monthly allowance. Is paying \(\$20\) each month a realistic plan? Explain.

Hints

- First subtract the amount Lucas has saved from the tablet’s price. - How many monthly payments fit into the remaining balance? - Find how much allowance remains after one payment.

Solution

1. Find the remaining amount: \(\$240 - \$60 = \$180\). 2. Find the number of payments: \(\$180 \div \$20 = 9\), so repayment takes \(9\) months. 3. After each payment, Lucas has \(\$25 - \$20 = \$5\) left. The plan is mathematically possible, but it leaves little money for other expenses.

Answer

a) Lucas still needs to repay \(\$180\). b) He must make payments for \(9\) months. c) The plan is possible, but it leaves him only \(\$5\) each month for anything else.
5167804
Mia is saving for a violin that costs \(\$400\). She saves \(\$15\) each month and plans to buy it after exactly \(2\) years. a) How much will she save in \(24\) months? b) Will she have enough? If not, how much more does she need? c) How much would she need to save each month to have \(\$400\) after exactly \(20\) months?

Hints

- Multiply the monthly savings by the number of months. - Compare the result with the violin’s price. - Find the monthly amount that, when multiplied by \(20\), equals the goal.

Solution

1. Find the amount saved in \(24\) months: \(24 \times \$15 = \$360\). 2. Compare it with the goal: \(\$400 - \$360 = \$40\), so she is short by \(\$40\). 3. Use an inverse multiplication fact for \(20\) months: \(20 \times \$20 = \$400\), so she needs to save \(\$20\) per month.

Answer

a) Mia will save \(\$360\). b) No. She will need \(\$40\) more. c) She would need to save \(\$20\) per month.
5168044
A warehouse has \(8\) pallets of copy paper. Each pallet holds \(25\) cartons. Each carton contains \(10\) reams, and each ream contains \(500\) sheets. How many sheets of paper are in the warehouse altogether?

Hints

- First find the number of cartons on all the pallets. - Then find the total number of reams. - Use place-value reasoning when multiplying by numbers with zeros. - Check how many zeros the final product should contain.

Solution

1. Find the total number of cartons: \(8\times 25=200\). 2. Find the total number of reams: \(200\times 10=2000\). 3. Find the total number of sheets: \(2000\times 500=1{,}000{,}000\).

Answer

The warehouse contains \(1{,}000{,}000\) sheets of paper.
5168864
A family of two adults and three children is taking a ferry to an island. The fares are: - Adult: \(\$22\) - Child: \(\$11\) A family special says that when two adults pay full fare, each child’s fare is only \(\$8\). a) Find the family’s total fare with the special. b) How much does the family save compared with the regular fares?

Hints

- Find the adult and child costs under the family special separately. - To find the savings, compare the special price with the total using regular fares. - Keep track of which child fare applies in each calculation.

Solution

1. With the special, the two adult fares cost \(2 \times \$22 = \$44\), and the three child fares cost \(3 \times \$8 = \$24\). 2. The family’s total with the special is \(\$44 + \$24 = \$68\). 3. At the regular fares, the adults cost \(\$44\), and the children cost \(3 \times \$11 = \$33\), for a total of \(\$77\). 4. The savings are \(\$77 - \$68 = \$9\).

Answer

a) The total fare with the family special is \(\$68\). b) The family saves \(\$9\).
5168874
Three adults and two children, ages 10 and 12, are taking a round-trip train ride. A one-way adult ticket costs \(\$84\). The train company’s family promotion has these rules: - Children under age 14 ride free with adult relatives. - The first adult pays full price. - Each additional adult pays half price. a) Find the total cost for the group’s one-way trip. b) Find the total cost for the round trip.

Hints

- Decide which travelers pay full price, half price, or no fare. - “Half price” means divide the full fare by \(2\). - A round trip includes two one-way trips.

Solution

1. The first adult pays \(\$84\). 2. Each of the other two adults pays half of \(\$84\): \(\$84 \div 2 = \$42\). 3. The children ride free, so the one-way total is \(\$84 + \$42 + \$42 = \$168\). 4. The round trip costs twice the one-way total: \(2 \times \$168 = \$336\).

Answer

a) The one-way trip costs \(\$168\). b) The round trip costs \(\$336\).
5168884
The Rivera family is visiting an amusement park. The family has two adults and four children. The ticket prices are: - Adult ticket: \(\$45\) - Child ticket: \(\$30\) - Family pass: \(\$130\), covering two adults and up to three children Any additional child needs a regular child ticket. Find the least expensive total admission cost for the Rivera family.

Hints

- Calculate the cost of buying all individual tickets. - Then calculate the cost of one family pass plus the extra child ticket. - Compare the two totals.

Solution

1. Buying individual tickets would cost \(2 \times \$45 = \$90\) for the adults and \(4 \times \$30 = \$120\) for the children. 2. The individual-ticket total is \(\$90 + \$120 = \$210\). 3. A family pass covers the two adults and three children. One additional child ticket is needed, so this option costs \(\$130 + \$30 = \$160\). 4. Since \(\$160 < \$210\), the family pass plus one child ticket is less expensive.

Answer

The least expensive total admission cost is \(\$160\).
5168924
Two people are taking a round-trip train ride together. A one-way ticket costs \(\$50\) per person. The first traveler pays full price, and the second traveler pays half price for each one-way ticket. What is the total cost of all four one-way tickets?

Hints

- Find the round-trip cost for the first traveler. - Find half of the one-way fare for the second traveler. - Remember that each traveler needs a ticket in both directions.

Solution

1. The first traveler’s round-trip cost is \(2 \times \$50 = \$100\). 2. The second traveler’s one-way fare is half of \(\$50\): \(\$50 \div 2 = \$25\). 3. The second traveler’s round-trip cost is \(2 \times \$25 = \$50\). 4. The total cost is \(\$100 + \$50 = \$150\).

Answer

The tickets cost \(\$150\) in all.
5168934
A one-way coach-class train ticket costs \(\$48\). A one-way first-class ticket costs the coach fare plus half of the coach fare. How much does one round-trip first-class ticket cost?

Hints

- Add half of the coach fare to the coach fare. - First find the one-way first-class fare. - A round trip includes two one-way tickets.

Solution

1. Half of the coach fare is \(\$48 \div 2 = \$24\). 2. A one-way first-class ticket costs \(\$48 + \$24 = \$72\). 3. A round-trip ticket costs \(2 \times \$72 = \$144\).

Answer

One round-trip first-class ticket costs \(\$144\).
5168944
Compare the costs of these two one-way train trips: - Trip A: A coach ticket costs \(\$58\). A first-class ticket costs the coach fare plus half of the coach fare. - Trip B: A coach ticket costs \(\$79\). Which trip costs more, and what is the difference in price?

Hints

- First find half of Trip A’s coach fare. - Add that amount to the coach fare to find the first-class fare. - Compare the two final fares, then subtract to find the difference.

Solution

1. Half of Trip A’s coach fare is \(\$58 \div 2 = \$29\). 2. Trip A’s first-class fare is \(\$58 + \$29 = \$87\). 3. Since \(\$87 > \$79\), Trip A costs more. 4. The difference is \(\$87 - \$79 = \$8\).

Answer

Trip A costs more by \(\$8\).
5168954
Admission to a community pool costs \(\$8.40\) for the first person in a group. Each additional person pays half price. Find the total admission cost for a group of: a) \(2\) people b) \(3\) people c) \(5\) people

Hints

- First find half of the regular admission price. - In each group, one person pays full price. - Count how many additional people pay the half-price admission.

Solution

1. Each additional person pays \(\$8.40 \div 2 = \$4.20\). 2. For \(2\) people, the total is \(\$8.40 + \$4.20 = \$12.60\). 3. For \(3\) people, the total is \(\$8.40 + 2 \times \$4.20 = \$16.80\). 4. For \(5\) people, the total is \(\$8.40 + 4 \times \$4.20 = \$25.20\).

Answer

a) \(\$12.60\) b) \(\$16.80\) c) \(\$25.20\)
5168964
At a historic fort, the first ticket in a family group costs \(\$14\). Each additional ticket costs half as much. A family pays \(\$42\) in all. How many people are in the family?

Hints

- Find the price of each additional ticket. - Subtract the first ticket’s price from the total. - Determine how many additional tickets the remaining money buys, then include the first person.

Solution

1. Each additional ticket costs \(\$14 \div 2 = \$7\). 2. After paying \(\$14\) for the first ticket, \(\$42 - \$14 = \$28\) remains. 3. The remaining amount pays for \(\$28 \div \$7 = 4\) additional tickets. 4. Including the first ticket, the family has \(1 + 4 = 5\) people.

Answer

The family has \(5\) people.
5168974
A group of four people is comparing two bus-fare offers: - Offer A: Each person pays \(\$12\). - Offer B: The first person pays \(\$18\), and each additional person pays half of that fare. Which offer costs less for the group, and how much less does it cost?

Hints

- Find the total for Offer A. - For Offer B, find the half-price fare and count the additional travelers. - Compare the totals and subtract to find the difference.

Solution

1. Offer A costs \(4 \times \$12 = \$48\). 2. Under Offer B, each additional person pays \(\$18 \div 2 = \$9\). 3. Offer B costs \(\$18 + 3 \times \$9 = \$45\). 4. Since \(\$45 < \$48\), Offer B costs less. 5. The difference is \(\$48 - \$45 = \$3\).

Answer

Offer B costs less by \(\$3\).
5168994
A rail company compares two trains for a new route: - Mountain Express: \(7\) cars with \(64\) seats in each car - Coastal Limited: \(5\) cars with \(92\) seats in each car Which train has more seats altogether, and what is the difference?

Hints

- Find the total capacity of each train separately. - Multiply the number of cars by the seats in each car. - Subtract the smaller capacity from the larger capacity.

Solution

1. Mountain Express has \(7\times 64=448\) seats. 2. Coastal Limited has \(5\times 92=460\) seats. 3. Since \(460>448\), Coastal Limited has more seats. 4. The difference is \(460-448=12\) seats.

Answer

Coastal Limited has more seats. It has \(12\) more seats than Mountain Express.
5169034
A circus convoy includes: - \(15\) standard trailers, each \(20\,\text{ft}\) long with sleeping space for \(4\) people - \(8\) deluxe trailers, each \(30\,\text{ft}\) long with sleeping space for \(7\) people - \(4\) equipment trailers, each \(40\,\text{ft}\) long with no sleeping spaces How long is the convoy if the trailers are lined up end to end? How many people can sleep in the trailers?

Hints

- All three trailer types count toward the total length, but only two provide sleeping spaces. - Read the capacity of each trailer type carefully. - Find each subtotal before adding the lengths and sleeping spaces.

Solution

1. Find the length of each type: \(15 \times 20\,\text{ft} = 300\,\text{ft}\), \(8 \times 30\,\text{ft} = 240\,\text{ft}\), and \(4 \times 40\,\text{ft} = 160\,\text{ft}\). 2. Add the lengths: \(300\,\text{ft} + 240\,\text{ft} + 160\,\text{ft} = 700\,\text{ft}\). 3. Find the sleeping spaces: \(15 \times 4 = 60\) and \(8 \times 7 = 56\). 4. Add the sleeping spaces: \(60 + 56 = 116\). The equipment trailers do not add sleeping spaces.

Answer

The convoy is \(700\,\text{ft}\) long, and \(116\) people can sleep in the trailers.
5169044
During a break, Felix eats a fruit bar containing \(12\,\text{g}\) of sugar and drinks a carton of chocolate milk containing \(18\,\text{g}\) of sugar. One sugar cube has a mass of \(3\,\text{g}\). The total sugar is equal to how many sugar cubes?

Hints

- Add the sugar from the food and drink. - Determine how many groups of \(3\,\text{g}\) are in the total amount.

Solution

1. Find the total amount of sugar: \(12\,\text{g} + 18\,\text{g} = 30\,\text{g}\). 2. Divide by the mass of one sugar cube: \(30\,\text{g} \div 3\,\text{g} = 10\).

Answer

The total sugar is equal to \(10\) sugar cubes.
5169054
Two soft drinks are compared. Super Sweet contains \(48\,\text{g}\) of sugar per bottle, and Fruit Mix contains \(36\,\text{g}\) per bottle. One sugar cube has a mass of \(3\,\text{g}\). What is the difference between the drinks in equivalent numbers of sugar cubes?

Hints

- Find the equivalent number of sugar cubes for each drink. - Then subtract the two results. - Another method is to find the difference in grams first.

Solution

1. Super Sweet contains the equivalent of \(48\,\text{g} \div 3\,\text{g} = 16\) sugar cubes. 2. Fruit Mix contains the equivalent of \(36\,\text{g} \div 3\,\text{g} = 12\) sugar cubes. 3. Find the difference: \(16 - 12 = 4\) sugar cubes.

Answer

The drinks differ by the equivalent of \(4\) sugar cubes.
5169684
A town-square improvement project costs \(\$450{,}000\). A state grant pays \(\$360{,}000\). Three neighboring towns will share the remaining cost equally. How much must each town pay?

Hints

- First subtract the grant from the total project cost. - Then divide the remaining amount equally among the three towns. - Check that the three equal shares add to the remaining cost.

Solution

1. Find the cost remaining after the grant: \(\$450{,}000 - \$360{,}000 = \$90{,}000\). 2. Divide the remaining cost equally among the three towns: \(\$90{,}000 \div 3 = \$30{,}000\).

Answer

Each town must pay \(\$30{,}000\).
5169704
Four families share the yearly costs of a community garden equally. a) Find the total yearly cost and each family’s share. b) Each family deposited \(\$19\) per month into a shared account. How much money should each family receive back at the end of the year? <table> <tr><td>Expense</td><td>Yearly cost</td></tr> <tr><td>Fence repair</td><td>\(\$456\)</td></tr> <tr><td>Soil</td><td>\(\$192\)</td></tr> <tr><td>Seeds</td><td>\(\$74\)</td></tr> <tr><td>Garden tools</td><td>\(\$142\)</td></tr> </table>

Hints

- Add all four expenses first. - Divide the total by the number of families. - Use \(12\) months to find each family’s total deposits. - Compare the deposits with the actual share.

Solution

1. Add the yearly costs: \(\$456 + \$192 + \$74 + \$142 = \$864\). 2. Divide the total equally: \(\$864 \div 4 = \$216\) per family. 3. Each family deposited \(12 \times \$19 = \$228\) during the year. 4. The refund is \(\$228 - \$216 = \$12\) per family.

Answer

a) The total cost is \(\$864\), and each family’s share is \(\$216\). b) Each family receives \(\$12\) back.
5169804
Five classes are sharing the costs of a school festival equally. <table> <tr><td>Drinks</td><td>\(\$115\)</td></tr> <tr><td>Baking ingredients</td><td>\(\$240\)</td></tr> <tr><td>Paper plates and napkins</td><td>\(\$35\)</td></tr> <tr><td>Decorations</td><td>\(\$60\)</td></tr> </table> How much must each class pay?

Hints

- Add every amount in the table. - Then divide the total by the number of classes. - Check that five equal shares make the full total.

Solution

1. Add all the festival costs: \(\$115 + \$240 + \$35 + \$60 = \$450\). 2. Divide the total equally among the five classes: \(\$450 \div 5 = \$90\).

Answer

Each class must pay \(\$90\).
5169924
My number is a multiple of \(25\) between \(300\) and \(400\). If \(150\) is subtracted from the number and the result is divided by \(2\), the answer is \(100\). What is my number?

Hints

- Solve the number riddle backward. - Use the inverse operations for dividing by \(2\) and subtracting \(150\). - Check the interval and the multiple-of-\(25\) condition.

Solution

1. Work backward by undoing division by \(2\): \(100 \times 2 = 200\). 2. Undo subtraction of \(150\): \(200 + 150 = 350\). 3. Check the conditions: \(350\) is between \(300\) and \(400\), and \(14 \times 25 = 350\), so it is a multiple of \(25\).

Answer

\(350\)
5170034
The Schmidt family pays \(\$52\) each month toward its electricity bill. At the end of the year, the utility company determines that the actual yearly cost was \(\$595\). How much money should the family receive back?

Hints

- How many months are in one year? - Find the total amount paid over all \(12\) months. - Subtract the actual yearly cost from the amount paid.

Solution

1. Find the total paid during the year: \(12 \times \$52 = \$624\). 2. Subtract the actual cost: \(\$624 - \$595 = \$29\).

Answer

The family should receive a refund of \(\$29\).
5170044
A gym offers two payment options: \(\$26\) each month or one yearly payment of \(\$285\). How much money is saved in one year by choosing the yearly payment?

Hints

- How many monthly payments are made in one year? - Find the total cost of paying monthly. - Compare the two yearly totals.

Solution

1. Find the total cost of \(12\) monthly payments: \(12 \times \$26 = \$312\). 2. Subtract the yearly price: \(\$312 - \$285 = \$27\).

Answer

The yearly payment saves \(\$27\).
5171004
A square meter of a cornfield has about \(8\) corn plants. Each plant has one ear with about \(500\) kernels. A group of \(1000\) kernels weighs \(380\,\text{g}\). About how many grams of corn are harvested from one square meter?

Hints

- How many kernels grow on all the plants together? - How many groups of \(1000\) kernels are in that total? - Multiply the number of groups by the weight of one group.

Solution

1. Find the approximate total number of kernels: \(8 \times 500 \approx 4000\). 2. The total contains about \(4000 \div 1000 \approx 4\) groups of \(1000\) kernels. 3. Find the approximate weight: \(4 \times 380\,\text{g} \approx 1520\,\text{g}\).

Answer

About \(1520\,\text{g}\) of corn is harvested per square meter.
5171274
An elephant eats \(150\,\text{kg}\) of plants each day and spends \(18\) hours per day eating. a) How many kilograms does it eat in \(7\) days? b) How many hours per day does it not spend eating? c) The elephant produces \(75\,\text{kg}\) of waste each day. How many kilograms is that in \(7\) days?

Hints

- How many hours are in one full day? - Multiply each daily amount by \(7\) to find a weekly amount. - Break a product such as \(150 \times 7\) into easier parts if needed.

Solution

1. For a), \(150\,\text{kg} \times 7 = 1050\,\text{kg}\). 2. For b), \(24\,\text{h} - 18\,\text{h} = 6\,\text{h}\). 3. For c), \(75\,\text{kg} \times 7 = 525\,\text{kg}\).

Answer

a) \(1050\,\text{kg}\) b) \(6\) hours c) \(525\,\text{kg}\)
5171324
A wildlife reserve in Africa has \(4870\) white rhinos and \(2450\) black rhinos. During the year, \(156\) calves were born and \(89\) rhinos died. How many rhinos are now living in the reserve?

Hints

- First find the total starting population. - Decide which change increases the population and which decreases it. - Work through the changes one step at a time.

Solution

1. Find the starting population: \(4870+2450=7320\). 2. Add the calves born: \(7320+156=7476\). 3. Subtract the rhinos that died: \(7476-89=7387\).

Answer

There are now \(7387\) rhinos living in the reserve.
5171344
A conservation program tracks \(3600\) rhinos in its Asian reserves. Of these, \(2750\) are greater one-horned rhinos, \(760\) are Sumatran rhinos, and the rest are Javan rhinos. The program tracks \(18{,}500\) rhinos in its African reserves. a) How many Javan rhinos are in the Asian reserves? b) How many more rhinos are in the African reserves than in the Asian reserves?

Hints

- In part a, find the amount left after accounting for the two known groups. - In part b, find the difference between the two regional totals. - Subtraction can be used to find both a remainder and a difference.

Solution

1. Add the known groups in the Asian reserves: \(2750+760=3510\). 2. Subtract from the Asian total: \(3600-3510=90\). There are \(90\) Javan rhinos. 3. Compare the regional totals: \(18{,}500-3600=14{,}900\).

Answer

a) \(90\) Javan rhinos b) \(14{,}900\) more rhinos
5171354
A large forest has \(34{,}000\) fir trees and \(16{,}000\) spruce trees. Together, these conifer trees are exactly one-tenth of the number of deciduous trees in the forest. How many conifer and deciduous trees are in the forest altogether?

Hints

- First find how many conifer trees there are altogether. - If the conifer count is one-tenth of the deciduous count, how many times as large is the deciduous count? - Find the number of deciduous trees. - What final operation combines the two groups?

Solution

1. Find the total number of conifer trees: \(34{,}000 + 16{,}000 = 50{,}000\). 2. The conifer count is one-tenth of the deciduous count, so the deciduous count is \(10\) times as great: \(10 \times 50{,}000 = 500{,}000\). 3. Add both groups: \(50{,}000 + 500{,}000 = 550{,}000\).

Answer

There are \(550{,}000\) trees in the forest altogether.
5171424
Lucas and Maya are making wire-frame models. Lucas makes a cube with edge length \(8\,\text{cm}\). Maya makes a rectangular prism with length \(10\,\text{cm}\), width \(6\,\text{cm}\), and height \(8\,\text{cm}\). a) Find the total length of wire Lucas needs. b) Find the total length of wire Maya needs. c) Compare the two amounts.

Hints

- A cube has \(12\) edges of equal length. - A rectangular prism has four edges matching each of its three dimensions. - Add the lengths of all the edges for each model.

Solution

1. A cube has \(12\) equal edges, so Lucas needs \(12 \times 8\,\text{cm}=96\,\text{cm}\). 2. A rectangular prism has four edges of each dimension. Maya needs \(4 \times 10\,\text{cm}+4 \times 6\,\text{cm}+4 \times 8\,\text{cm}=40\,\text{cm}+24\,\text{cm}+32\,\text{cm}=96\,\text{cm}\). 3. Both models require the same total length of wire.

Answer

a) \(96\,\text{cm}\) b) \(96\,\text{cm}\) c) The two models require the same amount of wire.
5171444
A horse eats \(12\,\text{lb}\) of hay each day. A large bale weighs \(300\,\text{lb}\) and costs \(\$45\). a) How many pounds of hay does the horse eat in \(100\) days? b) How many bales are needed for \(100\) days? c) What is the total cost of the hay?

Hints

- Start with the horse’s daily amount. - Determine how many bale weights fit in the total amount. - Multiply the number of bales by the price per bale.

Solution

1. The horse eats \(12\,\text{lb} \times 100 = 1200\,\text{lb}\). 2. Since \(4 \times 300\,\text{lb} = 1200\,\text{lb}\), \(4\) bales are needed. 3. The cost is \(4 \times \$45 = \$180\).

Answer

a) \(1200\,\text{lb}\) b) \(4\) bales c) \(\$180\)
5171464
A class has \(25\) students. Each student uses \(2\) sheets of paper per school day. A package of \(500\) sheets costs \(\$6\). a) How many sheets does the class use in one school day? b) How many sheets does the class use in \(20\) school weeks if each week has \(5\) school days? c) How many packages are needed for the \(20\) weeks, and what is their total cost?

Hints

- First find the paper used by the entire class in one day. - Find the total number of school days in \(20\) weeks. - Find how many groups of \(500\) sheets make the total by using multiplication.

Solution

1. The class uses \(25 \times 2 = 50\) sheets per day. 2. Twenty school weeks contain \(20 \times 5 = 100\) school days. 3. The class uses \(50 \times 100 = 5000\) sheets. 4. Since \(10 \times 500 = 5000\), the class needs \(10\) packages. 5. The total cost is \(10 \times \$6 = \$60\).

Answer

a) The class uses \(50\) sheets per day. b) The class uses \(5000\) sheets in \(20\) school weeks. c) The class needs \(10\) packages, costing \(\$60\) in all.
5171474
A school studies water use. During one school day, one student uses about \(2\,\text{L}\) for drinking, \(4\,\text{L}\) for handwashing, and \(18\,\text{L}\) for toilet flushing. a) About how many liters of water does one student use in a school day? b) A class has \(22\) students. About how many liters does the class use in one day? c) About how many liters does the class use in a \(5\)-day school week?

Hints

- Add the three uses for one student. - Multiply by \(22\) students. - Multiply the daily class amount by \(5\) days. - Use approximation language because the per-student amounts are averages.

Solution

1. Estimate one student's daily use: \(2\,\text{L} + 4\,\text{L} + 18\,\text{L} \approx 24\,\text{L}\). 2. Estimate the class's daily use: \(22 \times 24\,\text{L} \approx 528\,\text{L}\). 3. Estimate the class's weekly use: \(5 \times 528\,\text{L} \approx 2640\,\text{L}\).

Answer

a) About \(24\,\text{L}\) b) About \(528\,\text{L}\) c) About \(2640\,\text{L}\)
5171484
A four-person family uses \(1100\,\text{L}\) of water over a two-day weekend. On average, each person uses about \(45\,\text{L}\) per day for personal care and \(30\,\text{L}\) per day for toilet flushing. About how many liters does the family use for all other purposes during the weekend?

Hints

- Multiply each daily per-person amount by \(4\) people and \(2\) days. - Add the two known categories. - Subtract the known use from the weekend total. - Use approximation language because the category amounts are averages.

Solution

1. Estimate the personal-care use: \(4 \times 2 \times 45\,\text{L} \approx 360\,\text{L}\). 2. Estimate the toilet-flushing use: \(4 \times 2 \times 30\,\text{L} \approx 240\,\text{L}\). 3. Estimate the known use: \(360\,\text{L} + 240\,\text{L} \approx 600\,\text{L}\). 4. Estimate the water used for other purposes: \(1100\,\text{L} - 600\,\text{L} \approx 500\,\text{L}\).

Answer

The family uses about \(500\,\text{L}\) of water for other purposes during the weekend.
5171574
A bath uses about \(150\,\text{L}\) of water, while a shower uses about \(40\,\text{L}\). 1. About how many liters does one person save by showering instead of taking a bath? 2. A family of \(4\) showers once per day instead of taking baths. About how many liters does the family save in one day? 3. About how many liters does the family save in \(7\) days?

Hints

- Subtract the shower amount from the bath amount. - Multiply by \(4\) people for one day. - Multiply the daily family savings by \(7\) days. - Use approximation language because the starting amounts are estimates.

Solution

1. Estimate the savings for one person: \(150\,\text{L} - 40\,\text{L} \approx 110\,\text{L}\). 2. Estimate the family's daily savings: \(4 \times 110\,\text{L} \approx 440\,\text{L}\). 3. Estimate the weekly savings: \(440\,\text{L} \times 7 \approx 3080\,\text{L}\).

Answer

1. About \(110\,\text{L}\) 2. About \(440\,\text{L}\) 3. About \(3080\,\text{L}\)
5171584
Here is Liam's daily water use: <table> <tr><td>Brushing teeth:</td><td>\(3\,\text{L}\)</td></tr> <tr><td>Showering:</td><td>\(45\,\text{L}\)</td></tr> <tr><td>Toilet flushing:</td><td>\(32\,\text{L}\)</td></tr> <tr><td>Food and drinks:</td><td>\(6\,\text{L}\)</td></tr> <tr><td>Other uses:</td><td>\(4\,\text{L}\)</td></tr> </table> 1. Find Liam's total water use for one day. 2. How many liters does Liam use during the entire month of June, which has \(30\) days? 3. Liam's older sister uses \(15\,\text{L}\) more than Liam each day. How many liters does she use in one week, which has \(7\) days?

Hints

- Add all the entries in the table for one day. - Multiply the daily amount by \(30\) for June. - For the last part, find the sister's daily amount before multiplying by \(7\).

Solution

1. Add Liam's daily amounts: \(3\,\text{L} + 45\,\text{L} + 32\,\text{L} + 6\,\text{L} + 4\,\text{L} = 90\,\text{L}\). 2. Find Liam's June use: \(90\,\text{L} \times 30 = 2700\,\text{L}\). 3. Find his sister's daily use: \(90\,\text{L} + 15\,\text{L} = 105\,\text{L}\). Then find her weekly use: \(105\,\text{L} \times 7 = 735\,\text{L}\).

Answer

1. Liam uses \(90\,\text{L}\) per day. 2. Liam uses \(2700\,\text{L}\) in June. 3. His sister uses \(735\,\text{L}\) in one week.
5172374
A soccer team has \(22\) players and \(3\) coaches. After practice, everyone will eat pizza. An individual pizza costs \(\$8\). A family-size pizza feeds exactly \(5\) people and costs \(\$35\). a) Find the total cost if each person gets an individual pizza. b) How many family-size pizzas are needed for all \(25\) people? Find their total cost. c) Which option costs less, and how much money does it save? d) For the less expensive option, what is the cost per person?

Hints

- First find the total number of people. - Compare the total cost of the two ordering plans. - Find the savings by subtracting the lesser total from the greater total. - Use the cost and number served by one family-size pizza to find the per-person cost.

Solution

1. There are \(22+3=25\) people. 2. a) Individual pizzas cost \(25\times\$8=\$200\). 3. b) The team needs \(25\div5=5\) family-size pizzas. They cost \(5\times\$35=\$175\). 4. c) Family-size pizzas cost less. The savings are \(\$200-\$175=\$25\). 5. d) Each family-size pizza costs \(\$35\) for \(5\) people, so the cost per person is \(\$35\div5=\$7\).

Answer

a) \(\$200\) b) \(5\) family-size pizzas for \(\$175\) c) Family-size pizzas cost less and save \(\$25\). d) \(\$7\) per person
5172384
A teacher is comparing admission options for a group of \(30\) people visiting a science museum. An individual ticket costs \(\$6\). A group ticket for up to \(10\) people costs \(\$52\). a) Find the cost of \(30\) individual tickets. b) Find the total cost of \(2\) group tickets and individual tickets for the remaining people. c) Find the cost of exactly \(3\) group tickets. d) Which option is least expensive?

Hints

- Find each plan's total cost separately. - For part b, determine how many people are not covered by the two group tickets. - Compare the three totals after calculating them.

Solution

1. a) Thirty individual tickets cost \(30\times\$6=\$180\). 2. b) Two group tickets cover \(20\) people and cost \(2\times\$52=\$104\). The remaining \(10\) individual tickets cost \(10\times\$6=\$60\). The total is \(\$104+\$60=\$164\). 3. c) Three group tickets cost \(3\times\$52=\$156\). 4. d) Compare \(\$180, \$164\), and \(\$156\). Three group tickets are least expensive.

Answer

a) \(\$180\) b) \(\$164\) c) \(\$156\) d) Three group tickets are least expensive.
5173904
A school has \(22\) classes with an average of \(26\) students per class. 1. Find the total number of students if the average applies exactly. 2. Give a reasonable rounded total for a news report. State the place to which you rounded and explain your choice.

Hints

- Multiply the number of equal groups by the size of each group. - Decide whether the report needs a broad estimate or a more precise one. - Name the rounding place you choose.

Solution

1. Multiply the number of classes by the students per class: \(22\times26=572\). 2. A reasonable broad estimate is \(600\), rounded to the nearest hundred. A more precise report could use \(570\), rounded to the nearest ten. Either choice is reasonable when its intended precision is explained.

Answer

1. \(572\) students 2. One reasonable answer is about \(600\) students, rounded to the nearest hundred. About \(570\) students, rounded to the nearest ten, is also reasonable with an appropriate explanation.
5174474
A bakery has \(250\,\text{kg}\) of flour. It uses \(40\,\text{kg}\) on Monday morning. The remaining flour will be divided equally among the next \(7\) days. How many kilograms will the bakery use each day?

Hints

- First find the amount remaining after Monday morning. - Equal sharing among days indicates division. - Solve the problem in two steps.

Solution

1. Find the flour remaining: \(250\,\text{kg} - 40\,\text{kg} = 210\,\text{kg}\). 2. Divide equally among seven days: \(210\,\text{kg} \div 7 = 30\,\text{kg}\).

Answer

The bakery will use \(30\,\text{kg}\) each day.
5174484
A beekeeper has \(180\) jars of honey and keeps \(12\) jars for the family. The remaining jars are packed into boxes holding \(8\) jars each. How many full boxes can the beekeeper pack?

Hints

- First find the number of jars left for sale. - Then determine how many groups of \(8\) can be made. - Break apart \(168\) into convenient multiples of \(8\) if helpful.

Solution

1. Find the number of jars available for sale: \(180 - 12 = 168\). 2. Divide by the number of jars per box: \(168 \div 8 = 21\).

Answer

The beekeeper can pack \(21\) full boxes.
5174694
A school library has \(120\) new nonfiction books on one shelf. One-fourth of the books are about animals. There are exactly \(15\) more books about space than books about animals. How many animal and space books are on the shelf altogether?

Hints

- First think about what one-fourth of a number means. - What operation can you use to find a fraction of a set? - Find the number of books in each category one at a time. - Read carefully to see how the number of space books is related to the number of animal books.

Solution

1. Find the number of animal books: \(\frac{1}{4} \times 120 = 30\). 2. Find the number of space books: \(30 + 15 = 45\). 3. Add the two categories: \(30 + 45 = 75\).

Answer

There are \(75\) animal and space books altogether.
5174704
Lucas and Anna are working together on a \(500\)-piece puzzle. Lucas has already placed one-fifth of all the pieces correctly. Anna has placed \(35\) more pieces than Lucas. How many puzzle pieces still need to be placed?

Hints

- How many pieces did Lucas place if he completed one-fifth of the puzzle? - How many pieces did Anna place compared with Lucas? - How many pieces have they placed altogether? - What operation finds the number still missing from the whole puzzle?

Solution

1. Find the number of pieces Lucas placed: \(\frac{1}{5} \times 500 = 100\). 2. Find the number of pieces Anna placed: \(100 + 35 = 135\). 3. Find the total number already placed: \(100 + 135 = 235\). 4. Subtract from the total number of pieces: \(500 - 235 = 265\).

Answer

There are \(265\) puzzle pieces left to place.
5174754
Two caterpillars crawl directly toward each other along a straight garden path. They begin \(120\,\text{cm}\) apart. After one hour, the first caterpillar has crawled \(48\,\text{cm}\), and the second has crawled \(55\,\text{cm}\). How far apart are they now?

Hints

- First find how far the two caterpillars have crawled altogether. - Subtract that total from their starting distance. - A sketch of the path may help.

Solution

1. Add the distances the caterpillars have crawled toward each other: \(48 + 55 = 103\,\text{cm}\). 2. Subtract that distance from the starting distance: \(120 - 103 = 17\,\text{cm}\).

Answer

The caterpillars are \(17\,\text{cm}\) apart.
5174784
A fourth-grade class has collected \(\$60\) for a field trip. The class spends one-half of the money on the bus ride and one-fourth of the original total on museum tickets. The rest of the money will be used for ice cream. How much money is available for ice cream?

Hints

- First find one-half of \(\$60\). - Then find one-fourth of the original amount. - How much was spent on the bus and museum altogether? - What remains after subtracting those expenses from \(\$60\)?

Solution

1. Find the cost of the bus ride: \(\frac{1}{2} \times \$60 = \$30\). 2. Find the cost of the museum tickets: \(\frac{1}{4} \times \$60 = \$15\). 3. Subtract both expenses from the total: \(\$60 - \$30 - \$15 = \$15\).

Answer

The class has \(\$15\) available for ice cream.
5174794
A bakery plans to make \(150\) dinner rolls in one day. In the morning, the baker makes one-fifth of the planned amount. In the afternoon, the baker makes \(12\) more rolls than in the morning. How many more rolls must be made to reach the goal of \(150\)?

Hints

- First find how many rolls were made in the morning. - Was the afternoon amount greater or less than the morning amount? - How many rolls were made altogether? - What operation finds how many more are needed to reach the goal?

Solution

1. Find the number of rolls made in the morning: \(\frac{1}{5} \times 150 = 30\). 2. Find the number made in the afternoon: \(30 + 12 = 42\). 3. Find the total number made: \(30 + 42 = 72\). 4. Subtract from the goal: \(150 - 72 = 78\).

Answer

The baker must make \(78\) more rolls.
5175084
A sports club plans to buy \(8\) soccer balls for \(\$30\) each. The coach changes the plan and uses the same total amount to buy \(6\) higher-quality basketballs instead. How much does one basketball cost?

Hints

- What amount stays the same when the type of ball changes? - First find the total amount the club plans to spend. - Divide that amount equally among the six basketballs.

Solution

1. Find the total planned budget: \(8 \times \$30 = \$240\). 2. Divide the budget equally among six basketballs: \(\$240 \div 6 = \$40\).

Answer

One basketball costs \(\$40\).
5175294
One crate contains \(40\) apples, and another crate contains \(30\) pears. One-half of the apples are used for a large pie. One-third of the pears are used for fruit sauce. How many pieces of fruit remain in the two crates altogether?

Hints

- First find how many apples and pears remain separately. - What does it mean mathematically to use one-half of a group? - What fraction remains after one-third is used? - How can you combine the remaining amounts from both crates?

Solution

1. Find the number of apples used: \(\frac{1}{2} \times 40 = 20\). 2. Find the number of pears used: \(\frac{1}{3} \times 30 = 10\). 3. Find the amounts remaining: \(40 - 20 = 20\) apples and \(30 - 10 = 20\) pears. 4. Add the remaining amounts: \(20 + 20 = 40\).

Answer

There are \(40\) pieces of fruit remaining altogether.
5175304
Lucas and Sofia are saving for a gift. Lucas has \(\$120\), and Sofia has \(\$160\). Lucas contributes one-fourth of his money toward the gift. Sofia contributes one-eighth of her money. How much money do they have left altogether?

Hints

- How much money does each person keep after contributing to the gift? - Find each contribution separately. - What operation can you use to find one-fourth or one-eighth of an amount? - Remember to add the two remaining amounts at the end.

Solution

1. Find Lucas's contribution: \(\frac{1}{4} \times \$120 = \$30\). 2. Find Lucas's remaining money: \(\$120 - \$30 = \$90\). 3. Find Sofia's contribution: \(\frac{1}{8} \times \$160 = \$20\). 4. Find Sofia's remaining money: \(\$160 - \$20 = \$140\). 5. Add the remaining amounts: \(\$90 + \$140 = \$230\).

Answer

Together, they have \(\$230\) left.
5175314
An elementary school has two fourth-grade classes. Class 4A has \(24\) students, and Class 4B has \(28\) students. One-fourth of the students in each class enter an art contest. The other students go to the gym for physical education. How many students go to the gym altogether?

Hints

- First find how many students from each class enter the contest. - How many students remain in each class after subtracting the contestants? - Could you also begin by finding the total number of students in both classes? - What operation finds one-fourth of a group?

Solution

1. In Class 4A, \(\frac{1}{4} \times 24 = 6\) students enter the contest, so \(24 - 6 = 18\) students go to the gym. 2. In Class 4B, \(\frac{1}{4} \times 28 = 7\) students enter the contest, so \(28 - 7 = 21\) students go to the gym. 3. Add the numbers going to the gym: \(18 + 21 = 39\).

Answer

A total of \(39\) students go to the gym.
5175344
A school orders \(120\) drawing pads for \(6\) classes, with each class receiving the same number. How many drawing pads are needed to supply \(15\) classes at the same rate?

Hints

- How many pads does one class receive? - First divide the original total equally among the classes. - Once you know the amount for one class, how can you find the amount for fifteen classes?

Solution

1. Find the number of drawing pads for one class: \(120 \div 6 = 20\). 2. Find the number needed for fifteen classes: \(20 \times 15 = 300\).

Answer

The school needs \(300\) drawing pads.
5175464
A bag contains \(15\) blue marbles. The number of red marbles is four times the number of blue marbles. One-fifth of all the marbles in the bag are glass marbles. How many glass marbles are in the bag?

Hints

- First find the total number of red marbles. - How many marbles are in the bag altogether? - What operation can you use to find one-fifth of a set?

Solution

1. Find the number of red marbles: \(4 \times 15 = 60\). 2. Find the total number of marbles: \(15 + 60 = 75\). 3. Find one-fifth of the total: \(\frac{1}{5} \times 75 = 15\).

Answer

There are \(15\) glass marbles in the bag.
5175474
For a school celebration, \(6\,\text{gal}\) of apple juice is delivered. Four times as much sparkling water as apple juice is provided. One-third of the total liquid is mixed in a large dispenser to make a fruit drink. How many gallons of fruit drink are in the dispenser? Is this amount greater or less than the amount of apple juice?

Hints

- First find the amount of sparkling water. - How much apple juice and sparkling water are available altogether? - How can you find one-third of the total amount? - Compare your result with the original amount of apple juice.

Solution

1. Find the amount of sparkling water: \(4 \times 6\,\text{gal} = 24\,\text{gal}\). 2. Find the total amount of liquid: \(6\,\text{gal} + 24\,\text{gal} = 30\,\text{gal}\). 3. Find one-third of the total: \(\frac{1}{3} \times 30\,\text{gal} = 10\,\text{gal}\). 4. Compare the amounts: \(10\,\text{gal} > 6\,\text{gal}\), so the fruit drink amount is greater.

Answer

There are \(10\,\text{gal}\) of fruit drink in the dispenser. This is greater than the \(6\,\text{gal}\) of apple juice.
5175604
A store has \(18\) blue notebooks and \(12\) fewer green notebooks than blue notebooks. Each green notebook costs \(60\) cents. How much do all the green notebooks cost?

Hints

- First find the number of green notebooks. - Multiply that number by the cost of one notebook. - Express the total in dollars and cents.

Solution

1. Find the number of green notebooks: \(18 - 12 = 6\). 2. Find the total cost in cents: \(6 \times 60 = 360\) cents. 3. Convert to dollars: \(360\) cents is \(\$3.60\).

Answer

All the green notebooks cost \(\$3.60\).
5175644
A school library has \(18\) shelves of adventure books. It has \(7\) fewer shelves of nonfiction books. Each nonfiction shelf holds \(40\) books. How many nonfiction books are in the library?

Hints

- First find the number of nonfiction shelves. - Then multiply the number of shelves by the number of books on each shelf. - Pay attention to which operation must come first.

Solution

1. Find the number of nonfiction shelves: \(18 - 7 = 11\). 2. Find the number of nonfiction books: \(11 \times 40 = 440\).

Answer

The library has \(440\) nonfiction books.
5175654
Tim plants \(15\) rows of carrots with \(20\) carrots in each row. Lisa plants \(6\) fewer rows than Tim, with \(40\) carrots in each of her rows. Who has more carrots? Justify your answer with calculations.

Hints

- First find Lisa's number of rows. - Find each person's total number of carrots. - Compare the two totals.

Solution

1. Find Lisa's number of rows: \(15 - 6 = 9\). 2. Find Lisa's number of carrots: \(9 \times 40 = 360\). 3. Find Tim's number of carrots: \(15 \times 20 = 300\). 4. Compare: \(360 > 300\), so Lisa has more carrots.

Answer

Lisa has more carrots. She has \(360\) carrots, while Tim has \(300\).
5175784
A farmer harvests \(120\,\text{kg}\) of potatoes and fills bags that each hold \(10\,\text{kg}\). The farmer sells each bag for \(\$7\). How much money does the farmer earn altogether?

Hints

- First find how many bags can be filled from the harvest. - Once you know the number of bags, how can you find the total revenue?

Solution

1. Find the number of full bags: \(120\,\text{kg} \div 10\,\text{kg} = 12\). 2. Find the total revenue: \(12 \times \$7 = \$84\).

Answer

The farmer earns \(\$84\) altogether.
5175794
A class makes \(200\) greeting cards and bundles them into sets of \(5\) cards. The class sells each set for \(\$4\), then donates \(\$65\) of the money to an animal shelter. How much money remains for the class fund?

Hints

- How many sets are made when every set contains five cards? - First find the money earned from selling all the sets. - Which operation finds the amount left after the donation?

Solution

1. Find the number of card sets: \(200 \div 5 = 40\). 2. Find the total sales: \(40 \times \$4 = \$160\). 3. Subtract the donation: \(\$160 - \$65 = \$95\).

Answer

The class has \(\$95\) left for the class fund.
5175944
Mia and Ben plant \(135\) flower bulbs in the school garden. Mia works for \(4\) hours, and Ben works for \(5\) hours. They plant bulbs at the same rate. How many bulbs does each student plant?

Hints

- How many hours do Mia and Ben work altogether? - Use the total bulbs and total hours to find the shared hourly rate. - Multiply the hourly rate by each student’s work time.

Solution

1. Find the total work time: \(4 + 5 = 9\) hours. 2. Find the number of bulbs planted per hour: \(135 \div 9 = 15\) bulbs per hour. 3. Find Mia’s share: \(4 \times 15 = 60\) bulbs. 4. Find Ben’s share: \(5 \times 15 = 75\) bulbs.

Answer

Mia plants \(60\) bulbs, and Ben plants \(75\) bulbs.
5176014
Lucas collects seashells at the beach. He finds \(125\) shells in the morning. In the afternoon, he finds \(30\) more shells than he found in the morning. He divides all the shells equally among \(4\) bags. How many shells does he put in each bag?

Hints

- First find how many shells Lucas collects in the afternoon. - How many shells does he collect during the whole day? - What does “equally” tell you about the final operation? - Which operation separates a total into equal groups?

Solution

1. Find the number of shells collected in the afternoon: \(125 + 30 = 155\). 2. Find the total number of shells: \(125 + 155 = 280\). 3. Divide the shells equally among four bags: \(280 \div 4 = 70\).

Answer

Lucas puts \(70\) shells in each bag.
5176024
A school cafeteria receives three crates of apples. The first crate contains \(240\) apples. The second crate contains \(60\) fewer apples than the first crate. The third crate contains \(30\) more apples than the second crate. All the apples are divided equally among \(3\) shelves. How many apples are placed on each shelf?

Hints

- Find the number of apples in each crate one at a time. - Pay attention to which crate is used for each comparison. - How many apples are there altogether? - To divide \(630\) by \(3\), consider the hundreds and tens separately.

Solution

1. Find the number of apples in the second crate: \(240 - 60 = 180\). 2. Find the number of apples in the third crate: \(180 + 30 = 210\). 3. Find the total number of apples: \(240 + 180 + 210 = 630\). 4. Divide equally among the three shelves: \(630 \div 3 = 210\).

Answer

Each shelf holds \(210\) apples.
5176054
A school library receives \(150\) new nonfiction books. The librarian will place them in \(10\) shelf bins. The first \(4\) bins each receive \(12\) books. The remaining books will be divided equally among the other bins. How many books will go in each remaining bin?

Hints

- First find how many books are already in the first four bins. - Subtract to find how many books remain. - Determine how many empty bins are left. - Use division to share the remaining books equally.

Solution

1. Find the number of books placed in the first four bins: \(4\times 12=48\). 2. Find the number of books left: \(150-48=102\). 3. Find the number of remaining bins: \(10-4=6\). 4. Divide the remaining books equally: \(102\div 6=17\).

Answer

Each remaining bin will hold \(17\) books.
5176064
A baker made \(160\) soft pretzels and plans to place them on \(10\) baking sheets. The first \(6\) sheets each hold \(15\) pretzels. The baker plans to put \(18\) pretzels on each of the remaining \(4\) sheets. Are there enough pretzels for this plan? Find how many pretzels are missing or left over.

Hints

- Find how many pretzels remain after filling the first six sheets. - Find how many pretzels the last four sheets require. - Compare the amount available with the amount needed. - If the amount needed is greater, the difference is the shortage.

Solution

1. The first six sheets use \(6\times 15=90\) pretzels. 2. The number left is \(160-90=70\). 3. The remaining four sheets would require \(4\times 18=72\) pretzels. 4. Since \(72-70=2\), the baker is short by \(2\) pretzels.

Answer

No. The baker needs \(2\) more pretzels.
5176074
A bakery divides \(120\,\text{kg}\) of flour equally among \(3\) large bins. In the morning, the baker takes \(15\,\text{kg}\) of flour from the first bin. How many kilograms of flour remain in that bin?

Hints

- First find how much flour was in each bin. - What happens to the amount in one bin when some flour is used? - Remember that flour is taken from only one bin.

Solution

1. Find the amount of flour in each bin: \(120\,\text{kg} \div 3 = 40\,\text{kg}\). 2. Subtract the amount taken from the first bin: \(40\,\text{kg} - 15\,\text{kg} = 25\,\text{kg}\).

Answer

The first bin has \(25\,\text{kg}\) of flour remaining.
5176084
A water tank contains \(450\,\text{L}\) of water. The water is divided equally among \(5\) troughs in a horse pasture. Two horses drink from the first trough. One drinks \(22\,\text{L}\), and the other drinks \(18\,\text{L}\). How many liters of water remain in the first trough?

Hints

- How much water is placed in each trough at the beginning? - How much water do the two horses drink altogether? - Subtract the total amount they drink from the amount in the first trough.

Solution

1. Find the amount of water in each trough: \(450\,\text{L} \div 5 = 90\,\text{L}\). 2. Find the total amount the horses drink: \(22\,\text{L} + 18\,\text{L} = 40\,\text{L}\). 3. Subtract the amount they drink: \(90\,\text{L} - 40\,\text{L} = 50\,\text{L}\).

Answer

The first trough has \(50\,\text{L}\) of water remaining.
5176124
A school library receives \(150\) new nonfiction books and divides them equally among \(5\) shelves. On the first day, students check out \(12\) books from the top shelf. Then \(7\) donated books are added to the bottom shelf. How many nonfiction books are now on the top shelf, and how many are on the bottom shelf?

Hints

- How many books were on each shelf at the beginning? - For each shelf, decide whether books are added or removed. - The two shelves will have different answers.

Solution

1. Find the original number of books on each shelf: \(150 \div 5 = 30\). 2. Find the number remaining on the top shelf: \(30 - 12 = 18\). 3. Find the new number on the bottom shelf: \(30 + 7 = 37\).

Answer

The top shelf has \(18\) nonfiction books, and the bottom shelf has \(37\) nonfiction books.
5176454
Two fourth-grade classes buy zoo tickets. - Class 4A pays \(\$120\) for \(20\) students. - Class 4B uses a group rate and pays \(\$100\) for \(25\) students. Find the price per student for each class. How much does each Class 4B student save compared with each Class 4A student?

Hints

- Use multiplication facts to find each class’s price per student. - Then subtract the smaller per-student price from the larger one. - Check each unit price by multiplying by the class size.

Solution

1. Since \(20 \times \$6 = \$120\), Class 4A pays \(\$6\) per student. 2. Since \(25 \times \$4 = \$100\), Class 4B pays \(\$4\) per student. 3. Each Class 4B student saves \(\$6 - \$4 = \$2\).

Answer

Class 4A pays \(\$6\) per student, and Class 4B pays \(\$4\) per student. Each Class 4B student saves \(\$2\).
5176534
Leon and Sophie each saved \(\$25\) and go to a school-supply store. Leon buys \(3\) notebooks for \(\$4\) each. Sophie buys \(5\) glitter pens for \(\$2\) each. Who has more money left, and how much more?

Hints

- Find how much each person spends. - Subtract each spending amount from \(\$25\). - Compare the amounts left and find their difference.

Solution

1. Leon spends \(3 \times \$4 = \$12\), so he has \(\$25 - \$12 = \$13\) left. 2. Sophie spends \(5 \times \$2 = \$10\), so she has \(\$25 - \$10 = \$15\) left. 3. Since \(\$15 > \$13\), Sophie has more money left. 4. The difference is \(\$15 - \$13 = \$2\).

Answer

Sophie has more money left. She has \(\$2\) more than Leon.
5176544
A fruit seller begins the day with \(60\,\text{kg}\) of apples. Before lunch, the seller sells \(8\) small crates containing \(4\,\text{kg}\) each. The remaining apples are packed into bags holding \(3\,\text{kg}\) each. How many bags can be filled completely, and how many kilograms remain?

Hints

- Find how many kilograms were sold first. - Subtract that amount from the starting amount. - Divide the remaining apples into \(3\)-kilogram bags and interpret the remainder.

Solution

1. Find the amount sold: \(8 \times 4\,\text{kg} = 32\,\text{kg}\). 2. Find the amount remaining: \(60\,\text{kg} - 32\,\text{kg} = 28\,\text{kg}\). 3. Divide with a remainder: \(28\,\text{kg} \div 3\,\text{kg} = 9\) remainder \(1\). 4. Nine bags can be filled, with \(1\,\text{kg}\) left over.

Answer

The seller can fill \(9\) bags, and \(1\,\text{kg}\) remains.
5176594
A zookeeper orders \(8\) bags of special elephant feed for \(\$45\) each. The zookeeper could spend the same total amount on bales of hay that cost \(\$9\) each. How many bales of hay could be bought?

Hints

- First find the total cost of the special feed. - How many times does the cost of one bale of hay fit into the total amount? - A related basic multiplication fact may help with \(360 \div 9\).

Solution

1. Find the total cost of the special feed: \(8 \times \$45 = \$360\). 2. Divide by the cost of one bale of hay: \(\$360 \div \$9 = 40\).

Answer

The zookeeper could buy \(40\) bales of hay.
5176624
A set of \(6\) jump ropes costs \(\$24\). A teacher wants to buy \(9\) jump ropes at the same price per rope. How much will \(9\) jump ropes cost?

Hints

- First find the cost of one jump rope. - Then multiply the unit price by \(9\). - Check that your unit price gives the original cost for \(6\) ropes.

Solution

1. One jump rope costs \(\$24 \div 6 = \$4\). 2. Nine jump ropes cost \(9 \times \$4 = \$36\).

Answer

\(9\) jump ropes cost \(\$36\).
5176634
A gardener used \(48\) seedlings to plant \(4\) equal-sized garden beds. The gardener wants to plant \(3\) more beds of the same size. How many seedlings are needed for all \(7\) beds?

Hints

- Find the number of seedlings used in one bed. - The final garden will contain \(7\) beds. - You may instead find the seedlings for the \(3\) new beds and add them to \(48\).

Solution

1. Method 1: Find the number of seedlings per bed: \(48\div 4=12\). Then find the total for seven beds: \(7\times 12=84\). 2. Method 2: Again, each bed needs \(12\) seedlings. The three additional beds need \(3\times 12=36\) seedlings. Then \(48+36=84\). 3. Both methods give the same total.

Answer

The gardener needs \(84\) seedlings altogether.
5176654
A school fair has \(8\,\text{gal}\) of lemonade. During the first break, students drink one-fourth of the lemonade. During the second break, they drink one-eighth of the original amount. How many gallons of lemonade remain after the second break?

Hints

- Find the amount drunk during each break separately. - Both fractions refer to the original \(8\,\text{gal}\). - How much lemonade was drunk altogether? - Subtract the total amount drunk from the original amount.

Solution

1. Find the amount drunk during the first break: \(\frac{1}{4} \times 8\,\text{gal} = 2\,\text{gal}\). 2. Find the amount drunk during the second break: \(\frac{1}{8} \times 8\,\text{gal} = 1\,\text{gal}\). 3. Find the total amount drunk: \(2\,\text{gal} + 1\,\text{gal} = 3\,\text{gal}\). 4. Subtract from the original amount: \(8\,\text{gal} - 3\,\text{gal} = 5\,\text{gal}\).

Answer

There are \(5\,\text{gal}\) of lemonade remaining.
5176674
Paul and Sarah pack apples into bags. Paul packs \(8\) bags per hour, and Sarah packs \(12\) bags per hour. They must pack \(100\) bags and begin at \(8{:}00\) a.m. Sarah says, “If we work without a break, we will finish by \(12{:}00\) p.m.” Is Sarah correct? Justify your answer with calculations.

Hints

- Find how many bags they pack together in \(1\) hour. - Use that rate to find how many hours \(100\) bags take. - Add the needed hours to the start time and compare with noon.

Solution

1. Find their combined rate: \(8 + 12 = 20\) bags per hour. 2. Find the time needed: \(100 \div 20 = 5\) hours. 3. Add \(5\) hours to the start time: \(8{:}00\) a.m. plus \(5\) hours is \(1{:}00\) p.m. 4. Since \(1{:}00\) p.m. is later than \(12{:}00\) p.m., Sarah is not correct.

Answer

No. Together they pack \(20\) bags per hour, so \(100\) bags take \(5\) hours. They finish at \(1{:}00\) p.m.
5176724
A gardener buys \(8\) young shrubs for a park for \(\$72\) altogether. Later, the gardener orders more shrubs of the same kind for another section of the park and spends \(\$117\). How many shrubs does the gardener order altogether?

Hints

- How much does one shrub cost? - Find how many shrubs can be bought for \(\$117\). - Remember that the question asks for the total from both orders.

Solution

1. Find the cost of one shrub: \(\$72 \div 8 = \$9\). 2. Find the number of shrubs in the second order: \(\$117 \div \$9 = 13\). 3. Add the shrubs from both orders: \(8 + 13 = 21\).

Answer

The gardener orders \(21\) shrubs altogether.
5176924
A gardener has a \(32\,\text{ft}\) length of rope. First, she cuts off one-half of the rope. Then she cuts off one-fourth of the rope that remains. a) How long is the second piece she cuts off? b) How many feet of rope remain at the end?

Hints

- Does the second fraction refer to the original rope or only to the amount that remains? - Work one step at a time and record how much rope remains after each cut. - A drawing of a line divided into parts may help.

Solution

1. Find the length of the first piece: \(\frac{1}{2} \times 32\,\text{ft} = 16\,\text{ft}\). 2. After the first cut, \(32\,\text{ft} - 16\,\text{ft} = 16\,\text{ft}\) remain. 3. Find one-fourth of the remaining rope: \(\frac{1}{4} \times 16\,\text{ft} = 4\,\text{ft}\). 4. Subtract the second piece: \(16\,\text{ft} - 4\,\text{ft} = 12\,\text{ft}\).

Answer

a) The second piece is \(4\,\text{ft}\) long. b) \(12\,\text{ft}\) of rope remain.
5177104
At a sporting goods store, \(4\) basketballs cost \(\$32\) altogether. Park Elementary School wants to buy \(9\) of the basketballs for physical education classes. A teacher pays with a \(\$100\) bill. How much change does the teacher receive?

Hints

- First find the cost of one basketball. - Once you know the unit price, find the cost of nine basketballs. - Subtract the total cost from the amount paid to find the change.

Solution

1. Find the cost of one basketball: \(\$32 \div 4 = \$8\). 2. Find the cost of nine basketballs: \(9 \times \$8 = \$72\). 3. Subtract the cost from the amount paid: \(\$100 - \$72 = \$28\).

Answer

The teacher receives \(\$28\) in change.
5177354
Julia and Mark have \(72\) trading cards altogether. If Julia is given \(8\) additional cards, the two children will have the same number of cards. How many cards did each child have at the beginning?

Hints

- First find the new total after Julia receives \(8\) cards. - Divide the new total equally between the two children. - Reverse Julia’s increase to find her starting amount.

Solution

1. After Julia receives the cards, they will have \(72+8=80\) cards altogether. 2. At that point, each child will have \(80\div 2=40\) cards. 3. Mark’s number did not change, so he began with \(40\) cards. 4. Julia began with \(40-8=32\) cards.

Answer

Julia began with \(32\) cards, and Mark began with \(40\) cards.
5177864
Class 3A collects \(145\,\text{kg}\) of paper for a recycling project. Class 3B collects \(20\,\text{kg}\) more than Class 3A. Class 3C collects \(15\,\text{kg}\) less than Classes 3A and 3B combined. How many kilograms of paper does Class 3C collect?

Hints

- First find how much Class 3B collects. - Then combine the amounts for Classes 3A and 3B. - Use “less than” to determine the final operation.

Solution

1. Find the amount collected by Class 3B: \(145 + 20 = 165\,\text{kg}\). 2. Find the combined amount for Classes 3A and 3B: \(145 + 165 = 310\,\text{kg}\). 3. Subtract \(15\,\text{kg}\): \(310 - 15 = 295\,\text{kg}\).

Answer

Class 3C collects \(295\,\text{kg}\) of paper.
5177884
Birch Trail is \(185\,\text{m}\) long and is extended by \(55\,\text{m}\) at each end. Alder Trail is \(230\,\text{m}\) long and is extended by \(40\,\text{m}\) at one end and \(35\,\text{m}\) at the other. Which trail is longer after the extensions, and by how many meters?

Hints

- Find each trail's new length separately. - Remember that Birch Trail is extended at both ends by the same amount. - Subtract the shorter final length from the longer final length.

Solution

1. Find the new length of Birch Trail: \(185 + 55 + 55 = 295\,\text{m}\). 2. Find the new length of Alder Trail: \(230 + 40 + 35 = 305\,\text{m}\). 3. Compare: \(305 > 295\), so Alder Trail is longer. 4. Find the difference: \(305 - 295 = 10\,\text{m}\).

Answer

Alder Trail is longer by \(10\,\text{m}\).
5177984
A bakery begins with \(150\,\text{kg}\) of flour. It uses \(45\,\text{kg}\) for bread. The bakery uses \(20\,\text{kg}\) more flour for rolls than for bread. How much flour remains?

Hints

- First find the amount of flour used for rolls. - Add the amounts used for bread and rolls. - Subtract the total used from the starting amount.

Solution

1. Find the flour used for rolls: \(45 + 20 = 65\,\text{kg}\). 2. Find the total amount used: \(45 + 65 = 110\,\text{kg}\). 3. Subtract from the starting amount: \(150 - 110 = 40\,\text{kg}\).

Answer

\(40\,\text{kg}\) of flour remains.
5177994
A school event begins with \(350\,\text{L}\) of apple juice. Students drink \(115\,\text{L}\) during the first hour. During the second hour, they drink \(30\,\text{L}\) less than during the first hour. How much juice remains after two hours?

Hints

- First find how much juice is consumed during the second hour. - Add the amounts consumed during both hours. - Subtract the total consumed from the amount delivered.

Solution

1. Find the amount consumed during the second hour: \(115 - 30 = 85\,\text{L}\). 2. Find the total amount consumed: \(115 + 85 = 200\,\text{L}\). 3. Subtract from the starting amount: \(350 - 200 = 150\,\text{L}\).

Answer

\(150\,\text{L}\) of apple juice remains.
5178184
A bakery makes \(250\) muffins for a school event. Of these, \(70\) muffins have chocolate decorations. The remaining muffins are divided equally among \(6\) large serving trays. How many muffins are placed on each tray?

Hints

- First find how many muffins remain after removing the chocolate-decorated muffins. - Which operation divides a quantity equally among several trays? - Break the problem into two steps.

Solution

1. Find the number of muffins without chocolate decorations: \(250 - 70 = 180\). 2. Divide the remaining muffins equally among the trays: \(180 \div 6 = 30\).

Answer

Each serving tray holds \(30\) muffins.
5178194
A toy bin contains \(640\) building blocks. Of these, \(40\) blocks are broken and removed. Then \(120\) of the remaining blocks are used to build a large castle. All the blocks left are divided equally among \(8\) small bins. How many blocks are placed in each small bin?

Hints

- How many blocks remain after the broken ones are removed? - Next subtract the blocks used for the castle. - How many blocks are left for the small bins? - Divide that amount equally among the bins.

Solution

1. Find the number of unbroken blocks: \(640 - 40 = 600\). 2. Subtract the blocks used for the castle: \(600 - 120 = 480\). 3. Divide the remaining blocks equally among eight bins: \(480 \div 8 = 60\).

Answer

Each small bin holds \(60\) building blocks.
5178274
Two whole numbers are \(64\) units apart on a number line. They are the same distance from \(150\). Find the two numbers.

Hints

- Picture \(150\) halfway between the two unknown numbers. - Find half of the total distance. - Move that distance left and right from \(150\).

Solution

1. Because the two numbers are the same distance from \(150\), \(150\) is their midpoint. 2. Half of the total distance is \(64\div2=32\). 3. The smaller number is \(150-32=118\), and the larger number is \(150+32=182\).

Answer

The numbers are \(118\) and \(182\).
5178304
A crate contains \(60\) pieces of fruit. One-half are apples, one-fifth are bananas, and the rest are pears. How many pears are in the crate?

Hints

- First find one-half of \(60\). - How can you find one-fifth of a number? - After finding the numbers of apples and bananas, how can you find the rest?

Solution

1. Find the number of apples: \(\frac{1}{2} \times 60 = 30\). 2. Find the number of bananas: \(\frac{1}{5} \times 60 = 12\). 3. Subtract the apples and bananas from the total: \(60 - 30 - 12 = 18\).

Answer

There are \(18\) pears in the crate.
5178314
At an elementary school, \(120\) students take part in a charity run. One-fourth of the students run \(1\,\text{mi}\). One-third run \(2\,\text{mi}\). All the other students run \(3\,\text{mi}\). How many students run the \(3\,\text{mi}\) distance?

Hints

- How many students are one-fourth of the whole group? - Find the size of each group whose fraction is given. - What remains after subtracting those groups from the total? - The question asks for the number of students, not the distance.

Solution

1. Find the number who run \(1\,\text{mi}\): \(\frac{1}{4} \times 120 = 30\). 2. Find the number who run \(2\,\text{mi}\): \(\frac{1}{3} \times 120 = 40\). 3. Subtract the first two groups from the total: \(120 - 30 - 40 = 50\).

Answer

\(50\) students run the \(3\,\text{mi}\) distance.
5178324
A garden center has \(120\) seedlings in a row. One-fourth of the seedlings are sunflowers, one-tenth are roses, and all the others are tulips. How many tulip seedlings are in the row?

Hints

- First find one-fourth of \(120\). - How can you find one-tenth of a number? - After finding the sunflower and rose seedlings, how can you find what remains?

Solution

1. Find the number of sunflower seedlings: \(\frac{1}{4} \times 120 = 30\). 2. Find the number of rose seedlings: \(\frac{1}{10} \times 120 = 12\). 3. Subtract both groups from the total: \(120 - 30 - 12 = 78\).

Answer

There are \(78\) tulip seedlings.
5178334
Julia has a sticker album with \(240\) stickers. - One-eighth of the stickers show space scenes. - One-third show animals. - One-sixth show sports. - All the other stickers show smiley faces. How many smiley-face stickers does Julia have? Which of the four designs appears most often in her collection?

Hints

- Find the exact number in each of the three groups whose fractions are given. - What operation helps you find a fraction such as one-sixth of a set? - Compare all four group sizes to identify the design that appears most often.

Solution

1. Find the number of space stickers: \(\frac{1}{8} \times 240 = 30\). 2. Find the number of animal stickers: \(\frac{1}{3} \times 240 = 80\). 3. Find the number of sports stickers: \(\frac{1}{6} \times 240 = 40\). 4. Find the number of smiley-face stickers: \(240 - 30 - 80 - 40 = 90\). 5. Compare the groups: \(90 > 80 > 40 > 30\), so smiley faces appear most often.

Answer

Julia has \(90\) smiley-face stickers. Smiley faces appear most often.
5178344
An observation tower stands on a hill whose base is \(485\,\text{m}\) above sea level. a) The lower observation deck is \(156\,\text{m}\) above the tower's base. The upper deck is \(68\,\text{m}\) above the lower deck. The top of the tower is another \(92\,\text{m}\) above the upper deck. Find the elevation of the tower's top above sea level. b) The top of a nearby bell tower is \(572\,\text{m}\) above sea level. How much higher is the observation tower?

Hints

- Begin with the elevation of the hill's base. - Add each vertical distance in order. - For part b, compare the two elevations by subtraction.

Solution

1. The lower deck is at \(485\,\text{m}+156\,\text{m}=641\,\text{m}\). 2. The upper deck is at \(641\,\text{m}+68\,\text{m}=709\,\text{m}\). 3. The top is at \(709\,\text{m}+92\,\text{m}=801\,\text{m}\). 4. The difference is \(801\,\text{m}-572\,\text{m}=229\,\text{m}\).

Answer

a) \(801\,\text{m}\) above sea level b) \(229\,\text{m}\) higher
5178384
An ice cream shop has freezer space for \(100\) packages. The owner adds \(4\) boxes with \(12\) packages of vanilla ice cream in each box and \(3\) boxes with \(15\) packages of chocolate ice cream in each box. How many more packages can fit in the freezer?

Hints

- Find the number of packages of each flavor. - Add to find how many packages are in the freezer. - Subtract from the freezer's capacity.

Solution

1. Find the number of vanilla packages: \(4 \times 12 = 48\). 2. Find the number of chocolate packages: \(3 \times 15 = 45\). 3. Find the total number added: \(48 + 45 = 93\). 4. Find the remaining space: \(100 - 93 = 7\).

Answer

The freezer can hold \(7\) more packages.
5178524
At a stationery store, \(4\) special notebooks cost \(\$12\) altogether. Ms. Weber wants to buy \(7\) of the notebooks. She wonders, “If I pay with \(\$25\), will I receive more or less than \(\$5\) in change?” Justify your answer with calculations.

Hints

- First find the cost of one notebook. - How much do all seven notebooks cost? - Subtract the total cost from the amount paid. - Compare the change with \(\$5\).

Solution

1. Find the cost of one notebook: \(\$12 \div 4 = \$3\). 2. Find the cost of seven notebooks: \(7 \times \$3 = \$21\). 3. Find the change: \(\$25 - \$21 = \$4\). 4. Compare the change with five dollars: \(\$4 < \$5\), so the change is less than \(\$5\).

Answer

Ms. Weber receives less than \(\$5\) in change because she receives exactly \(\$4\).
5178534
A sports club orders new equipment. Four soccer balls cost \(\$40\) altogether, and the club orders \(12\) soccer balls. The club also buys \(5\) basketballs. Each basketball costs \(\$5\) more than one soccer ball. What is the total cost of the order?

Hints

- First determine the cost of one soccer ball. - How much more does one basketball cost? - Find the costs of the soccer balls and basketballs separately. - Add the two costs to find the total.

Solution

1. Find the cost of one soccer ball: \(\$40 \div 4 = \$10\). 2. Find the cost of one basketball: \(\$10 + \$5 = \$15\). 3. Find the cost of twelve soccer balls: \(12 \times \$10 = \$120\). 4. Find the cost of five basketballs: \(5 \times \$15 = \$75\). 5. Add the two costs: \(\$120 + \$75 = \$195\).

Answer

The total cost of the order is \(\$195\).
5178554
An older machine packs \(198\) bags of cookies per hour. A new machine packs \(1648\) bags during an \(8\)-hour shift. How many more bags per hour does the new machine pack than the older machine?

Hints

- First find how many bags the new machine packs in one hour. - Then compare the two hourly rates. - The question asks for a difference, not the new machine’s full rate.

Solution

1. Find the new machine’s hourly rate: \(1648 \div 8 = 206\) bags per hour. 2. Find the difference between the hourly rates: \(206 - 198 = 8\) bags per hour.

Answer

The new machine packs \(8\) more bags per hour.
5178704
For a class trip, students pack \(5\) boxes with \(12\) apples in each box and \(3\) bags with \(20\) oranges in each bag. How many pieces of fruit do they pack altogether?

Hints

- Find the total number of apples. - Find the total number of oranges. - Add the two totals.

Solution

1. Find the number of apples: \(5 \times 12 = 60\). 2. Find the number of oranges: \(3 \times 20 = 60\). 3. Add the two amounts: \(60 + 60 = 120\).

Answer

They pack \(120\) pieces of fruit altogether.
5178714
A gym has \(8\) rows with \(15\) blue chairs in each row and \(6\) rows with \(14\) red chairs in each row. How many chairs are in the gym altogether?

Hints

- Find the total number of blue chairs. - Find the total number of red chairs. - Add the two totals.

Solution

1. Find the number of blue chairs: \(8 \times 15 = 120\). 2. Find the number of red chairs: \(6 \times 14 = 84\). 3. Add the two amounts: \(120 + 84 = 204\).

Answer

There are \(204\) chairs altogether.
5178834
Three classes buy \(600\) sheets of craft paper for \(\$24\). Class 4A takes \(200\) sheets, Class 4B takes \(300\) sheets, and Class 4C takes the rest. How much should each class pay for its share?

Hints

- Find the cost of \(100\) sheets. - Determine how many sheets remain for Class 4C. - Express each class’s share as groups of \(100\) sheets.

Solution

1. The paper costs \(\$24 \div 6 = \$4\) for each group of \(100\) sheets. 2. Class 4A takes two groups of \(100\), so it pays \(2 \times \$4 = \$8\). 3. Class 4B takes three groups of \(100\), so it pays \(3 \times \$4 = \$12\). 4. Class 4C receives \(600 - 200 - 300 = 100\) sheets, so it pays \(\$4\).

Answer

Class 4A pays \(\$8\), Class 4B pays \(\$12\), and Class 4C pays \(\$4\).
5178944
A school receives \(144\) new laptops. First, each of the school’s \(4\) computer labs receives \(16\) laptops. All the remaining laptops are divided equally among \(8\) classrooms. How many laptops does each classroom receive?

Hints

- How many laptops are placed in the computer labs altogether? - Find how many laptops remain after the labs receive theirs. - Which operation divides the remaining laptops equally among the classrooms?

Solution

1. Find the number of laptops placed in the computer labs: \(4 \times 16 = 64\). 2. Find the number of laptops remaining: \(144 - 64 = 80\). 3. Divide the remaining laptops equally among eight classrooms: \(80 \div 8 = 10\).

Answer

Each classroom receives \(10\) laptops.
5178954
One library shelf has \(148\) adventure books and \(165\) animal books. Another shelf has \(152\) adventure books and \(159\) animal books. Which shelf has more books in all, and what is the difference?

Hints

- Find the total number of books on each shelf. - Compare the two totals. - Subtract the smaller total from the larger total.

Solution

1. Find the total on the first shelf: \(148 + 165 = 313\). 2. Find the total on the second shelf: \(152 + 159 = 311\). 3. Compare the totals: \(313 > 311\), so the first shelf has more books. 4. Find the difference: \(313 - 311 = 2\).

Answer

The first shelf has more books. It has \(2\) more books than the second shelf.
5178964
At a school event, \(125\) cups of apple juice and \(140\) cups of orange juice are sold in the morning. In the afternoon, \(30\) more cups of apple juice are sold than in the morning, but \(45\) fewer cups of orange juice are sold. When are more cups of juice sold in all, and what is the difference?

Hints

- Find the afternoon amount for each kind of juice. - Find the total for each time of day. - Compare the totals and subtract to find the difference.

Solution

1. Find the morning total: \(125 + 140 = 265\) cups. 2. Find the afternoon apple juice sales: \(125 + 30 = 155\) cups. 3. Find the afternoon orange juice sales: \(140 - 45 = 95\) cups. 4. Find the afternoon total: \(155 + 95 = 250\) cups. 5. Compare and subtract: \(265 > 250\) and \(265 - 250 = 15\).

Answer

More juice is sold in the morning. The difference is \(15\) cups.
5179064
A fruit stand sells apples in two bags. The small bag weighs \(4\,\text{kg}\) and costs \(\$12\). The large bag weighs \(10\,\text{kg}\) and costs \(\$20\). Lucas claims, “Because the large bag costs more altogether, its price per kilogram is also higher.” Is Lucas correct? Calculate the price of \(1\,\text{kg}\) of apples in each bag.

Hints

- How can you find the price of one kilogram in each bag? - Does a higher total price always mean a higher unit price? - Find each unit price separately and compare them.

Solution

1. Find the price per kilogram for the small bag: \(\$12 \div 4 = \$3\) per kilogram. 2. Find the price per kilogram for the large bag: \(\$20 \div 10 = \$2\) per kilogram. 3. Since \(\$2 < \$3\), the large bag has the lower price per kilogram. Lucas is not correct.

Answer

Lucas is not correct. Apples cost \(\$3\) per kilogram in the small bag and \(\$2\) per kilogram in the large bag.
5179654
A collection tank must hold \(15{,}000\,\text{L}\) of water. It receives \(4250\,\text{L}\) on Monday, \(5180\,\text{L}\) on Tuesday, and \(3940\,\text{L}\) on Wednesday. A leak lets \(285\,\text{L}\) drain out. How much more water is needed to fill the tank?

Hints

- First find the total amount that flowed into the tank. - The leak decreases the amount in the tank. - Compare the remaining amount with the tank's target capacity.

Solution

1. Add the water received: \(4250\,\text{L}+5180\,\text{L}+3940\,\text{L}=13{,}370\,\text{L}\). 2. Subtract the leaked water: \(13{,}370\,\text{L}-285\,\text{L}=13{,}085\,\text{L}\). 3. Find the amount still needed: \(15{,}000\,\text{L}-13{,}085\,\text{L}=1915\,\text{L}\).

Answer

\(1915\,\text{L}\)
5179904
A nursery ships flower seedlings in equal-sized packages. Four packages contain \(56\) seedlings altogether. A customer wants \(126\) seedlings and says, “I will need exactly twice as many packages as I would for \(56\) seedlings.” Is the customer correct? Support your answer with calculations.

Hints

- Identify what the customer is trying to compare. - First find the number of seedlings in one package. - Find twice the original number of packages. - Calculate the actual packages needed for \(126\) seedlings and compare.

Solution

1. Find the number of seedlings per package: \(56\div 4=14\). 2. Find the packages needed for \(126\) seedlings: \(126\div 14=9\). 3. Twice the original \(4\) packages would be \(4\times 2=8\) packages. 4. Since \(9\neq 8\), the customer is incorrect.

Answer

No. The customer needs \(9\) packages, but twice \(4\) packages is only \(8\) packages.
5180174
Class 3A harvests \(56\,\text{kg}\) of apples on Monday. On Tuesday, the class harvests \(16\,\text{kg}\) fewer than on Monday. The entire harvest is packed into crates that hold \(8\,\text{kg}\) each. a) How many crates are filled altogether? b) Would more or fewer crates be needed if each crate held \(12\,\text{kg}\)? Explain briefly without calculating again.

Hints

- How many kilograms of apples are harvested on Tuesday? - What is the total harvest for both days? - If each crate holds more apples, are more or fewer crates needed?

Solution

1. Find Tuesday's harvest: \(56\,\text{kg} - 16\,\text{kg} = 40\,\text{kg}\). 2. Find the total harvest: \(56\,\text{kg} + 40\,\text{kg} = 96\,\text{kg}\). 3. Find the number of eight-kilogram crates: \(96\,\text{kg} \div 8\,\text{kg} = 12\). 4. Fewer crates would be needed if each crate held \(12\,\text{kg}\), because each crate would hold more apples.

Answer

a) \(12\) crates are filled altogether. b) Fewer crates would be needed because each crate would hold more apples.
5180184
A bakery packs cookies into bags of exactly \(9\). Three trays contain different numbers of cookies: Tray A: \(85\) cookies Tray B: \(78\) cookies Tray C: \(92\) cookies The baker wants to use the tray that will leave the fewest cookies after filling as many complete bags as possible. Which tray should the baker choose?

Hints

- For each tray, find the greatest multiple of \(9\) that does not exceed the number of cookies. - Subtract that multiple from the number of cookies to find the remainder. - Compare the three remainders.

Solution

1. Find the remainder when the number of cookies on each tray is divided by \(9\). 2. Tray A: \(85 \div 9 = 9\) remainder \(4\). 3. Tray B: \(78 \div 9 = 8\) remainder \(6\). 4. Tray C: \(92 \div 9 = 10\) remainder \(2\). 5. Compare the remainders: \(2 < 4 < 6\). Tray C leaves the fewest cookies.

Answer

The baker should choose Tray C because it leaves only \(2\) cookies.
5180194
A teacher needs to divide \(46\) students into equal-size groups. The teacher is considering groups of \(4\), \(5\), or \(6\). Which group size leaves the fewest students without a complete group?

Hints

- Find the remainder for each possible group size. - Use nearby multiples of \(4\), \(5\), and \(6\). - Compare the three remainders.

Solution

1. Divide \(46\) by each possible group size and identify the remainder. 2. Groups of \(4\): \(46 \div 4 = 11\) remainder \(2\). 3. Groups of \(5\): \(46 \div 5 = 9\) remainder \(1\). 4. Groups of \(6\): \(46 \div 6 = 7\) remainder \(4\). 5. The smallest remainder is \(1\), so groups of \(5\) leave the fewest students without a complete group.

Answer

Groups of \(5\) leave the fewest students without a complete group.
5180214
A gardener buys young rose bushes for a park. Four bushes cost \(\$36\) altogether. The gardener needs \(15\) bushes for a new flower bed and pays with one \(\$100\) bill and one \(\$50\) bill. How much change does the gardener receive?

Hints

- First find the cost of one bush. - How much do all fifteen bushes cost? - Find the total amount the gardener pays. - Subtract the cost from the amount paid to find the change.

Solution

1. Find the cost of one rose bush: \(\$36 \div 4 = \$9\). 2. Find the cost of fifteen bushes: \(15 \times \$9 = \$135\). 3. Find the amount paid: \(\$100 + \$50 = \$150\). 4. Find the change: \(\$150 - \$135 = \$15\).

Answer

The gardener receives \(\$15\) in change.
5180284
A baker made \(135\) dinner rolls and packs \(7\) rolls in each bag. a) How many bags can the baker fill completely? b) How many rolls will be left over? c) How many more rolls would the baker need to fill one additional bag?

Hints

- Find the quotient and remainder when \(135\) is divided by \(7\). - What do the quotient and remainder represent? - Subtract the remainder from \(7\) to find how many more rolls are needed.

Solution

1. Divide the number of rolls by the number in each bag: \(135 \div 7\). 2. Since \(7 \times 19 = 133\), \(135 \div 7 = 19\) remainder \(2\). 3. The baker can fill \(19\) bags, with \(2\) rolls left over. 4. Subtract the remainder from the bag size: \(7 - 2 = 5\). The baker needs \(5\) more rolls for another full bag.

Answer

a) The baker can fill \(19\) bags completely. b) \(2\) rolls will be left over. c) The baker needs \(5\) more rolls to fill another bag.
5180334
A bag containing \(6\) apples of equal weight weighs \(900\,\text{g}\). How much do \(4\) of the apples weigh altogether?

Hints

- First find the weight of one apple. - To divide \(900\) by \(6\), you can break \(900\) into easier parts. - Once you know one apple's weight, how can you find the weight of four apples?

Solution

1. Find the weight of one apple: \(900\,\text{g} \div 6 = 150\,\text{g}\). 2. Find the weight of four apples: \(4 \times 150\,\text{g} = 600\,\text{g}\).

Answer

Four apples weigh \(600\,\text{g}\) altogether.
5180434
A gardener uses \(32\,\text{L}\) of soil to fill \(4\) flower boxes. How many liters of soil are needed to fill \(12\) identical flower boxes?

Hints

- First find the amount of soil for one flower box. - How many groups of four boxes are in twelve boxes? - How does the soil amount change when the number of boxes increases?

Solution

1. Find the amount of soil for one flower box: \(32\,\text{L} \div 4 = 8\,\text{L}\). 2. Find the amount for twelve flower boxes: \(12 \times 8\,\text{L} = 96\,\text{L}\).

Answer

The gardener needs \(96\,\text{L}\) of soil.
5180444
A machine packages \(45\) boxes in \(5\) minutes. If it works at a constant rate, how many boxes does it package in \(20\) minutes?

Hints

- First find how many boxes the machine packages in one minute. - How many groups of five minutes are in twenty minutes? - You can also scale the original amount by that factor.

Solution

1. Find the number of boxes packaged per minute: \(45 \div 5 = 9\). 2. Find the number packaged in twenty minutes: \(20 \times 9 = 180\).

Answer

The machine packages \(180\) boxes in \(20\) minutes.
5180454
A soccer stadium has \(2500\) seats. During the first ticket sale, \(845\) tickets are sold. During the second sale, \(125\) more tickets are sold than during the first sale. Sponsors also receive \(210\) complimentary tickets. How many seats remain available?

Hints

- First find the number of tickets sold during the second sale. - Add all sold and complimentary tickets. - Subtract the assigned seats from the stadium capacity.

Solution

1. The second sale includes \(845+125=970\) tickets. 2. The total number of assigned seats is \(845+970+210=2025\). 3. The number of available seats is \(2500-2025=475\).

Answer

\(475\) seats
5180534
A box contains \(120\) building blocks. One-third of the blocks are red, one-eighth are blue, and the rest are yellow. Find the number of yellow blocks. Are there more red blocks or yellow blocks? Support your answer with a comparison.

Hints

- First find the number of blocks in each given color group. - How many blocks remain after subtracting the red and blue blocks? - Compare the two numbers named in the question.

Solution

1. Find the number of red blocks: \(\frac{1}{3} \times 120 = 40\). 2. Find the number of blue blocks: \(\frac{1}{8} \times 120 = 15\). 3. Add the red and blue blocks: \(40 + 15 = 55\). 4. Subtract from the total to find the yellow blocks: \(120 - 55 = 65\). 5. Since \(65 > 40\), there are more yellow blocks than red blocks.

Answer

There are \(65\) yellow blocks. Because \(65 > 40\), there are more yellow blocks than red blocks.
5180554
A nursery plants \(22\) large flower boxes with \(12\) pansies in each box and \(18\) small boxes with \(8\) pansies in each box. After planting, \(25\) pansies remain on the supply cart. How many pansies were available at the beginning?

Hints

- First find how many flowers were planted in all the boxes. - Include the flowers that remained on the cart. - Calculate the two box sizes separately before combining them.

Solution

1. The large boxes use \(22\times 12=264\) pansies. 2. The small boxes use \(18\times 8=144\) pansies. 3. The boxes use \(264+144=408\) pansies altogether. 4. Add the remaining plants: \(408+25=433\).

Answer

There were \(433\) pansies at the beginning.
5180564
Drinks are delivered for a school fair. The delivery includes \(16\) cases of water with \(12\) bottles in each case and \(14\) cases of juice with \(15\) bottles in each case. a) How many bottles were delivered altogether? b) A volunteer says, “If we had ordered \(20\) cases with \(20\) bottles in each case, we would have received more bottles.” Is the volunteer correct? Compare the two totals.

Hints

- Find the number of bottles for each type of drink. - Find the total number of bottles in the proposed order. - Compare the proposed total with the answer from part a.

Solution

1. The water cases contain \(16\times 12=192\) bottles. 2. The juice cases contain \(14\times 15=210\) bottles. 3. The actual delivery contains \(192+210=402\) bottles. 4. The proposed order would contain \(20\times 20=400\) bottles. 5. Since \(402>400\), the volunteer is incorrect. The actual delivery has \(2\) more bottles.

Answer

a) \(402\) bottles b) No. The proposed order would have \(400\) bottles, which is \(2\) fewer than the actual delivery.
5180574
A class buys drinks for a school party: \(15\) bottles of apple juice at \(\$2\) each and \(24\) bottles of sparkling water at \(\$1\) each. After the purchase, \(\$13\) remains in the class fund. How much money was in the fund before the purchase?

Hints

- Find the total cost of each kind of drink. - Add those costs to find the amount spent. - Combine the amount spent with the money remaining.

Solution

1. The apple juice costs \(15 \times \$2 = \$30\). 2. The sparkling water costs \(24 \times \$1 = \$24\). 3. The class spends \(\$30 + \$24 = \$54\). 4. The starting amount was \(\$54 + \$13 = \$67\).

Answer

The class fund originally contained \(\$67\).
5180584
A teacher cuts pieces from a long wooden strip for a classroom project. She cuts \(18\) pieces that are each \(50\,\text{cm}\) long and \(25\) pieces that are each \(30\,\text{cm}\) long. A piece \(85\,\text{cm}\) long remains. How long was the original wooden strip in centimeters?

Hints

- Find the combined length of all the \(50\)-centimeter pieces. - Find the combined length of all the \(30\)-centimeter pieces. - The original strip included both groups of pieces and the leftover piece.

Solution

1. Find the total length of the \(50\)-centimeter pieces: \(18 \times 50 = 900\,\text{cm}\). 2. Find the total length of the \(30\)-centimeter pieces: \(25 \times 30 = 750\,\text{cm}\). 3. Add the lengths of all the cut pieces: \(900 + 750 = 1650\,\text{cm}\). 4. Add the remaining piece: \(1650 + 85 = 1735\,\text{cm}\).

Answer

The original wooden strip was \(1735\,\text{cm}\) long.
5180624
A gardener plans to plant \(120\) seedlings. After working for \(4\) hours, \(32\) seedlings remain. If the gardener planted the same number each hour, how many seedlings were planted per hour?

Hints

- First find how many seedlings have already been planted. - If the same number were planted each hour, divide the total planted among the four hours. - Break the problem into two operations.

Solution

1. Find the number of seedlings already planted: \(120 - 32 = 88\). 2. Divide by the number of hours: \(88 \div 4 = 22\).

Answer

The gardener planted \(22\) seedlings per hour.
5180634
A toy factory plans to package \(400\) toy cars during the morning. After \(3\) hours, \(160\) cars remain. On average, how many cars were packaged in each of the \(3\) hours?

Hints

- How many cars have already been packaged? - If that amount was completed in three hours, how many were packaged per hour? - Think about hundreds and tens when dividing.

Solution

1. Find the number of cars already packaged: \(400 - 160 = 240\). 2. Divide by the number of hours: \(240 \div 3 = 80\).

Answer

An average of \(80\) cars were packaged per hour.
5180724
A beekeeper fills \(380\) jars with honey. At a farmers market, the beekeeper sells \(80\) jars in the morning. The remaining jars will be divided equally among \(5\) boxes for delivery to a grocery store. A helper claims, “Each box will contain exactly \(50\) jars.” Is the helper correct? Calculate how many jars should actually be packed in each box.

Hints

- First find how many jars remain after the market sale. - Divide the remaining jars equally among the boxes. - Compare your result with the helper's claim.

Solution

1. Find the number of jars remaining after the sale: \(380 - 80 = 300\). 2. Divide the remaining jars equally among five boxes: \(300 \div 5 = 60\). 3. Compare the result with the helper's claim: \(60 \ne 50\). 4. The helper is not correct. Each box should contain \(60\) jars.

Answer

No. The helper is not correct; each box should contain \(60\) jars.
5180744
Three jars contain \(90\) marbles altogether. To make the amounts equal, Leo moves \(5\) marbles from the first jar to the second jar. Then he moves \(7\) marbles from the first jar to the third jar. Each jar now contains the same number of marbles. How many marbles were originally in each jar?

Hints

- First find the equal number of marbles in each jar at the end. - For each jar, decide whether marbles were added or removed. - Reverse all the moves made from the first jar. - A before-and-after table may help organize the changes.

Solution

1. After the moves, each jar contains \(90\div 3=30\) marbles. 2. The second jar received \(5\) marbles, so it began with \(30-5=25\). 3. The third jar received \(7\) marbles, so it began with \(30-7=23\). 4. The first jar gave away \(5+7=12\) marbles, so it began with \(30+12=42\).

Answer

The first jar had \(42\) marbles, the second had \(25\), and the third had \(23\).
5180794
A bookshelf holds \(600\) books when full. Students have already unpacked \(8\) boxes containing the same number of books. The shelf still has room for \(120\) more books. How many books were in each box?

Hints

- How many books are already on the shelf if there is room for \(120\) more? - Those books came from eight equal boxes. How can you find the number in one box? - First find the total number already unpacked.

Solution

1. Find the number of books already on the shelf: \(600 - 120 = 480\). 2. Divide by the number of boxes: \(480 \div 8 = 60\).

Answer

Each box contained \(60\) books.
5180804
Three crates contain \(75\) apples altogether. Six apples are removed from the first crate, and \(3\) apples are removed from the second crate. The three crates then contain equal numbers of apples. How many apples were in each crate at the beginning?

Hints

- Find how many apples remain after the removals. - Divide the remaining total equally among the three crates. - Add the removed apples back to the appropriate starting amounts.

Solution

1. A total of \(6+3=9\) apples are removed. 2. The crates then contain \(75-9=66\) apples altogether. 3. Each crate has \(66\div 3=22\) apples after the removals. 4. Reverse the changes: the first crate began with \(22+6=28\), the second with \(22+3=25\), and the third with \(22\).

Answer

The first crate had \(28\) apples, the second had \(25\), and the third had \(22\).
5180814
Four children collected \(100\) acorns altogether. Ethan loses \(8\) acorns, and Maya finds \(4\) more. The other two children’s amounts do not change. Afterward, all four children have the same number of acorns. How many acorns did each child have at the beginning?

Hints

- First find the new total after Ethan’s and Maya’s changes. - Divide the new total equally among four children. - Reverse each change to recover Ethan’s and Maya’s starting amounts.

Solution

1. After the changes, the total is \(100-8+4=96\). 2. Each child then has \(96\div 4=24\) acorns. 3. Reverse Ethan’s change: he began with \(24+8=32\) acorns. 4. Reverse Maya’s change: she began with \(24-4=20\) acorns. 5. The other two children each began with \(24\) acorns because their amounts did not change.

Answer

Ethan had \(32\) acorns, Maya had \(20\), and each of the other two children had \(24\).
5180954
Max wants to buy sticker packs. Large packs cost \(\$7\), and small packs cost \(\$4\). Max is exactly \(\$5\) short of having enough money for \(4\) large packs. How many small packs can he buy instead, and how much money will he have left?

Hints

- Find the cost of four large packs. - Use the amount he is short to find how much money he has. - Divide that amount by the small-pack price and interpret the remainder.

Solution

1. Four large packs cost \(4 \times \$7 = \$28\). 2. Max has \(\$28 - \$5 = \$23\). 3. Divide by the price of a small pack: \(\$23 \div \$4 = 5\) remainder \(\$3\). 4. Max can buy \(5\) small packs and have \(\$3\) left.

Answer

Max can buy \(5\) small packs and will have \(\$3\) left.
5180994
A gardener has \(450\) flower bulbs in \(6\) boxes. After removing \(30\) rotten bulbs, the gardener redistributes the remaining bulbs equally among the \(6\) boxes. How many flower bulbs are now in each box?

Hints

- How many bulbs remain after the rotten ones are removed? - The remaining bulbs are equally distributed among all six boxes. Which operation finds the number in each box?

Solution

1. Find the total number of bulbs remaining: \(450 - 30 = 420\). 2. Divide the remaining bulbs equally among the six boxes: \(420 \div 6 = 70\).

Answer

Each box now contains \(70\) flower bulbs.
5181144
Three friends share the cost of a large bag of marbles that costs \(\$12\). Lucas takes \(15\) marbles, Tim takes \(25\), and Sarah takes the remaining \(20\). Each person will pay for the number of marbles taken. How much should each person pay?

Hints

- Add the numbers of marbles to find the total. - Convert the total price to cents and use multiplication to find the cost per marble. - Multiply the cost per marble by each person’s number of marbles.

Solution

1. The bag contains \(15 + 25 + 20 = 60\) marbles. 2. The total cost is \(1200\) cents. Since \(60 \times 20 = 1200\), each marble costs \(20\) cents. 3. Lucas pays \(15 \times 20 = 300\) cents, or \(\$3\). 4. Tim pays \(25 \times 20 = 500\) cents, or \(\$5\). 5. Sarah pays \(20 \times 20 = 400\) cents, or \(\$4\).

Answer

Lucas should pay \(\$3\), Tim should pay \(\$5\), and Sarah should pay \(\$4\).
5181224
A package of \(5\) glitter stickers costs \(60\) cents. Lucas has saved \(\$3\) and wants to buy exactly \(30\) stickers. Does he have enough money? Support your answer with a calculation.

Hints

- Find how many packages are needed for \(30\) stickers. - Multiply the number of packages by the price per package. - Compare the total cost with \(\$3\).

Solution

1. Thirty stickers require \(30 \div 5 = 6\) packages. 2. The six packages cost \(6 \times 60 = 360\) cents, or \(\$3.60\). 3. Since \(\$3.60 > \$3.00\), Lucas does not have enough money.

Answer

No. The \(30\) stickers cost \(\$3.60\), so Lucas is \(60\) cents short.
5181284
Three classes collect paper for recycling. Class A collects \(14\,\text{kg}\). Class B collects four times as much as Class A. Class C collects \(20\,\text{kg}\) less than Class B. 1) How many kilograms does Class C collect? 2) How many kilograms do the three classes collect altogether?

Hints

- Find the amount for each class in order. - “Four times as much” indicates multiplication. - Add all three amounts for the final total.

Solution

1. Find the amount collected by Class B: \(14 \times 4 = 56\,\text{kg}\). 2. Find the amount collected by Class C: \(56\,\text{kg} - 20\,\text{kg} = 36\,\text{kg}\). 3. Add all three amounts: \(14\,\text{kg} + 56\,\text{kg} + 36\,\text{kg} = 106\,\text{kg}\).

Answer

1) Class C collects \(36\,\text{kg}\). 2) The three classes collect \(106\,\text{kg}\) altogether.
5181304
Julia is spending \(7\) days at summer camp. During the first \(3\) days, she spends \(\$8\) each day. She then sees that the money she has left is \(\$12\) more than the total she has spent so far. If she divides her remaining money equally among the days left, how much can she spend each day?

Hints

- First find the total amount Julia spends during the first \(3\) days. - How many days remain in the week? - Interpret “\(\$12\) more than the total she has spent” carefully. - Organizing the amounts and days in a small table may help.

Solution

1. Find how much Julia spends during the first \(3\) days: \(3 \times \$8 = \$24\). 2. Find how much money she has left: \(\$24 + \$12 = \$36\). 3. Find the number of days remaining: \(7 - 3 = 4\) days. 4. Divide the remaining money equally: \(\$36 \div 4 = \$9\) per day.

Answer

Julia can spend \(\$9\) each day for the remaining \(4\) days.
5181314
A water tank is used to irrigate a sports field. During the first \(4\) days, \(60\,\text{gal}\) are used each day. The amount left in the tank is \(90\,\text{gal}\) less than the total amount used during those \(4\) days. The remaining water must last for the next \(5\) days. How many gallons can be used each day?

Hints

- How much water is used during the first \(4\) days? - The amount left is \(90\,\text{gal}\) less than that total. What amount remains? - Divide the remaining amount among the next \(5\) days.

Solution

1. Find the total amount used during the first \(4\) days: \(4 \times 60 = 240\,\text{gal}\). 2. Find the amount left in the tank: \(240 - 90 = 150\,\text{gal}\). 3. Divide the remaining water equally among \(5\) days: \(150 \div 5 = 30\,\text{gal}\) per day.

Answer

They can use \(30\,\text{gal}\) each day for the next \(5\) days.
5181324
A school flea-market booth earns \(\$84\). It sells \(4\) books for \(\$6\) each. The rest of the money comes from stuffed animals sold for \(\$5\) each. How many items were sold altogether?

Hints

- Find the money earned from books first. - Subtract that amount from the total earnings. - Use the remaining earnings to find the number of stuffed animals, then add both item counts.

Solution

1. The books bring in \(4 \times \$6 = \$24\). 2. The stuffed animals bring in \(\$84 - \$24 = \$60\). 3. The booth sells \(\$60 \div \$5 = 12\) stuffed animals. 4. The total number of items is \(4 + 12 = 16\).

Answer

The booth sold \(16\) items altogether.
5181334
A fruit seller receives \(152\,\text{lb}\) of apples. The delivery includes \(6\) large crates containing \(18\,\text{lb}\) each and several small crates containing \(11\,\text{lb}\) each. How many crates are in the delivery altogether?

Hints

- Find the total weight in the large crates. - Subtract that amount from the full delivery. - Determine how many \(11\)-pound crates make the remaining weight, then add both kinds of crates.

Solution

1. Find the apples in the large crates: \(6 \times 18\,\text{lb} = 108\,\text{lb}\). 2. Find the apples in the small crates: \(152\,\text{lb} - 108\,\text{lb} = 44\,\text{lb}\). 3. Four small crates contain \(4 \times 11\,\text{lb} = 44\,\text{lb}\), so there are \(4\) small crates. 4. Add all the crates: \(6 + 4 = 10\).

Answer

The delivery contains \(10\) crates altogether.
5181404
A garden center uses \(35\) quarts of potting soil to fill \(5\) equal planters. a) How many quarts are needed to fill \(8\) planters? b) Ms. Miller has \(60\) quarts of potting soil. How many planters can she fill completely, and how many quarts will remain?

Hints

- Find the amount of soil for one planter first. - Use that amount to find the soil needed for \(8\) planters. - For part b), divide and interpret both the quotient and the remainder.

Solution

1. Find the amount for one planter: \(35 \div 5 = 7\) quarts. 2. Eight planters need \(8 \times 7 = 56\) quarts. 3. Divide \(60\) by \(7\): \(60 = 7 \times 8 + 4\). 4. Ms. Miller can fill \(8\) planters, with \(4\) quarts remaining.

Answer

a) \(56\) quarts b) She can fill \(8\) planters completely, and \(4\) quarts will remain.
5181474
Seven fence posts are placed at equal distances along a straight \(24\,\text{m}\) section of road. The first post is at \(0\,\text{m}\), and the last post is at \(24\,\text{m}\). What is the distance between two neighboring posts?

Hints

- How many spaces are there between \(7\) posts? - Draw a shorter row of posts and count the spaces. - Divide the total length by the number of equal spaces.

Solution

1. Seven posts create \(7 - 1 = 6\) equal spaces between neighboring posts. 2. Divide the total length by the number of spaces: \(24 \div 6 = 4\,\text{m}\).

Answer

Neighboring posts are \(4\,\text{m}\) apart.
5181634
Max removes \(24\) marbles from a box. Sophie then removes three times as many marbles as Max. There are \(15\) marbles left in the box. How many marbles were in the box at the beginning?

Hints

- First find how many marbles Sophie removes. - Add the amounts removed by Max and Sophie. - The starting number includes the marbles removed and the marbles left.

Solution

1. Find the number Sophie removes: \(24 \times 3 = 72\). 2. Find the total number removed: \(24 + 72 = 96\). 3. Add the marbles left: \(96 + 15 = 111\).

Answer

There were \(111\) marbles in the box at the beginning.
5181644
Lucas is saving for a bicycle that costs \(\$240\). He saves \(\$15\) in January. In February, he saves three times as much as in January. In March, he saves \(\$20\) less than in February. How much more money does Lucas need at the end of March?

Hints

- Find the amount saved in each month in order. - Add the three monthly amounts. - Subtract the total savings from the bicycle's price.

Solution

1. Find the February savings: \(15 \times 3 = 45\), so he saves \(\$45\). 2. Find the March savings: \(\$45 - \$20 = \$25\). 3. Find the total saved: \(\$15 + \$45 + \$25 = \$85\). 4. Find the amount still needed: \(\$240 - \$85 = \$155\).

Answer

Lucas still needs \(\$155\).
5181654
A gardener plants \(21\) rosebushes in one straight row. Each rosebush is exactly \(4\,\text{m}\) from the next one. The first rosebush is at the beginning of the row, and the last is at the end. How long is the row?

Hints

- Think about the spaces between the rosebushes. - How many spaces are there between \(3\) plants? Between \(4\) plants? - Multiply the number of spaces, not the number of plants, by \(4\,\text{m}\).

Solution

1. There is one fewer space than rosebushes: \(21 - 1 = 20\) spaces. 2. Multiply the number of spaces by the distance between neighboring rosebushes: \(20 \times 4 = 80\,\text{m}\).

Answer

The row is \(80\,\text{m}\) long.
5181664
A \(30\,\text{m}\) string will be decorated with small flags for a school event. a) If a flag is placed every \(3\,\text{m}\), including at both ends of the string, how many flags are needed? b) If the spacing is changed to \(5\,\text{m}\), how many fewer flags are needed than in part a)?

Hints

- First find the number of equal spaces along the string. - Remember that flags are placed at both the beginning and the end. - Repeat the process for the second spacing, then subtract.

Solution

1. With \(3\)-meter spacing, there are \(30 \div 3 = 10\) spaces. Including both endpoints requires \(10 + 1 = 11\) flags. 2. With \(5\)-meter spacing, there are \(30 \div 5 = 6\) spaces. Including both endpoints requires \(6 + 1 = 7\) flags. 3. The number saved is \(11 - 7 = 4\) flags.

Answer

a) \(11\) flags are needed. b) The larger spacing uses \(4\) fewer flags.
5181974
A small robot travels \(320\,\text{m}\) in \(4\) minutes. A newer robot travels \(15\,\text{m}\) farther each minute. How many meters does the newer robot travel in one minute?

Hints

- First find how far the small robot travels in one minute. - Which operation does the word “farther” suggest? - Use place value and a related basic division fact.

Solution

1. Find the distance the first robot travels in one minute: \(320\,\text{m} \div 4 = 80\,\text{m}\). 2. Add the additional distance traveled by the newer robot: \(80\,\text{m} + 15\,\text{m} = 95\,\text{m}\).

Answer

The newer robot travels \(95\,\text{m}\) in one minute.
5182024
A custodian delivers \(6\) cases of bottled water containing \(72\) bottles altogether. For a large school celebration, the custodian orders \(15\) identical cases. How many bottles are delivered for the celebration?

Hints

- How many bottles are in one case? - Decompose a factor by place value if helpful. - Remember that every case contains the same number of bottles.

Solution

1. Find the number of bottles in one case: \(72 \div 6 = 12\). 2. Find the number of bottles in fifteen cases: \(15 \times 12 = 180\).

Answer

\(180\) bottles are delivered for the celebration.
5182854
At an amusement park, the Blue Bolt roller coaster is \(645\,\text{m}\) long. The new Red Dragon coaster is \(278\,\text{m}\) longer than the Blue Bolt. a) How long is the Red Dragon? b) Are \(1500\,\text{m}\) of track enough to build both coasters? Show a calculation to support your answer.

Hints

- Use the phrase “longer than” to find the second coaster's length. - Add the lengths of both coasters. - Compare the total with \(1500\,\text{m}\).

Solution

1. Find the length of the Red Dragon: \(645 + 278 = 923\,\text{m}\). 2. Find the combined length of both coasters: \(645 + 923 = 1568\,\text{m}\). 3. Since \(1568 > 1500\), the available track is not enough. 4. Find the shortage: \(1568 - 1500 = 68\,\text{m}\).

Answer

a) The Red Dragon is \(923\,\text{m}\) long. b) No. The two coasters require \(1568\,\text{m}\) of track, which is \(68\,\text{m}\) more than the amount available.
5182864
The Miller family is planning a vacation. A one-way trip to the coast is \(534\,\text{miles}\). A one-way trip to the mountains is \(218\,\text{miles}\) longer. How many miles would the family travel on a round trip to the mountains?

Hints

- First find the one-way distance to the mountains. - A round trip includes the trip there and the trip back. - How many times is the one-way distance traveled?

Solution

1. Find the one-way distance to the mountains: \(534 + 218 = 752\,\text{miles}\). 2. Double the one-way distance for the round trip: \(752 \times 2 = 1504\,\text{miles}\).

Answer

The family would travel \(1504\,\text{miles}\) on the round trip.
5182994
A school buys \(750\) sheets of craft paper for project week. Each of \(9\) groups receives \(65\) sheets. How many sheets remain in storage?

Hints

- First find how many sheets all nine groups receive. - Use the number of groups and the amount per group. - Subtract the distributed amount from \(750\).

Solution

1. Find the total number distributed: \(9 \times 65 = 585\). 2. Subtract from the amount purchased: \(750 - 585 = 165\).

Answer

There are \(165\) sheets left in storage.
5183074
A school library receives \(145\) books on Monday. On Tuesday, it receives \(67\) more books than on Monday. On Wednesday, it receives \(25\) fewer books than the Monday and Tuesday deliveries combined. How many books arrive on Wednesday?

Hints

- First find Tuesday’s delivery. - Then combine the Monday and Tuesday deliveries. - Read carefully whether Wednesday’s delivery is more or fewer than that combined amount.

Solution

1. Tuesday’s delivery is \(145+67=212\) books. 2. Monday and Tuesday together total \(145+212=357\) books. 3. Wednesday’s delivery is \(357-25=332\) books.

Answer

\(332\) books arrive on Wednesday.
5183084
Three fourth-grade classes collect aluminum cans for an environmental project. Class 4A collects \(326\) cans. Class 4B collects \(47\) fewer cans than Class 4A. Class 4C collects \(85\) more cans than Classes 4A and 4B combined. a) How many cans did Class 4C collect? b) Did the three classes collect more than \(1200\) cans altogether? Support your answer with a calculation.

Hints

- Find Class 4B’s amount first, paying attention to “fewer.” - Combine Classes 4A and 4B before finding Class 4C’s amount. - Add all three class amounts to answer part b.

Solution

1. Class 4B collected \(326-47=279\) cans. 2. Classes 4A and 4B collected \(326+279=605\) cans together. 3. Class 4C collected \(605+85=690\) cans. 4. The three classes collected \(605+690=1295\) cans altogether. 5. Since \(1295>1200\), they collected more than \(1200\) cans.

Answer

a) Class 4C collected \(690\) cans. b) Yes. The classes collected \(1295\) cans altogether.
5183254
A store sells small packages containing \(125\) colored pencils and large packages containing \(160\) colored pencils. Ms. Miller buys \(4\) small packages, and Mr. Smith buys \(3\) large packages. Who buys more pencils, and by how many?

Hints

- Find each person's total number of pencils. - Break apart the larger factors if helpful. - Compare the products and subtract to find the difference.

Solution

1. Find Ms. Miller's total: \(4 \times 125 = 500\). 2. Find Mr. Smith's total: \(3 \times 160 = 480\). 3. Compare: \(500 > 480\), so Ms. Miller buys more. 4. Find the difference: \(500 - 480 = 20\).

Answer

Ms. Miller buys \(20\) more colored pencils than Mr. Smith.
5183264
Lucas packs grapes for a school fair. Each bag needs exactly \(145\,\text{g}\) of grapes, and he has \(750\,\text{g}\) altogether. Does he have enough to fill \(5\) bags? How many grams will be left, or how many more grams will he need?

Hints

- Find the total amount needed for all five bags. - Break apart \(145\) to make the multiplication easier. - Compare the amount needed with \(750\,\text{g}\).

Solution

1. Find the amount needed for five bags: \(5 \times 145\,\text{g} = 725\,\text{g}\). 2. Since \(725\,\text{g} < 750\,\text{g}\), there are enough grapes. 3. Find the amount left: \(750\,\text{g} - 725\,\text{g} = 25\,\text{g}\).

Answer

Yes. Lucas can fill the five bags and will have \(25\,\text{g}\) of grapes left.
5183274
Lucas has \(8\) packs with \(15\) stickers in each pack. Mia has \(6\) packs with \(22\) stickers in each pack. Who has more stickers, and how many more?

Hints

- Find each child's total number of stickers. - Use multiplication for each set of equal groups. - Subtract the smaller total from the larger total.

Solution

1. Find Lucas's total: \(8 \times 15 = 120\). 2. Find Mia's total: \(6 \times 22 = 132\). 3. Since \(132 > 120\), Mia has more stickers. 4. Find the difference: \(132 - 120 = 12\).

Answer

Mia has \(12\) more stickers than Lucas.
5183384
A nursery harvests \(400\) tulips. The workers first make \(8\) large bouquets with \(15\) tulips in each bouquet. All the remaining tulips are divided into small bouquets with \(7\) tulips each. How many small bouquets can the workers make?

Hints

- How many tulips are used in the large bouquets altogether? - How many tulips remain for the small bouquets? - Which operation divides a quantity into equal groups? - Check that you have used every important number in the problem.

Solution

1. Find the number of tulips used in the large bouquets: \(8 \times 15 = 120\). 2. Find the number of tulips remaining: \(400 - 120 = 280\). 3. Divide the remaining tulips into groups of seven: \(280 \div 7 = 40\).

Answer

The workers can make \(40\) small bouquets.
5183394
A gardener buys \(8\) rose bushes for \(\$12\) each and several lavender plants for \(\$5\) each. The total cost is \(\$151\). How many lavender plants does the gardener buy?

Hints

- First find the amount spent on rose bushes. - Subtract that amount from the total to find the amount spent on lavender plants. - How many times does the price of one lavender plant fit into that amount? - Work through the problem one step at a time.

Solution

1. Find the total cost of the rose bushes: \(8 \times \$12 = \$96\). 2. Find the amount spent on lavender plants: \(\$151 - \$96 = \$55\). 3. Divide by the cost of one lavender plant: \(\$55 \div \$5 = 11\).

Answer

The gardener buys \(11\) lavender plants.
5183404
Sarah buys \(15\) pens for her class for \(\$4\) each and several drawing pads for \(\$3\) each. She pays with \(\$100\) and receives \(\$7\) in change. How many drawing pads does Sarah buy?

Hints

- How much does Sarah spend after receiving change? - How much do all the pens cost? - Subtract the pen cost to find the amount spent on drawing pads. - Which operation finds the number of pads from their total cost?

Solution

1. Find the total amount spent: \(\$100 - \$7 = \$93\). 2. Find the cost of the pens: \(15 \times \$4 = \$60\). 3. Find the amount spent on drawing pads: \(\$93 - \$60 = \$33\). 4. Divide by the cost of one drawing pad: \(\$33 \div \$3 = 11\).

Answer

Sarah buys \(11\) drawing pads.
5183724
The Johnson family buys \(4\) young trees for \(\$125\) each and one bag of fertilizer for \(\$38\). The Patel family buys \(6\) shrubs for \(\$82\) each and soil for \(\$54\). Which family spends more, and what is the difference?

Hints

- Find each family’s plant cost first. - Add the fertilizer or soil cost to each subtotal. - Compare the totals and subtract to find the difference.

Solution

1. The Johnson family spends \(4 \times \$125 + \$38 = \$500 + \$38 = \$538\). 2. The Patel family spends \(6 \times \$82 + \$54 = \$492 + \$54 = \$546\). 3. Since \(\$546 > \$538\), the Patel family spends more. 4. The difference is \(\$546 - \$538 = \$8\).

Answer

The Patel family spends more by \(\$8\).
5183734
A fourth-grade class of \(24\) students is comparing two museum-trip offers. - Offer A: Admission is \(\$12\) per student, plus a one-time bus fee of \(\$55\). - Offer B: Admission and the bus are included for a flat price of \(\$310\). Which offer costs less, and how much does the class save?

Hints

- Find the total admission cost under Offer A. - Add the one-time bus fee to that amount. - Compare the result with the flat price and find the difference.

Solution

1. Offer A costs \(24 \times \$12 + \$55 = \$288 + \$55 = \$343\). 2. Offer B costs \(\$310\). 3. Since \(\$310 < \$343\), Offer B costs less. 4. The class saves \(\$343 - \$310 = \$33\).

Answer

Offer B costs less. The class saves \(\$33\).
5183744
Two school classes collect acorns in the fall for a wildlife habitat project. Class 3A fills \(8\) bags containing \(72\,\text{kg}\) of acorns altogether. Class 3B fills \(6\) bags containing \(36\,\text{kg}\) altogether. Every bag filled by the same class has an equal weight. a) How many kilograms are in one Class 3A bag? b) How many kilograms are in one Class 3B bag? c) Which class collects more per bag?

Hints

- How can you find the weight in one bag for each class? - Which operation divides a total equally among several bags? - Compare the two per-bag weights at the end.

Solution

1. Find the weight in each Class 3A bag: \(72\,\text{kg} \div 8 = 9\,\text{kg}\). 2. Find the weight in each Class 3B bag: \(36\,\text{kg} \div 6 = 6\,\text{kg}\). 3. Compare the results: \(9\,\text{kg} > 6\,\text{kg}\), so Class 3A collects more per bag.

Answer

a) Each Class 3A bag contains \(9\,\text{kg}\). b) Each Class 3B bag contains \(6\,\text{kg}\). c) Class 3A collects more per bag.
5183754
A nursery places young plants on tables. In the first section, \(450\) plants are divided equally among \(9\) tables. In the second section, \(490\) plants are divided equally among \(7\) tables. How many more plants are on each table in the second section than on each table in the first section?

Hints

- How many plants are on one table in the first section? - How many plants are on one table in the second section? - Once you know both values, find their difference. - Related basic division facts may help.

Solution

1. Find the number of plants on each table in the first section: \(450 \div 9 = 50\). 2. Find the number of plants on each table in the second section: \(490 \div 7 = 70\). 3. Find the difference: \(70 - 50 = 20\).

Answer

Each table in the second section has \(20\) more plants.
5183944
At the beginning of the semester, each of the \(24\) students in Class 4B receives \(4\) lined notebooks and \(3\) graph-paper notebooks. How many notebooks must be ordered for the class altogether?

Hints

- First find the number of notebooks one student receives. - Multiply that amount by the number of students. - Another method is to find the total for each notebook type separately and then add.

Solution

1. One student receives \(4+3=7\) notebooks. 2. For \(24\) students, the total is \(24\times 7=168\) notebooks. 3. Another method is to calculate \(24\times 4=96\) lined notebooks and \(24\times 3=72\) graph-paper notebooks, then add \(96+72=168\).

Answer

\(168\) notebooks must be ordered.
5183954
A small animal sanctuary cares for \(6\) ponies. Each pony eats \(10\,\text{lb}\) of hay per day. The owner has \(900\,\text{lb}\) of hay. Is that enough to feed all the ponies for exactly \(2\) weeks, or \(14\) days? Show your work and state how much hay will be left or how much more is needed.

Hints

- How much hay do all the ponies eat in one day? - How many days are in \(2\) weeks? - Compare the total amount needed with the available supply.

Solution

1. Find the hay needed by all \(6\) ponies in one day: \(6 \times 10 = 60\,\text{lb}\). 2. Find the hay needed for \(14\) days: \(60 \times 14 = 840\,\text{lb}\). 3. Since \(840 < 900\), the supply is enough. 4. Find the amount left: \(900 - 840 = 60\,\text{lb}\). Alternative method: 1. Find the hay needed by one pony for \(14\) days: \(10 \times 14 = 140\,\text{lb}\). 2. Find the hay needed by \(6\) ponies: \(140 \times 6 = 840\,\text{lb}\). 3. Compare \(840\,\text{lb}\) with \(900\,\text{lb}\), and find the difference.

Answer

Yes. The ponies need \(840\,\text{lb}\) of hay, so \(60\,\text{lb}\) will be left.
5183964
Leonie has \(12\) seashells. Her mother has six times as many seashells as Leonie. Her cousin has \(50\) fewer seashells than her mother. How many more seashells does the cousin have than Leonie?

Hints

- First find the mother's number of seashells. - Use that amount to find the cousin's number. - Then compare the cousin's amount with Leonie's.

Solution

1. Find the mother's number of seashells: \(12 \times 6 = 72\). 2. Find the cousin's number: \(72 - 50 = 22\). 3. Find the difference between the cousin and Leonie: \(22 - 12 = 10\).

Answer

The cousin has \(10\) more seashells than Leonie.
5183994
A sports club has \(\$450\) to buy equipment. A coach buys \(6\) basketballs for \(\$24\) each and \(4\) identical sports bags. After the purchase, \(\$194\) remains. What is the price of one sports bag?

Hints

- First find the total amount the coach spends. - Find the cost of all the basketballs. - The remaining part of the spending is the cost of the sports bags. - Divide that amount equally among the four bags.

Solution

1. Find the total amount spent: \(\$450 - \$194 = \$256\). 2. Find the cost of the basketballs: \(6 \times \$24 = \$144\). 3. Find the amount spent on sports bags: \(\$256 - \$144 = \$112\). 4. Divide by the number of sports bags: \(\$112 \div 4 = \$28\).

Answer

One sports bag costs \(\$28\).
5184434
A store has \(8\) shelves with \(42\) notebooks on each shelf. During a sale, the store sells \(175\) notebooks. On Friday evening, a delivery adds \(60\) new notebooks. How many notebooks are on the shelves Saturday morning?

Hints

- Find the starting total on all eight shelves. - Subtract the notebooks sold. - Then add the new delivery.

Solution

1. Find the starting number of notebooks: \(8 \times 42 = 336\). 2. Subtract the notebooks sold: \(336 - 175 = 161\). 3. Add the delivery: \(161 + 60 = 221\).

Answer

There are \(221\) notebooks on the shelves Saturday morning.
5184444
A worker at a strawberry farm earns \(\$125\) per day. The owner hires \(6\) workers for \(4\) days. How much will the owner pay all the workers in total?

Hints

- First find what all the workers earn in one day. - Then account for all \(4\) days. - Another approach is to find the total number of worker-days first.

Solution

1. Find what all \(6\) workers earn in one day: \(\$125 \times 6 = \$750\). 2. Find the total for \(4\) days: \(\$750 \times 4 = \$3000\). Alternative method: 1. Find the total number of worker-days: \(6 \times 4 = 24\) worker-days. 2. Find the total pay: \(24 \times \$125 = \$3000\).

Answer

The owner will pay \(\$3000\) in total.
5184454
A school festival committee buys \(3\) cases of apple juice for each of \(6\) classes. Each case costs \(\$14\). The committee budgeted \(\$300\) for the juice. How much money will remain after the purchase?

Hints

- Find the total number of cases first. - Multiply the number of cases by the cost per case. - Subtract the purchase cost from the budget.

Solution

1. The committee buys \(6 \times 3 = 18\) cases. 2. The cases cost \(18 \times \$14 = \$252\). 3. The amount remaining is \(\$300 - \$252 = \$48\).

Answer

\(\$48\) will remain.
5184504
A farmer packs potatoes into \(24\) small sacks weighing \(15\,\text{lb}\) each and \(16\) large sacks weighing \(45\,\text{lb}\) each. What is the total weight of the potato harvest?

Hints

- Find the total weight of the small sacks. - Find the total weight of the large sacks. - Add the two results.

Solution

1. Find the weight of the small sacks: \(24 \times 15\,\text{lb} = 360\,\text{lb}\). 2. Find the weight of the large sacks: \(16 \times 45\,\text{lb} = 720\,\text{lb}\). 3. Add the two amounts: \(360\,\text{lb} + 720\,\text{lb} = 1080\,\text{lb}\).

Answer

The potato harvest weighs \(1080\,\text{lb}\).
5184514
A movie theater has two auditoriums. One auditorium has \(14\) rows with \(18\) seats in each row. The other has \(12\) rows with \(22\) seats in each row. How many seats are in the two auditoriums altogether?

Hints

- Find the number of seats in each auditorium separately. - Multiply the rows by the seats in each row. - Add the two auditorium totals.

Solution

1. The first auditorium has \(14\times 18=252\) seats. 2. The second auditorium has \(12\times 22=264\) seats. 3. Together, they have \(252+264=516\) seats.

Answer

The two auditoriums have \(516\) seats altogether.
5184524
A gardener is planting flower beds in a city park. Eight tulip beds each hold \(55\) bulbs, and \(12\) daffodil beds each hold \(45\) bulbs. Before planting, the gardener finds that \(15\) tulip bulbs and \(20\) daffodil bulbs are damaged and must be discarded. How many healthy bulbs will be planted altogether?

Hints

- Find how many bulbs would be planted if none were damaged. - Find the total number of damaged bulbs. - You may subtract the damaged bulbs by type or from the combined total. - Organize the tulip and daffodil calculations separately.

Solution

1. The tulip beds were planned for \(8\times 55=440\) bulbs. 2. The daffodil beds were planned for \(12\times 45=540\) bulbs. 3. The total planned number is \(440+540=980\) bulbs. 4. The number of damaged bulbs is \(15+20=35\). 5. The number of healthy bulbs is \(980-35=945\). 6. Alternatively, \(440-15=425\) healthy tulip bulbs and \(540-20=520\) healthy daffodil bulbs, and \(425+520=945\).

Answer

\(945\) healthy bulbs will be planted.
5184564
A gardener buys \(12\) red rose plants for \(\$5\) each and several white rose plants for \(\$7\) each. The total cost is \(\$95\). How many white rose plants does the gardener buy? How much more do all the red rose plants cost than all the white rose plants?

Hints

- First find the total cost of the type of rose whose quantity is known. - How much of the total cost remains for the other roses? - How can you find the number of roses bought with that amount? - Finally, compare the two total costs.

Solution

1. Find the total cost of the red rose plants: \(12 \times \$5 = \$60\). 2. Find the amount spent on white rose plants: \(\$95 - \$60 = \$35\). 3. Find the number of white rose plants: \(\$35 \div \$7 = 5\). 4. Find the difference between the two total costs: \(\$60 - \$35 = \$25\).

Answer

The gardener buys \(5\) white rose plants. The red rose plants cost \(\$25\) more altogether than the white rose plants.
5184644
An elementary school attends a theater performance. A child’s ticket costs \(\$6\), and an adult ticket costs \(\$9\). The school spends \(\$78\) on children’s tickets and \(\$54\) on adult tickets. How many people attend the performance altogether?

Hints

- First find how many children attend. - How many adults attend if their tickets cost \(\$54\) altogether? - To divide \(78\) by \(6\), you can break \(78\) into easier parts. - Add the numbers of children and adults at the end.

Solution

1. Find the number of children: \(\$78 \div \$6 = 13\). 2. Find the number of adults: \(\$54 \div \$9 = 6\). 3. Find the total number of people: \(13 + 6 = 19\).

Answer

\(19\) people attend the theater performance altogether.
5184694
A school garden buys \(20\) rose bushes for \(\$16\) each and \(20\) lavender plants. Each lavender plant costs \(\$9\) less than one rose bush. The school has a budget of \(\$500\). Is the budget enough? Find the total cost and the amount left over or still needed.

Hints

- Find the price of one lavender plant. - Use the equal quantities to find the total cost efficiently. - Compare the total with the budget and find the difference.

Solution

1. One lavender plant costs \(\$16 - \$9 = \$7\). 2. One rose bush and one lavender plant cost \(\$16 + \$7 = \$23\). 3. Twenty pairs cost \(20 \times \$23 = \$460\). 4. Since \(\$460 < \$500\), the budget is enough. 5. The amount left is \(\$500 - \$460 = \$40\).

Answer

Yes. The total cost is \(\$460\), and \(\$40\) remains.
5184744
A school library has \(64\) nonfiction books. It has \(5\) times as many children’s books as nonfiction books. It has twice as many comic books as the nonfiction books and children’s books combined. How many books are there altogether?

Hints

- First find the number of children’s books. - Combine the nonfiction and children’s books. - Use that combined amount to find the comic books. - Include all three types in the final total.

Solution

1. The number of children’s books is \(64\times 5=320\). 2. The nonfiction and children’s books total \(64+320=384\). 3. The number of comic books is \(384\times 2=768\). 4. The library has \(384+768=1152\) books altogether.

Answer

The library has \(1152\) books altogether.
5184754
A bakery makes \(132\) rolls on Monday. On Tuesday, it makes \(3\) times as many rolls as on Monday. On Wednesday, it makes \(4\) times as many rolls as the Monday and Tuesday amounts combined. How many rolls does the bakery make over the three days altogether?

Hints

- Find Tuesday’s amount first. - Combine Monday and Tuesday before finding Wednesday’s amount. - Use that intermediate sum to calculate Wednesday’s production. - Include all three days in the final total.

Solution

1. Tuesday’s amount is \(132\times 3=396\) rolls. 2. Monday and Tuesday together total \(132+396=528\) rolls. 3. Wednesday’s amount is \(528\times 4=2112\) rolls. 4. The three-day total is \(528+2112=2640\) rolls.

Answer

The bakery makes \(2640\) rolls over the three days.
5184794
A baker has a \(50\,\text{kg}\) bag of flour. For the first \(4\) days, the baker uses \(6\,\text{kg}\) each day to make bread. After that, the baker uses \(5\,\text{kg}\) each day to make rolls. For how many complete days does the flour last altogether, and how much flour remains?

Hints

- First find the total amount of flour used during the first four days. - How much flour remains after that? - How many complete groups of \(5\,\text{kg}\) fit into the remaining amount? - Report both the total number of days and the amount left over.

Solution

1. Find the flour used during the first four days: \(4 \times 6\,\text{kg} = 24\,\text{kg}\). 2. Find the flour remaining: \(50\,\text{kg} - 24\,\text{kg} = 26\,\text{kg}\). 3. The remaining flour lasts for \(5\) complete days at \(5\,\text{kg}\) per day, with \(1\,\text{kg}\) left over. 4. Find the total number of complete days: \(4 + 5 = 9\).

Answer

The flour lasts for \(9\) complete days altogether, and \(1\,\text{kg}\) remains.
5184804
Leonie has collected \(135\) stickers. She places \(15\) stickers on each of the first \(4\) pages of her album. She puts \(9\) stickers on each additional page. How many additional pages can she fill, and how many stickers remain?

Hints

- How many stickers are used on the first four pages? - How many stickers remain for the additional pages? - Which multiplication fact helps you find the number of groups of nine? - Does the division have a remainder?

Solution

1. Find the number of stickers used on the first four pages: \(4 \times 15 = 60\). 2. Find the number of stickers remaining: \(135 - 60 = 75\). 3. Divide the remaining stickers into groups of nine: \(75 \div 9 = 8\) remainder \(3\).

Answer

Leonie can fill \(8\) additional pages, and \(3\) stickers remain.
5184834
An elementary school orders fruit for a school event. The order includes \(14\) crates of apples weighing \(12\,\text{lb}\) each and \(16\) crates of pears weighing \(9\,\text{lb}\) each. How many more pounds do the apples weigh than the pears?

Hints

- Find the total weight of all the apple crates. - Find the total weight of all the pear crates. - Subtract to find the difference.

Solution

1. Find the total weight of the apples: \(14 \times 12\,\text{lb} = 168\,\text{lb}\). 2. Find the total weight of the pears: \(16 \times 9\,\text{lb} = 144\,\text{lb}\). 3. Find the difference: \(168\,\text{lb} - 144\,\text{lb} = 24\,\text{lb}\).

Answer

The apples weigh \(24\,\text{lb}\) more than the pears.
5184844
A truck carries \(25\) bags of playground sand weighing \(20\,\text{lb}\) each and \(18\) buckets of gravel weighing \(15\,\text{lb}\) each. The truck may carry at most \(1000\,\text{lb}\) of cargo. How many more pounds can be added?

Hints

- Find the weight of each kind of material. - Add the two cargo weights. - Subtract the current cargo from the maximum capacity.

Solution

1. Find the sand's weight: \(25 \times 20\,\text{lb} = 500\,\text{lb}\). 2. Find the gravel's weight: \(18 \times 15\,\text{lb} = 270\,\text{lb}\). 3. Find the current cargo weight: \(500\,\text{lb} + 270\,\text{lb} = 770\,\text{lb}\). 4. Find the remaining capacity: \(1000\,\text{lb} - 770\,\text{lb} = 230\,\text{lb}\).

Answer

The truck can carry \(230\,\text{lb}\) more.
5185074
Lucas has saved \(\$220\) for skateboarding gear. A skateboard costs \(\$84\). Knee pads cost half as much as the skateboard. A helmet costs \(\$15\) more than the knee pads. How much money will Lucas have left after buying all three items?

Hints

- Find the knee-pad price first. - Use that price to find the helmet price. - Add all three costs, then subtract from the amount saved.

Solution

1. The knee pads cost \(\$84 \div 2 = \$42\). 2. The helmet costs \(\$42 + \$15 = \$57\). 3. The three items cost \(\$84 + \$42 + \$57 = \$183\). 4. Lucas has \(\$220 - \$183 = \$37\) left.

Answer

Lucas will have \(\$37\) left.
5185084
A fourth-grade class has raised \(\$350\) for an end-of-year picnic. The picnic-site rental costs \(\$120\). Drinks cost one-third of the rental fee, and food costs three times as much as the drinks. The remaining money will be used for decorations. How much money is available for decorations?

Hints

- Find one-third of the rental fee. - Use the drink cost to find the food cost. - Add all planned expenses, then subtract from the amount raised.

Solution

1. The drinks cost \(\$120 \div 3 = \$40\). 2. The food costs \(3 \times \$40 = \$120\). 3. The rental, drinks, and food cost \(\$120 + \$40 + \$120 = \$280\). 4. The amount available for decorations is \(\$350 - \$280 = \$70\).

Answer

\(\$70\) is available for decorations.
5185094
The Miller family uses \(145\,\text{gal}\) of water per day. The Schmidt family uses \(112\,\text{gal}\) per day. Does the Miller family use more or less water in \(7\) days than the Schmidt family uses in \(9\) days? Find the exact difference in gallons.

Hints

- A week has \(7\) days. - Find each family’s total water use separately. - Compare the two totals. - Subtract to find the difference.

Solution

1. Find the Miller family’s use in \(7\) days: \(145 \times 7 = 1015\,\text{gal}\). 2. Find the Schmidt family’s use in \(9\) days: \(112 \times 9 = 1008\,\text{gal}\). 3. Since \(1015 > 1008\), the Miller family uses more water. 4. Find the difference: \(1015 - 1008 = 7\,\text{gal}\).

Answer

The Miller family uses \(1015\,\text{gal}\), which is \(7\,\text{gal}\) more than the Schmidt family’s \(1008\,\text{gal}\).
5185104
A baker uses two sizes of baking sheets. In the morning, each sheet holds \(18\) rolls, and the baker fills \(14\) sheets. In the afternoon, each smaller sheet holds \(15\) rolls, and the baker fills \(16\) sheets. The baker says, “I filled more sheets in the afternoon, so I must have baked more rolls then.” Is the baker correct? Compare the two totals.

Hints

- Find the total number of rolls for each part of the day. - The number of sheets alone does not determine the total. - Compare the two products.

Solution

1. In the morning, the baker makes \(14\times 18=252\) rolls. 2. In the afternoon, the baker makes \(16\times 15=240\) rolls. 3. Since \(252>240\), the baker is incorrect. 4. The morning total is \(252-240=12\) rolls greater.

Answer

No. The baker made \(252\) rolls in the morning and \(240\) in the afternoon, so the morning total was \(12\) greater.
5185214
A baker makes \(245\) pretzels on Saturday morning. In the afternoon, the baker makes \(85\) fewer pretzels than in the morning. All the pretzels will be packed in bags that hold exactly \(8\) pretzels each. The baker has \(50\) empty bags. Are there enough bags? Justify your answer with calculations.

Hints

- How many pretzels are made in the afternoon? - How many pretzels are made altogether? - How many pretzels can the fifty bags hold? - Compare the bag capacity with the total number of pretzels.

Solution

1. Find the number of pretzels made in the afternoon: \(245 - 85 = 160\). 2. Find the total number of pretzels: \(245 + 160 = 405\). 3. Find the capacity of fifty bags: \(50 \times 8 = 400\) pretzels. 4. Since \(405 > 400\), fifty bags are not enough. One additional bag is needed, for a total of \(51\) bags.

Answer

No. The baker makes \(405\) pretzels, but \(50\) bags hold only \(400\) pretzels. One more bag is needed.
5185234
Six packages of tile weigh \(240\,\text{kg}\) altogether. All the packages have the same weight. a) How much does one package weigh? b) A tile installer needs only \(4\) of the packages for a small bathroom. How much do those \(4\) packages weigh altogether?

Hints

- First find the weight of one package. - Once you know one package's weight, how can you find the weight of four packages? - Break the problem into two steps.

Solution

1. Find the weight of one package: \(240\,\text{kg} \div 6 = 40\,\text{kg}\). 2. Find the weight of four packages: \(4 \times 40\,\text{kg} = 160\,\text{kg}\).

Answer

a) One package weighs \(40\,\text{kg}\). b) Four packages weigh \(160\,\text{kg}\) altogether.
5185584
A stork flies \(450\,\text{km}\) in \(3\) days during migration. A crane flies \(560\,\text{km}\) in \(4\) days. If each bird flies the same distance each day, which bird travels farther in one day?

Hints

- First find each bird's distance for one day. - Break the larger numbers into parts that are easier to divide by three or four. - Compare the two daily distances.

Solution

1. Find the stork's daily distance: \(450\,\text{km} \div 3 = 150\,\text{km}\). 2. Find the crane's daily distance: \(560\,\text{km} \div 4 = 140\,\text{km}\). 3. Compare the daily distances: \(150\,\text{km} > 140\,\text{km}\). 4. The stork travels farther in one day.

Answer

The stork travels farther in one day.
5185674
A bus has \(14\) passengers. At the first stop, enough people board to triple the number of passengers. At the second stop, \(8\) people leave and \(5\) people board. How many passengers are now on the bus?

Hints

- “Triple” means multiply by \(3\). - Follow the events at the stops in order. - Subtract people who leave and add people who board.

Solution

1. Find the number after the first stop: \(14 \times 3 = 42\). 2. Subtract the passengers who leave: \(42 - 8 = 34\). 3. Add the passengers who board: \(34 + 5 = 39\).

Answer

There are \(39\) passengers on the bus.
5185814
For a school event, Class 3A buys lemonade. Eight cases cost \(\$72\) altogether. The teacher decides that the class needs \(12\) cases. How much do the \(12\) cases cost altogether?

Hints

- First find the cost of one case. - Once you know the unit price, how can you find the cost of twelve cases?

Solution

1. Find the cost of one case: \(\$72 \div 8 = \$9\). 2. Find the cost of twelve cases: \(12 \times \$9 = \$108\).

Answer

Twelve cases of lemonade cost \(\$108\) altogether.
5185894
A stationery store has three containers of pencils. The first contains \(120\) pencils. The second contains exactly half as many as the first. The third contains \(20\) fewer pencils than the second. All the pencils are combined and divided equally among \(4\) new boxes. How many pencils are placed in each box?

Hints

- Find the number of pencils in each container one at a time. - What operation finds one-half of a quantity? - How many pencils are there altogether? - Divide the total equally among the new boxes.

Solution

1. Find the number of pencils in the second container: \(120 \div 2 = 60\). 2. Find the number in the third container: \(60 - 20 = 40\). 3. Find the total number of pencils: \(120 + 60 + 40 = 220\). 4. Divide equally among four boxes: \(220 \div 4 = 55\).

Answer

Each box contains \(55\) pencils.
5185964
For a school event, \(420\) apples are packed into bags. a) How many bags are needed if each bag holds \(6\) apples? b) How many bags are needed if each bag holds \(7\) apples? c) Compare your answers. Explain why part b needs fewer bags.

Hints

- Divide \(420\) by each bag size. - Compare the two quotients. - Think about what happens to the number of bags when each bag holds more apples.

Solution

1. For part a, divide by \(6\): \(420 \div 6 = 70\). 2. For part b, divide by \(7\): \(420 \div 7 = 60\). 3. Compare the results: \(60 < 70\), so part b uses \(10\) fewer bags. 4. When each bag holds more apples, fewer bags are needed for the same total number of apples.

Answer

a) \(70\) bags are needed. b) \(60\) bags are needed. c) Part b needs fewer bags because each bag holds more apples.
5185994
Four friends buy \(3\) new books for \(\$12\) each and a large board game for \(\$32\) for their classroom game area. They divide the total cost equally. How much does each child pay?

Hints

- How much do all three books cost? - What is the total cost including the board game? - Which operation divides the total cost equally among four children? - You can break the total into parts that are easy to divide by four.

Solution

1. Find the cost of the three books: \(3 \times \$12 = \$36\). 2. Find the total cost: \(\$36 + \$32 = \$68\). 3. Divide the cost equally among four friends: \(\$68 \div 4 = \$17\).

Answer

Each child pays \(\$17\).
5186004
A group of \(9\) children plans a trip. Their tickets cost \(\$108\) altogether, and they spend another \(\$45\) on food. They use a \(\$9\) coupon toward the total cost. If the remaining cost is divided equally, how much does each child pay?

Hints

- What do the tickets and food cost altogether? - Does the coupon increase or decrease the amount owed? - How much does the group still need to pay after using the coupon? - Divide that amount equally among nine children.

Solution

1. Add the ticket and food costs: \(\$108 + \$45 = \$153\). 2. Subtract the coupon: \(\$153 - \$9 = \$144\). 3. Divide the remaining cost equally among nine children: \(\$144 \div 9 = \$16\).

Answer

Each child pays \(\$16\).
5186014
The table shows distances between cities in Colorado, in miles. <table> <thead> <tr><th>Distances in miles</th><th>Denver</th><th>Fort Collins</th><th>Boulder</th><th>Colorado Springs</th><th>Pueblo</th></tr> </thead> <tbody> <tr><th>Denver</th><td>---</td><td>65</td><td>30</td><td>70</td><td>115</td></tr> <tr><th>Fort Collins</th><td>65</td><td>---</td><td>50</td><td>130</td><td>175</td></tr> <tr><th>Boulder</th><td>30</td><td>50</td><td>---</td><td>100</td><td>145</td></tr> <tr><th>Colorado Springs</th><td>70</td><td>130</td><td>100</td><td>---</td><td>45</td></tr> <tr><th>Pueblo</th><td>115</td><td>175</td><td>145</td><td>45</td><td>---</td></tr> </tbody> </table> A group travels along this route: Pueblo \(\rightarrow\) Colorado Springs \(\rightarrow\) Boulder \(\rightarrow\) Denver \(\rightarrow\) Fort Collins. How many miles does the group travel altogether?

Hints

- Find the distance for each pair of consecutive cities in the route. - Check the correct row and column for each distance. - Add all four route segments.

Solution

1. Read each leg of the route from the table: Pueblo to Colorado Springs is \(45\) miles, Colorado Springs to Boulder is \(100\) miles, Boulder to Denver is \(30\) miles, and Denver to Fort Collins is \(65\) miles. 2. Add the distances: \(45 + 100 + 30 + 65 = 240\).

Answer

The group travels \(240\) miles altogether.
5186024
A delivery van must travel from Seattle to Portland with two stops. The table shows distances in miles. <table> <thead> <tr><th>Distances in miles</th><th>Seattle</th><th>Tacoma</th><th>Olympia</th><th>Everett</th><th>Portland</th></tr> </thead> <tbody> <tr><th>Seattle</th><td>---</td><td>35</td><td>60</td><td>30</td><td>175</td></tr> <tr><th>Tacoma</th><td>35</td><td>---</td><td>30</td><td>65</td><td>145</td></tr> <tr><th>Olympia</th><td>60</td><td>30</td><td>---</td><td>90</td><td>115</td></tr> <tr><th>Everett</th><td>30</td><td>65</td><td>90</td><td>---</td><td>205</td></tr> <tr><th>Portland</th><td>175</td><td>145</td><td>115</td><td>205</td><td>---</td></tr> </tbody> </table> Route A: Seattle \(\rightarrow\) Tacoma \(\rightarrow\) Olympia \(\rightarrow\) Portland Route B: Seattle \(\rightarrow\) Everett \(\rightarrow\) Olympia \(\rightarrow\) Portland Which route is shorter, and by how many miles?

Hints

- Find the total distance for each route separately. - Compare the two totals. - Subtract the shorter distance from the longer distance.

Solution

1. Route A is \(35 + 30 + 115 = 180\) miles. 2. Route B is \(30 + 90 + 115 = 235\) miles. 3. Compare and subtract: \(235 - 180 = 55\).

Answer

Route A is shorter. It is \(55\) miles shorter than Route B.
5186044
An orchard harvests \(600\,\text{lb}\) of apples. Workers remove \(48\,\text{lb}\) of damaged apples. The remaining apples are divided equally among \(6\) large crates. How many pounds of apples are placed in each crate?

Hints

- Subtract the damaged apples first. - Divide the remaining amount equally among \(6\) crates. - Check by multiplication.

Solution

1. Find the amount remaining: \(600\,\text{lb} - 48\,\text{lb} = 552\,\text{lb}\). 2. Divide equally among the crates: \(552\,\text{lb} \div 6 = 92\,\text{lb}\).

Answer

Each crate contains \(92\,\text{lb}\) of apples.
5186054
Two fourth-grade classes raise money at a flea market for a charity. Class 4A raises \(\$245\), and Class 4B raises \(\$275\). They first pay \(\$50\) for booth and supply costs. They donate the remaining money equally to two animal shelters. How much does each shelter receive?

Hints

- Add the amounts raised by both classes. - Subtract the expenses. - Divide the remaining amount equally between the two shelters.

Solution

1. The classes raise \(\$245 + \$275 = \$520\). 2. After paying expenses, \(\$520 - \$50 = \$470\) remains. 3. Each shelter receives \(\$470 \div 2 = \$235\).

Answer

Each animal shelter receives \(\$235\).
5186084
\(5\) identical machines fill a total of \(900\,\text{L}\) of juice in \(6\) hours. Each machine fills the same amount each hour. How many liters does one machine fill in one hour?

Hints

- First find how much all the machines fill in one hour. - Then divide that amount equally among the \(5\) machines. - Keep track of the unit in each step.

Solution

1. Find how much juice all \(5\) machines fill in one hour: \(900 \div 6 = 150\). They fill \(150\,\text{L}\) per hour together. 2. Divide that hourly amount equally among the \(5\) machines: \(150 \div 5 = 30\). 3. One machine fills \(30\,\text{L}\) in one hour.

Answer

One machine fills \(30\,\text{L}\) in one hour.
5186174
There are \(22\) students on the playground. After two boys leave for lunch, the number of girls and boys on the playground is the same. How many girls and how many boys were on the playground at first? Briefly explain your reasoning.

Hints

- Find how many students remain after the two boys leave. - When two groups are equal and their total is known, divide the total by \(2\). - The number of girls does not change. - Check that your two starting groups add to \(22\).

Solution

1. After two boys leave, \(22 - 2 = 20\) students remain. 2. The remaining students are split equally between girls and boys: \(20 \div 2 = 10\). 3. The number of girls did not change, so there were \(10\) girls at first. 4. Add back the two boys who left: \(10 + 2 = 12\).

Answer

There were \(10\) girls and \(12\) boys at first.
5186184
A fruit bowl contains \(16\) apples and pears altogether. After Paul eats one pear, there are exactly twice as many apples as pears. How many apples and pears were in the bowl at first?

Hints

- Find how many fruits remain after one pear is eaten. - If the number of apples is twice the number of pears, how many equal parts are there altogether? - Divide the remaining fruits into those equal parts. - Add the eaten pear back at the end.

Solution

1. After one pear is eaten, \(16 - 1 = 15\) fruits remain. 2. The apples make up two equal parts and the pears make up one equal part, for \(3\) parts total. 3. Each part is \(15 \div 3 = 5\). Therefore, \(10\) apples and \(5\) pears remain. 4. Add back the pear that was eaten: there were originally \(5 + 1 = 6\) pears and \(10\) apples.

Answer

There were \(10\) apples and \(6\) pears at first.
5186224
A small playground ball costs \(\$3\). A leather soccer ball costs \(\$27\). a) How many times as much does the soccer ball cost as the playground ball? b) Ms. Weber buys two soccer balls for her school. How many playground balls could she have bought for the same amount of money?

Hints

- Use division to find how many times one price fits into the other. - For part b), find the total cost of two soccer balls. - Divide that total by the price of one playground ball.

Solution

1. Divide the prices: \(27 \div 3 = 9\). The soccer ball costs \(9\) times as much. 2. Two soccer balls cost \(2 \times 27 = 54\) dollars. 3. Divide by the price of a playground ball: \(54 \div 3 = 18\).

Answer

a) \(9\) times as much b) \(18\) playground balls
5186254
An elementary school receives \(8\) boxes of new notebooks. Each box contains \(50\) notebooks. The school will give each class a bundle of \(20\) notebooks. For how many classes are there enough notebooks?

Hints

- First find the total number of notebooks in all the boxes. - Divide the total by the number given to each class. - Work with the total supply before determining the number of classes.

Solution

1. Find the total number of notebooks: \(8\times 50=400\). 2. Divide the notebooks into bundles of \(20\): \(400\div 20=20\).

Answer

There are enough notebooks for \(20\) classes.
5186264
A fruit seller receives \(5\) crates of apples weighing \(24\,\text{lb}\) each. a) How many \(3\)-pound bags can be filled? b) How many fewer bags are needed if \(4\)-pound bags are used instead?

Hints

- Find the total weight of all the apples. - Divide by each bag size separately. - Subtract the two numbers of bags.

Solution

1. Find the total weight of the apples: \(5 \times 24\,\text{lb} = 120\,\text{lb}\). 2. With \(3\)-pound bags: \(120\,\text{lb} \div 3\,\text{lb} = 40\) bags. 3. With \(4\)-pound bags: \(120\,\text{lb} \div 4\,\text{lb} = 30\) bags. 4. Find the difference: \(40 - 30 = 10\) bags.

Answer

a) The seller can fill \(40\) three-pound bags. b) The seller needs \(10\) fewer bags when using four-pound bags.
5186494
A home improvement store sells \(6\) packages of paving stones for \(\$120\) altogether. One package of edging stones costs \(\$15\) more than one package of paving stones. How much do \(8\) packages of edging stones cost?

Hints

- First find the cost of one package of paving stones. - How much more does one package of edging stones cost? - Which operation finds the cost of eight packages?

Solution

1. Find the cost of one package of paving stones: \(\$120 \div 6 = \$20\). 2. Find the cost of one package of edging stones: \(\$20 + \$15 = \$35\). 3. Find the cost of eight packages of edging stones: \(8 \times \$35 = \$280\).

Answer

Eight packages of edging stones cost \(\$280\).
5186694
A teacher buys \(8\) packages of markers for \(\$32\). Another teacher wants \(12\) packages of the same markers and has a \(\$50\) bill. Is \(\$50\) enough? Support your answer with a calculation.

Hints

- Find the price of one package. - Use the unit price to find the cost of \(12\) packages. - Compare the cost with \(\$50\).

Solution

1. One package costs \(\$32 \div 8 = \$4\). 2. Twelve packages cost \(12 \times \$4 = \$48\). 3. Since \(\$48 < \$50\), the money is enough.

Answer

Yes. The \(12\) packages cost \(\$48\), so \(\$50\) is enough.
5186734
A school library plans to buy \(6\) bookcases. Each bookcase costs \(\$120\), and each one also needs a \(\$15\) bookend set. The library has a budget of \(\$850\). Is the budget enough for the entire order? Justify your answer.

Hints

- Find the cost of one bookcase together with its bookend set. - Multiply that cost by \(6\). - Compare the total cost with the budget.

Solution

1. Find the cost of one bookcase with its bookend set: \(\$120 + \$15 = \$135\). 2. Find the cost of six sets: \(6 \times \$135 = \$810\). 3. Compare with the budget: \(\$810 < \$850\), so the budget is enough.

Answer

Yes. The order costs \(\$810\), which is less than the \(\$850\) budget.
5186754
Six flower beds in a city park will be planted with bulbs. The delivery includes \(4\) bags with \(145\) bulbs each, \(3\) bags with \(120\) bulbs each, and one bag with \(110\) bulbs. If all the bulbs are divided equally among the beds, how many bulbs will be planted in each bed?

Hints

- First find the total number of bulbs delivered. - Calculate each type of bag separately before adding. - Include the single bag of \(110\) bulbs. - Use division to share the total equally.

Solution

1. The four larger bags contain \(4\times 145=580\) bulbs. 2. The three other bags contain \(3\times 120=360\) bulbs. 3. The total delivery is \(580+360+110=1050\) bulbs. 4. Divide equally among six beds: \(1050\div 6=175\).

Answer

Each flower bed will receive \(175\) bulbs.
5186854
A farm has \(8\) goats and a supply of \(960\,\text{lb}\) of hay. Each goat eats \(4\,\text{lb}\) of hay per day. How many days will the hay supply last for all the goats?

Hints

- One method is to divide the hay equally among the \(8\) goats first. - Then determine how many days each goat’s share will last. - Another method is to find how much hay all \(8\) goats eat in one day and use multiplication to find the number of days.

Solution

1. Divide the hay equally among the \(8\) goats: \(960 \div 8 = 120\,\text{lb}\) per goat. 2. Find how many days each goat’s share will last: \(120 \div 4 = 30\) days. Alternative method: 1. Find how much hay all \(8\) goats eat in one day: \(8 \times 4 = 32\,\text{lb}\). 2. Since \(30 \times 32\,\text{lb} = 960\,\text{lb}\), the full supply lasts \(30\) days.

Answer

The hay supply will last \(30\) days.
5186914
A class buys \(30\) plants for a garden project. Twelve are tomato plants costing \(\$2\) each. The remaining plants are herb plants costing \(\$3\) each. The class pays with \(\$100\). How much change does it receive?

Hints

- First find the number of herb plants. - Find the cost of each type of plant. - Add the costs, then subtract from \(\$100\).

Solution

1. Find the number of herb plants: \(30 - 12 = 18\). 2. Find the cost of the tomato plants: \(12 \times \$2 = \$24\). 3. Find the cost of the herb plants: \(18 \times \$3 = \$54\). 4. Find the total cost: \(\$24 + \$54 = \$78\). 5. Find the change: \(\$100 - \$78 = \$22\).

Answer

The class receives \(\$22\) in change.
5186924
A class of \(24\) students and \(2\) teachers visits a museum. Individual tickets cost \(\$5\) per student and \(\$8\) per teacher. A group ticket for everyone costs \(\$130\). Which option costs less, and what is the difference in price?

Hints

- Find the total cost for all student tickets. - Include the teacher tickets. - Compare the individual-ticket total with the group-ticket price.

Solution

1. Find the cost of the student tickets: \(24 \times \$5 = \$120\). 2. Find the cost of the teacher tickets: \(2 \times \$8 = \$16\). 3. Find the total for individual tickets: \(\$120 + \$16 = \$136\). 4. Compare with the group ticket: \(\$130 < \$136\). 5. Find the difference: \(\$136 - \$130 = \$6\).

Answer

The group ticket costs less by \(\$6\).
5186974
Two art classes order identical watercolor sets. Class 4A orders \(12\) sets, and Class 4B orders \(15\) sets. Class 4B’s bill is \(\$36\) more than Class 4A’s bill. a) How much does one watercolor set cost? b) How much does each class pay? c) How much do the two classes pay altogether?

Hints

- Find how many more sets Class 4B orders. - Use the bill difference to find the price of one set. - Multiply the unit price by each order, then add the two totals.

Solution

1. Class 4B orders \(15 - 12 = 3\) more sets. 2. One set costs \(\$36 \div 3 = \$12\). 3. Class 4A pays \(12 \times \$12 = \$144\). 4. Class 4B pays \(15 \times \$12 = \$180\). 5. Together, the classes pay \(\$144 + \$180 = \$324\).

Answer

a) One watercolor set costs \(\$12\). b) Class 4A pays \(\$144\), and Class 4B pays \(\$180\). c) Together, the classes pay \(\$324\).
5187074
A school buys art supplies. Class 4A receives \(7\) watercolor sets at \(\$14\) each. Class 4B receives \(9\) oil-paint sets. Each oil-paint set costs \(\$6\) more than a watercolor set. How much does the school pay altogether?

Hints

- Find the price of one oil-paint set. - Find the two class totals separately. - Add the class totals.

Solution

1. One oil-paint set costs \(\$14 + \$6 = \$20\). 2. The watercolor sets cost \(7 \times \$14 = \$98\). 3. The oil-paint sets cost \(9 \times \$20 = \$180\). 4. The total cost is \(\$98 + \$180 = \$278\).

Answer

The school pays \(\$278\) altogether.
5187084
An orchard sells fruit by the crate. A crate of apples costs \(\$18\), and the orchard sells \(15\) crates. It also sells \(12\) crates of pears. Each crate of pears costs twice as much as a crate of apples. How much greater is the revenue from the pear crates than the revenue from the apple crates?

Hints

- Find the apple-crate revenue. - Use “twice as much” to find the price of one pear crate. - Find the pear-crate revenue, then subtract the two totals.

Solution

1. The apple-crate revenue is \(15 \times \$18 = \$270\). 2. One pear crate costs \(2 \times \$18 = \$36\). 3. The pear-crate revenue is \(12 \times \$36 = \$432\). 4. The difference is \(\$432 - \$270 = \$162\).

Answer

The pear-crate revenue is \(\$162\) greater.
5187094
Two bags of flour weigh \(100\,\text{oz}\) altogether. After \(10\,\text{oz}\) of flour is moved from the first bag to the second bag, the bags weigh the same. How many ounces of flour were originally in each bag?

Hints

- Find how much each bag contains after the transfer. - Reverse the transfer to find the original amounts. - Check that the original amounts add to the total.

Solution

1. After the transfer, each bag has half of the total: \(100\,\text{oz} \div 2 = 50\,\text{oz}\). 2. The first bag lost \(10\,\text{oz}\), so it originally held \(50\,\text{oz} + 10\,\text{oz} = 60\,\text{oz}\). 3. The second bag gained \(10\,\text{oz}\), so it originally held \(50\,\text{oz} - 10\,\text{oz} = 40\,\text{oz}\).

Answer

The first bag originally held \(60\,\text{oz}\), and the second bag originally held \(40\,\text{oz}\).
5187194
A hiking group travels a total of \(12\,\text{mi}\). The group hikes one-fourth of the distance through a forest and one-third across a meadow. The rest of the route is a steep climb. How many miles long is the climbing section?

Hints

- You could draw a bar for the whole route and mark the known parts. - What do one-fourth and one-third mean as operations? - Find the lengths of the known sections first. - What remains after subtracting the known sections from the whole route?

Solution

1. Find the forest distance: \(\frac{1}{4} \times 12\,\text{mi} = 3\,\text{mi}\). 2. Find the meadow distance: \(\frac{1}{3} \times 12\,\text{mi} = 4\,\text{mi}\). 3. Add the known sections: \(3\,\text{mi} + 4\,\text{mi} = 7\,\text{mi}\). 4. Subtract from the total: \(12\,\text{mi} - 7\,\text{mi} = 5\,\text{mi}\).

Answer

The climbing section is \(5\,\text{mi}\) long.
5187254
A sports club buys \(6\) basketball sets for \(\$432\) altogether. It also buys \(9\) volleyball sets whose total cost is \(\$54\) more than the basketball-set total. Which type of ball set costs more per set, and by how much?

Hints

- Find the price per basketball set. - Use the difference between the two total bills to find the volleyball total. - Find the price per volleyball set and compare the unit prices.

Solution

1. One basketball set costs \(\$432 \div 6 = \$72\). 2. The nine volleyball sets cost \(\$432 + \$54 = \$486\). 3. One volleyball set costs \(\$486 \div 9 = \$54\). 4. Since \(\$72 > \$54\), a basketball set costs more. 5. The difference is \(\$72 - \$54 = \$18\).

Answer

A basketball set costs \(\$18\) more than a volleyball set.
5187264
A wildlife park has two rabbit areas. The first area has \(14\) hutches with \(4\) rabbits in each hutch. The second area has \(11\) hutches with \(6\) rabbits in each hutch. Which area has more rabbits, and how many more?

Hints

- Find the total number of rabbits in each area. - Use multiplication for each set of equal groups. - Subtract the smaller total from the larger total.

Solution

1. Find the number in the first area: \(14 \times 4 = 56\). 2. Find the number in the second area: \(11 \times 6 = 66\). 3. Find the difference: \(66 - 56 = 10\).

Answer

The second area has \(10\) more rabbits.
5187284
Animals are counted at a safari park. There are \(144\) zebras. There are four times as many antelopes as zebras. The number of lions is one-sixth the number of zebras. There are \(120\) fewer giraffes than zebras. How many animals of these four species live in the park altogether?

Hints

- Find the number of animals in each species one at a time. - Pay close attention to “four times as many,” “one-sixth as many,” and “120 fewer.” - Which operation matches each phrase? - Add all four groups at the end.

Solution

1. The number of zebras is \(144\). 2. Find the number of antelopes: \(4 \times 144 = 576\). 3. Find the number of lions: \(\frac{1}{6} \times 144 = 24\). 4. Find the number of giraffes: \(144 - 120 = 24\). 5. Add all four groups: \(144 + 576 + 24 + 24 = 768\).

Answer

There are \(768\) animals of these four species altogether.
5187364
A farmers market starts the day with: - \(30\) jars of honey at \(\$7\) each - \(45\) jars of jam at \(\$4\) each - \(60\) cartons of eggs at \(\$3\) each At the end of the day, \(12\) jars of honey, \(18\) jars of jam, and \(15\) cartons of eggs remain. Which product earns the most money, and what is the market’s total revenue from these three products?

Hints

- Find how many of each product were sold. - Find the revenue from each product separately. - Compare the three revenues, then add them.

Solution

1. The market sells \(30 - 12 = 18\) jars of honey, \(45 - 18 = 27\) jars of jam, and \(60 - 15 = 45\) cartons of eggs. 2. Honey earns \(18 \times \$7 = \$126\). 3. Jam earns \(27 \times \$4 = \$108\). 4. Eggs earn \(45 \times \$3 = \$135\). 5. The egg cartons earn the most money. 6. The total revenue is \(\$126 + \$108 + \$135 = \$369\).

Answer

The egg cartons earn the most, with \(\$135\). The total revenue is \(\$369\).
5187414
At a field day, \(6\) groups of \(12\) students participate. Each student receives one bottle of water and one apple. a) How many items are distributed altogether? b) Lucas uses \(6 \times 12 + 6 \times 12\). Julia uses \(6 \times 24\). Are both methods correct? Explain where Julia gets \(24\).

Hints

- Find the total number of students and the number of items each student receives. - Identify what each \(6 \times 12\) represents in Lucas's method. - Find the number of items received by one group.

Solution

1. There are \(6 \times 12 = 72\) students, and each receives \(2\) items, so \(72 \times 2 = 144\) items. 2. Lucas counts \(72\) water bottles and \(72\) apples: \(6 \times 12 + 6 \times 12 = 144\). 3. One group receives \(12 + 12 = 24\) items, so Julia counts \(6\) groups of \(24\): \(6 \times 24 = 144\). Both methods are correct.

Answer

a) \(144\) items are distributed. b) Yes. Julia's \(24\) is the \(12\) water bottles and \(12\) apples received by one group.
5187424
A school garden has two tulip beds. The first bed has \(4\) rows with \(18\) tulips in each row. The second bed has \(5\) rows with \(14\) tulips in each row. Which bed has more tulips, and how many more?

Hints

- Find the total number in each bed. - Compare the two products. - Subtract to find how many more.

Solution

1. Find the number in the first bed: \(4 \times 18 = 72\). 2. Find the number in the second bed: \(5 \times 14 = 70\). 3. Since \(72 > 70\), the first bed has more tulips. 4. Find the difference: \(72 - 70 = 2\).

Answer

The first bed has \(2\) more tulips than the second bed.
5187484
At a school snack bar, small pretzels cost \(40\) cents and large pretzels cost \(70\) cents. Felix buys \(6\) small pretzels for his friends and has \(60\) cents left. Did he have enough money at the start to buy \(5\) large pretzels instead? Explain.

Hints

- Find the cost of the six small pretzels. - Use the money left to find Felix's starting amount. - Find the cost of five large pretzels and compare.

Solution

1. Find the cost of the small pretzels: \(6 \times 40 = 240\) cents, or \(\$2.40\). 2. Find Felix's starting amount: \(\$2.40 + \$0.60 = \$3.00\). 3. Find the cost of five large pretzels: \(5 \times 70 = 350\) cents, or \(\$3.50\). 4. Since \(\$3.00 < \$3.50\), he would be \(\$0.50\) short.

Answer

No. Felix had \(\$3.00\), but five large pretzels cost \(\$3.50\), so he would be \(\$0.50\) short.
5187524
A fruit farm harvests \(456\,\text{lb}\) of apples. The pear harvest is one-half as large as the apple harvest. The plum harvest is \(85\,\text{lb}\) less than the pear harvest. How many pounds of fruit are harvested altogether?

Hints

- Put the information in the order needed for the calculations. - How can you find one-half of a quantity? - What operation combines all the fruit amounts at the end? - Notice which fruit amount depends on another amount.

Solution

1. Find the pear harvest: \(\frac{1}{2} \times 456\,\text{lb} = 228\,\text{lb}\). 2. Find the plum harvest: \(228\,\text{lb} - 85\,\text{lb} = 143\,\text{lb}\). 3. Add all three harvests: \(456\,\text{lb} + 228\,\text{lb} + 143\,\text{lb} = 827\,\text{lb}\).

Answer

The farm harvests \(827\,\text{lb}\) of fruit altogether.
5187534
A fourth-grade class wants to raise \(\$2000\) for a class trip. In the first week, a bake sale raises \(\$420\). In the second week, a flea market raises twice as much as the first week. In the third week, the class raises \(\$150\) less than in the second week. How much more money does the class need to reach its goal?

Hints

- Find the amount raised in each week. - Add the three weekly amounts. - Subtract the total raised from the goal.

Solution

1. The second week raises \(2 \times \$420 = \$840\). 2. The third week raises \(\$840 - \$150 = \$690\). 3. The class has raised \(\$420 + \$840 + \$690 = \$1950\). 4. The amount still needed is \(\$2000 - \$1950 = \$50\).

Answer

The class still needs \(\$50\).
5187564
Lucas has \(14\) bags with \(6\) marbles in each bag. His sister Marie has \(8\) bags with \(11\) marbles in each bag. Who has more marbles, and how many more?

Hints

- Find each child's total number of marbles. - Use multiplication for each set of equal groups. - Compare the totals and subtract.

Solution

1. Find Lucas's total: \(14 \times 6 = 84\). 2. Find Marie's total: \(8 \times 11 = 88\). 3. Since \(88 > 84\), Marie has more. 4. Find the difference: \(88 - 84 = 4\).

Answer

Marie has \(4\) more marbles than Lucas.
5187574
A gardener plants \(28\) rows with \(7\) tulips in each row in one flower bed. In a second bed, the gardener plants \(35\) rows with \(5\) daffodils in each row. Which bed has more flowers, and what is the difference?

Hints

- Find the total number of flowers in each bed. - Use a multiplication equation for each bed. - Compare the totals and find their difference.

Solution

1. Find the number of tulips: \(28 \times 7 = 196\). 2. Find the number of daffodils: \(35 \times 5 = 175\). 3. Since \(196 > 175\), the first bed has more flowers. 4. Find the difference: \(196 - 175 = 21\).

Answer

The first bed has \(21\) more flowers than the second bed.
5187614
Lucas has saved \(\$42\). He buys \(4\) card games for \(\$6\) each. With the remaining money, he buys as many comic books as possible for \(\$5\) each. How many comic books can Lucas buy, and how much money remains?

Hints

- First find how much Lucas spends on the card games. - How much money remains after that purchase? - How many times does the comic-book price fit into the remaining amount? - Does any money remain after buying the greatest possible number?

Solution

1. Find the cost of the card games: \(4 \times \$6 = \$24\). 2. Find the money remaining: \(\$42 - \$24 = \$18\). 3. Divide by the cost of one comic book: \(\$18 \div \$5 = 3\) remainder \(\$3\).

Answer

Lucas can buy \(3\) comic books, and \(\$3\) remains.
5187664
A gardener has enough tulip bulbs to plant \(12\) rows with \(9\) bulbs in each row and still have \(12\) bulbs left. The gardener changes the plan and puts \(8\) bulbs in each row so that no bulbs remain. How many rows can be planted under the new plan?

Hints

- How many tulip bulbs are there altogether? - First find the number used in twelve rows of nine. - Add the bulbs that were left over. - Divide the total by the new number in each row.

Solution

1. Find the number of bulbs in the original twelve rows: \(12 \times 9 = 108\). 2. Find the total number of bulbs: \(108 + 12 = 120\). 3. Divide by the new number of bulbs per row: \(120 \div 8 = 15\).

Answer

The gardener can plant \(15\) rows.
5187684
Four children are saving for a project. They currently have: - Lucas: \(\$12\) - Mia: \(\$17\) - Noah: \(\$9\) - Sara: \(\$22\) They will add money so that each child ends with the same amount and together they have exactly \(\$100\). How much must each child add? How much will they add altogether?

Hints

- Divide the final total equally among the four children. - Compare each current amount with that target. - Add the four missing amounts.

Solution

1. Each child’s target amount is \(\$100 \div 4 = \$25\). 2. Lucas must add \(\$25 - \$12 = \$13\). 3. Mia must add \(\$25 - \$17 = \$8\). 4. Noah must add \(\$25 - \$9 = \$16\). 5. Sara must add \(\$25 - \$22 = \$3\). 6. Altogether, they add \(\$13 + \$8 + \$16 + \$3 = \$40\).

Answer

Lucas must add \(\$13\), Mia \(\$8\), Noah \(\$16\), and Sara \(\$3\). Altogether, they add \(\$40\).
5187704
A nursery has \(45\) red roses. It has three times as many yellow roses as red roses. It has \(56\) fewer white roses than yellow roses. How many white roses are there?

Hints

- First find the number of yellow roses. - The number of white roses is compared with the yellow roses. - “Fewer than” indicates subtraction.

Solution

1. Find the number of yellow roses: \(45 \times 3 = 135\). 2. Find the number of white roses: \(135 - 56 = 79\).

Answer

There are \(79\) white roses.
5187844
A fruit seller receives \(180\,\text{kg}\) of fruit in crates. There are \(15\) crates of apples that each weigh \(4\,\text{kg}\). The remaining weight comes from crates of pears that each weigh \(8\,\text{kg}\). How many crates does the seller receive altogether?

Hints

- First find the total weight of all the apple crates. - How much of the total weight remains for the pear crates? - Once you know the pear crates' total weight, how can you find their number? - Add the two numbers of crates at the end.

Solution

1. Find the total weight of the apple crates: \(15 \times 4\,\text{kg} = 60\,\text{kg}\). 2. Find the total weight of the pear crates: \(180\,\text{kg} - 60\,\text{kg} = 120\,\text{kg}\). 3. Find the number of pear crates: \(120\,\text{kg} \div 8\,\text{kg} = 15\). 4. Find the total number of crates: \(15 + 15 = 30\).

Answer

The seller receives \(30\) crates altogether.
5188004
Admission to a museum field trip costs \(\$6\) per student. A class has \(32\) students. The museum also offers a class rate of \(\$180\). Which option costs less, and how much does the class save?

Hints

- Find the total cost if every student pays separately. - Compare that total with the class rate. - Subtract the lower cost from the higher cost to find the savings.

Solution

1. Find the cost of individual admission: \(32 \times 6 = 192\) dollars. 2. Compare the prices: \(\$180 < \$192\), so the class rate costs less. 3. Find the savings: \(192 - 180 = 12\) dollars.

Answer

The class rate costs less, and the class saves \(\$12\).
5188194
A small truck carries \(250\,\text{kg}\) of sand per trip and makes \(4\) trips each day. A large truck carries \(450\,\text{kg}\) per trip but makes only \(2\) trips each day. Which truck carries more sand in one day, and what is the difference in kilograms?

Hints

- Find each truck's total for one day. - Compare the two totals. - Subtract to find the difference. - Be sure to use the correct number of trips for each truck.

Solution

1. The small truck carries \(4 \times 250 = 1000\,\text{kg}\) per day. 2. The large truck carries \(2 \times 450 = 900\,\text{kg}\) per day. 3. Since \(1000 > 900\), the small truck carries more. 4. The difference is \(1000 - 900 = 100\,\text{kg}\).

Answer

The small truck carries more sand, by \(100\,\text{kg}\).
5188274
A bakery sells bags containing \(5\) rolls each. Jonas has \(7\) quarters, \(5\) dimes, and \(5\) nickels. This money is exactly enough to buy one bag. How many cents does one roll cost?

Hints

- First find the total value of all the coins. - Calculate the value of each kind of coin separately. - Once you know the price of the whole bag, divide by the number of rolls.

Solution

1. Find the value of each group of coins: \(7 \times 25 = 175\) cents, \(5 \times 10 = 50\) cents, and \(5 \times 5 = 25\) cents. 2. Find the total cost of the bag: \(175 + 50 + 25 = 250\) cents. 3. Divide by the number of rolls: \(250 \div 5 = 50\) cents.

Answer

One roll costs \(50\) cents.
5188334
Two third-grade classes collect pinecones for a craft project. Class 3A collects \(135\) pinecones, and Class 3B collects \(145\). Each craft figure requires exactly \(7\) pinecones. The teacher claims, “We can make exactly \(40\) figures from all our pinecones.” Use calculations to determine whether the teacher is correct.

Hints

- How many pinecones do the two classes collect altogether? - How many figures can be made from the total? - Divide the total by the number needed for one figure. - Does your result match the teacher's number?

Solution

1. Find the total number of pinecones: \(135 + 145 = 280\). 2. Find the number of figures that can be made: \(280 \div 7 = 40\). 3. The result matches the teacher's claim, so the claim is correct.

Answer

Yes. The classes have \(280\) pinecones altogether, and \(280 \div 7 = 40\).
5188354
An art teacher needs exactly \(24\) watercolor sets. One set costs \(\$5\), or a pack of \(4\) sets costs \(\$18\). How much does the teacher save by buying only packs of \(4\) instead of individual sets?

Hints

- Find the cost of buying all \(24\) sets individually. - Find how many packs of \(4\) are needed. - Compare the two total costs.

Solution

1. Buying \(24\) individual sets costs \(24 \times 5 = 120\) dollars. 2. The teacher needs \(24 \div 4 = 6\) packs. 3. Six packs cost \(6 \times 18 = 108\) dollars. 4. The savings are \(120 - 108 = 12\) dollars.

Answer

The teacher saves \(\$12\).
5188734
Lucas has saved \(95\) cents. Dinner rolls cost \(24\) cents each. a) Find the prices of \(1\), \(2\), \(3\), and \(4\) rolls. b) Can Lucas buy \(4\) rolls? Use your results from part a to explain. c) How much change will he receive if he buys \(3\) rolls?

Hints

- Build the prices by adding another \(24\) cents each time. - Compare the price of four rolls with \(95\) cents. - Change is the amount paid minus the cost.

Solution

1. Find the prices: \(1 \times 24 = 24\) cents, \(2 \times 24 = 48\) cents, \(3 \times 24 = 72\) cents, and \(4 \times 24 = 96\) cents. 2. Four rolls cost \(96\) cents. Since \(96 > 95\), Lucas cannot buy four rolls. 3. Three rolls cost \(72\) cents. The change is \(95 - 72 = 23\) cents.

Answer

a) The prices are \(\$0.24\), \(\$0.48\), \(\$0.72\), and \(\$0.96\). b) No. Four rolls cost \(\$0.96\), which is more than \(\$0.95\). c) Lucas receives \(\$0.23\) in change.
5188744
At a craft store, one glitter sticker costs \(35\) cents. A special pack of \(3\) stickers costs \(90\) cents. a) How much would \(3\) individual stickers cost? Which option costs less, and what is the difference? b) Ms. Meyer has \(\$2.00\). What is the greatest number of special packs she can buy, and how much money will remain?

Hints

- Find the cost of three individual stickers. - Compare that cost with the pack price. - Determine how many \(90\)-cent packs fit within \(\$2.00\), then find the remainder.

Solution

1. Three individual stickers cost \(3 \times 35 = 105\) cents, or \(\$1.05\). 2. The pack costs \(90\) cents, so it costs \(105 - 90 = 15\) cents less. 3. Two packs cost \(2 \times 90 = 180\) cents. Three packs would cost \(270\) cents, which is more than \(200\) cents. 4. After buying two packs, \(200 - 180 = 20\) cents remain.

Answer

a) Three individual stickers cost \(\$1.05\). The pack costs \(\$0.15\) less. b) Ms. Meyer can buy \(2\) packs and will have \(\$0.20\) left.
5188814
At a beverage store, \(5\) cases of lemonade cost \(\$40\) altogether. Each case contains \(4\) large bottles. How much does one bottle of lemonade cost?

Hints

- First find the total number of bottles. - How can you divide the total cost equally among all the bottles? - Another method is to find the cost of one case first.

Solution

1. Find the total number of bottles: \(5 \times 4 = 20\). 2. Divide the total cost by the number of bottles: \(\$40 \div 20 = \$2\).

Answer

One bottle of lemonade costs \(\$2\).
5188824
A fruit seller offers \(8\) crates of apples, each containing \(5\,\text{kg}\), for \(\$120\) altogether. At the same price per kilogram, how much do \(3\,\text{kg}\) of the apples cost?

Hints

- How many kilograms of apples are in all the crates altogether? - Once you know the total weight and total price, how can you find the price per kilogram? - Use the unit price to find the cost of three kilograms.

Solution

1. Find the total weight of the apples: \(8 \times 5\,\text{kg} = 40\,\text{kg}\). 2. Find the price per kilogram: \(\$120 \div 40 = \$3\) per kilogram. 3. Find the price of three kilograms: \(3 \times \$3 = \$9\).

Answer

Three kilograms of apples cost \(\$9\).
5189064
A school event has \(135\) muffins. Each serving tray holds exactly \(9\) muffins. A staff member says, “We need at least \(20\) trays to display all the muffins at once.” Is the staff member correct? Justify your answer with a calculation.

Hints

- First calculate the exact number of trays needed. - Break \(135\) into numbers that are easy to divide by \(9\). - Compare your result with \(20\).

Solution

1. Find the number of trays needed: \(135 \div 9\). 2. Break \(135\) into compatible numbers: \(135 = 90 + 45\). 3. Divide each part: \(90 \div 9 = 10\) and \(45 \div 9 = 5\). Then add: \(10 + 5 = 15\). 4. Only \(15\) trays are needed, so the statement that at least \(20\) are needed is false.

Answer

No. Only \(15\) trays are needed because \(135 \div 9 = 15\).
5189244
A large city library has \(125{,}450\) books in its children’s and young adult section and \(243{,}120\) books in its adult section. It buys \(15{,}230\) new children’s and young adult books and \(22{,}845\) new adult books. How many books does the library have after the purchases?

Hints

- You may update each section separately. - Another method is to find the original library total first. - Both methods should combine all original books and all new books.

Solution

1. The children’s and young adult section then has \(125{,}450+15{,}230=140{,}680\) books. 2. The adult section then has \(243{,}120+22{,}845=265{,}965\) books. 3. The new library total is \(140{,}680+265{,}965=406{,}645\). 4. Another method is to add the original total, \(125{,}450+243{,}120=368{,}570\), and the total purchases, \(15{,}230+22{,}845=38{,}075\). Then \(368{,}570+38{,}075=406{,}645\).

Answer

The library has \(406{,}645\) books after the purchases.
5189264
A bookseller packs heavy reference books in identical boxes. Six boxes weigh \(48\,\text{kg}\) altogether. a) How much does one box weigh? b) How much do \(15\) of these boxes weigh altogether?

Hints

- How much does one box weigh if six boxes weigh \(48\,\text{kg}\)? - Which operation divides the total weight equally among the boxes? - Once you know one box's weight, how can you find the weight of fifteen boxes?

Solution

1. Find the weight of one box: \(48\,\text{kg} \div 6 = 8\,\text{kg}\). 2. Find the weight of fifteen boxes: \(15 \times 8\,\text{kg} = 120\,\text{kg}\).

Answer

a) One box weighs \(8\,\text{kg}\). b) Fifteen boxes weigh \(120\,\text{kg}\) altogether.
5189674
A gardener plants flowers in rows of exactly \(7\). The gardener has \(80\) tulips, \(95\) daffodils, and \(110\) crocuses. Which type of flower leaves the greatest number of plants after as many complete rows as possible are made?

Hints

- Find the remainder when each number is divided by \(7\). - Use the greatest multiple of \(7\) that does not exceed each number. - Compare the three remainders.

Solution

1. Find the remainder for the tulips: \(80 \div 7 = 11\) remainder \(3\). 2. Find the remainder for the daffodils: \(95 \div 7 = 13\) remainder \(4\). 3. Find the remainder for the crocuses: \(110 \div 7 = 15\) remainder \(5\). 4. Compare the remainders: \(5 > 4 > 3\). The crocuses leave the greatest number of plants.

Answer

The crocuses leave the most plants, with \(5\) left over.
5189754
A concert sells \(38{,}450\) tickets in Chicago. It sells \(12{,}675\) more tickets in Dallas than in Chicago. It sells \(15{,}340\) fewer tickets in Seattle than in Dallas. How many tickets are sold in the three cities altogether?

Hints

- Translate each comparison into a calculation one step at a time. - Find the number of tickets in each city before finding the total. - Seattle’s amount is compared with Dallas’s amount. - A table may help organize the three values.

Solution

1. Dallas sells \(38{,}450+12{,}675=51{,}125\) tickets. 2. Seattle sells \(51{,}125-15{,}340=35{,}785\) tickets. 3. The total is \(38{,}450+51{,}125+35{,}785=125{,}360\) tickets.

Answer

The concert sells \(125{,}360\) tickets in the three cities altogether.
5189774
A truck driver records the distance driven over three days. On Monday, the driver travels \(865\,\text{miles}\). On Tuesday, the driver travels \(145\,\text{miles}\) more than on Monday. On Wednesday, the driver travels \(210\,\text{miles}\) less than on Tuesday. How many miles does the driver travel altogether?

Hints

- Find Tuesday's distance first. - Notice that Wednesday's distance is compared with Tuesday's distance. - Add the three daily distances after finding the missing values.

Solution

1. Find Tuesday's distance: \(865 + 145 = 1010\,\text{miles}\). 2. Find Wednesday's distance: \(1010 - 210 = 800\,\text{miles}\). 3. Add all three days: \(865 + 1010 + 800 = 2675\,\text{miles}\).

Answer

The driver travels \(2675\,\text{miles}\) altogether.
5189984
A large package contains \(240\) marbles. A small package contains one-fourth as many marbles as the large package. Paul buys one large package and two small packages. How many marbles does he buy altogether?

Hints

- First find the number of marbles in one small package. - Pay attention to how many small packages Paul buys. - Add the marbles from all the packages at the end.

Solution

1. Find the number of marbles in one small package: \(240 \div 4 = 60\). 2. Find the number in two small packages: \(2 \times 60 = 120\). 3. Add the marbles in all three packages: \(240 + 120 = 360\).

Answer

Paul buys \(360\) marbles altogether.
5190024
Tim is \(9\) years old. His sister Julia is \(3\) years older than Tim. Their father is three times as old as Julia. How old was their father when Tim was born?

Hints

- First find Julia's age. - Use Julia's age to find the father's current age. - Go back the same number of years as Tim's age.

Solution

1. Find Julia's age: \(9 + 3 = 12\). 2. Find the father's current age: \(12 \times 3 = 36\). 3. Tim was born \(9\) years ago, so the father was \(36 - 9 = 27\) years old.

Answer

Their father was \(27\) years old when Tim was born.
5190194
Tim and Lisa have \(64\) stickers altogether. Tim gives Lisa \(4\) stickers, and then they have the same number. How many stickers did each person have at first?

Hints

- First find how many stickers each person has when the amounts are equal. - Reverse the transfer of \(4\) stickers. - Decide who had more and who had fewer before the transfer. - Use a sketch or table to check that the starting amounts total \(64\).

Solution

1. After the transfer, each person has \(64 \div 2 = 32\) stickers. 2. Before giving away \(4\), Tim had \(32 + 4 = 36\) stickers. 3. Before receiving \(4\), Lisa had \(32 - 4 = 28\) stickers.

Answer

Tim had \(36\) stickers, and Lisa had \(28\) stickers.
5190294
A warehouse has \(4\) pallets of flour, with \(1245\,\text{lb}\) on each pallet. A bakery needs \(5500\,\text{lb}\) for a large order. Is there enough flour? Find how many pounds are missing or left over.

Hints

- Find the total amount on all four pallets. - Compare the available amount with the amount needed. - Subtract to find the shortage or excess.

Solution

1. Find the amount available: \(4 \times 1245\,\text{lb} = 4980\,\text{lb}\). 2. Since \(4980\,\text{lb} < 5500\,\text{lb}\), there is not enough flour. 3. Find the shortage: \(5500\,\text{lb} - 4980\,\text{lb} = 520\,\text{lb}\).

Answer

No. The bakery is short by \(520\,\text{lb}\).
5190364
Max has \(134\) marbles. His sister Lena has \(158\) marbles. How many marbles must Lena give Max so that they each have the same number?

Hints

- Find how many marbles they have altogether. - If they share the total equally, how many marbles will each person have? - Find how many marbles Lena has above that equal-share amount.

Solution

1. Find the total number of marbles: \(134 + 158 = 292\). 2. Divide the total equally: \(292 \div 2 = 146\). 3. Lena must give away the amount she has above \(146\): \(158 - 146 = 12\).

Answer

Lena must give Max \(12\) marbles.
5190374
Three friends collect pinecones in a park. Paul collects \(150\), Marie collects \(90\), and Jonas collects \(60\). They redistribute the pinecones so that each friend has the same number. Paul gives some of his pinecones to Marie and Jonas. How many pinecones does Marie receive, how many does Jonas receive, and how many does Paul give away altogether?

Hints

- First find the total number of pinecones. - How many should each friend have after an equal redistribution? - How many more do Marie and Jonas each need? - Who supplies the pinecones they need?

Solution

1. Find the total number of pinecones: \(150 + 90 + 60 = 300\). 2. Find the equal share for each person: \(300 \div 3 = 100\). 3. Find how many Marie receives: \(100 - 90 = 10\). 4. Find how many Jonas receives: \(100 - 60 = 40\). 5. Find how many Paul gives away altogether: \(10 + 40 = 50\).

Answer

Marie receives \(10\) pinecones, Jonas receives \(40\), and Paul gives away \(50\) pinecones altogether.
5190384
Two shelves hold \(92\) books altogether. The second shelf has \(18\) more books than the first shelf. How many books are on each shelf?

Hints

- Imagine first that both shelves had the same number of books. - Temporarily remove the difference of \(18\) from the total. - Divide the remaining total equally between the two shelves. - Add the \(18\) extra books back to the second shelf.

Solution

1. Remove the extra \(18\) books from the total: \(92 - 18 = 74\). 2. Divide the remaining books equally: \(74 \div 2 = 37\). This is the number on the first shelf. 3. The second shelf has \(37 + 18 = 55\) books.

Answer

The first shelf has \(37\) books, and the second shelf has \(55\) books.
5190394
Three classes collect \(450\,\text{kg}\) of paper for a recycling contest. Classes A and B collect \(310\,\text{kg}\) altogether, and Class B collects \(20\,\text{kg}\) more than Class A. How many kilograms does each class collect?

Hints

- Find Class C's amount first. - Then focus on the combined amount for Classes A and B. - Temporarily remove their difference before dividing equally. - Check that all three amounts total \(450\,\text{kg}\).

Solution

1. Class C collects \(450 - 310 = 140\), so it collects \(140\,\text{kg}\). 2. Remove the \(20\,\text{kg}\) difference from the combined amount for Classes A and B: \(310 - 20 = 290\). 3. Divide equally to find Class A's amount: \(290 \div 2 = 145\). 4. Class B collects \(145 + 20 = 165\), so it collects \(165\,\text{kg}\).

Answer

Class A collects \(145\,\text{kg}\), Class B collects \(165\,\text{kg}\), and Class C collects \(140\,\text{kg}\).
5190454
A sports club wants to buy new balls. A soccer ball costs \(\$12\), and a basketball costs \(\$8\). The club wants to buy \(7\) soccer balls but is \(\$11\) short. 1. How much money is in the club's account? 2. Can the club buy \(10\) basketballs with that money? Justify your answer. 3. What is the greatest number of basketballs the club can buy, and how much money will remain?

Hints

- First find the total cost of seven soccer balls. - What does being “short” mean for finding the available amount? - Compare the cost of ten basketballs with the available money. - Use division with a remainder to find the greatest possible number.

Solution

1. Find the cost of seven soccer balls: \(7 \times \$12 = \$84\). 2. Find the amount in the account: \(\$84 - \$11 = \$73\). 3. Check the cost of ten basketballs: \(10 \times \$8 = \$80\). Since \(\$80 > \$73\), the club cannot buy ten. 4. Divide the available money by the basketball price: \(\$73 \div \$8 = 9\) remainder \(\$1\).

Answer

1. The club has \(\$73\). 2. No. Ten basketballs would cost \(\$80\), which is more than \(\$73\). 3. The club can buy at most \(9\) basketballs, and \(\$1\) will remain.
5190724
I am thinking of a number. Half of the number is \(40\). What result do I get if I double the original number instead?

Hints

- Find the original number first. - Undo halving by doubling. - Then double the original number. - Consider the direct relationship between half a number and twice the number.

Solution

1. The original number is \(40 \times 2 = 80\). 2. Doubling the original number gives \(80 \times 2 = 160\).

Answer

The result is \(160\).
5191164
A crate of tile weighs \(18\,\text{lb}\). a) What is the total weight of \(30\) crates? b) A small trailer may carry at most \(1000\,\text{lb}\). May \(60\) crates be carried at once? Show a calculation.

Hints

- Multiply the weight of one crate by the number of crates. - For part b, compare the total with the trailer's maximum load. - Find the amount over or under the limit.

Solution

1. For part a, multiply: \(18\,\text{lb} \times 30 = 540\,\text{lb}\). 2. For part b, find the weight of \(60\) crates: \(18\,\text{lb} \times 60 = 1080\,\text{lb}\). 3. Since \(1080\,\text{lb} > 1000\,\text{lb}\), the load exceeds the trailer's limit. 4. Find the amount over the limit: \(1080\,\text{lb} - 1000\,\text{lb} = 80\,\text{lb}\).

Answer

a) \(30\) crates weigh \(540\,\text{lb}\). b) No. \(60\) crates weigh \(1080\,\text{lb}\), which is \(80\,\text{lb}\) over the limit.
5192014
A class of \(24\) students and \(2\) teachers visits a zoo. Student admission is \(\$6\), and adult admission is \(\$14\). What is the total admission cost for the group?

Hints

- Find the student and adult costs separately. - Use the correct ticket price for each group. - Add the two costs.

Solution

1. Student admission costs \(24\times\$6=\$144\). 2. Teacher admission costs \(2\times\$14=\$28\). 3. The total cost is \(\$144+\$28=\$172\).

Answer

\(\$172\)
5192024
A class has \(\$150\) in its field-trip fund. Twenty-eight students and one teacher are visiting a museum. A student ticket costs \(\$4\), and an adult ticket costs \(\$9\). Is there enough money? If so, how much remains?

Hints

- Find the total ticket cost first. - Compare that cost with the amount in the fund. - Subtract to find the amount remaining.

Solution

1. The student tickets cost \(28\times\$4=\$112\). 2. Add the teacher's ticket: \(\$112+\$9=\$121\). 3. Since \(\$121<\$150\), there is enough money. 4. The amount remaining is \(\$150-\$121=\$29\).

Answer

Yes. \(\$29\) remains.
5192034
A group of \(25\) teenagers and \(2\) chaperones is going to a movie theater. They can choose between two pricing plans: 1. Individual tickets: \(\$7\) for each teenager and \(\$12\) for each adult. 2. Group price: For groups of at least \(25\) people, every person pays \(\$6\). Which plan costs less, and how much does the group save?

Hints

- Find the individual-ticket total for the teenagers and adults. - Find the total number of people before using the group price. - Compare the two totals and subtract to find the savings.

Solution

1. Individual tickets cost \(25\times\$7+2\times\$12=\$175+\$24=\$199\). 2. The group has \(25+2=27\) people. 3. The group-price total is \(27\times\$6=\$162\). 4. The group price costs less and saves \(\$199-\$162=\$37\).

Answer

The group price costs less and saves \(\$37\).
5192524
A gardener buys \(14\) pallets of red geraniums and \(9\) pallets of white geraniums. Each pallet holds \(120\) pots. How many more red geraniums than white geraniums did the gardener buy? Show two different methods.

Hints

- You may first find how many more pallets of red geraniums there are. - Another method is to find each color’s total separately. - Because each pallet has the same number of pots, the pallet difference can be multiplied by \(120\).

Solution

1. Method 1: Find the difference in pallets: \(14-9=5\). Then \(5\times 120=600\) geraniums. 2. Method 2: Find each total: \(14\times 120=1680\) red geraniums and \(9\times 120=1080\) white geraniums. Then \(1680-1080=600\).

Answer

The gardener bought \(600\) more red geraniums than white geraniums.
5192534
Two fourth-grade classes collect acorns for animals at a wildlife center. Ms. Lee’s class fills \(18\) bags, and Mr. Green’s class fills \(23\) bags. Each bag weighs \(15\,\text{lb}\). a) How many more pounds of acorns does Mr. Green’s class collect? b) The center uses about \(3\,\text{lb}\) of acorns per day as supplemental feed for one deer. How many days will the extra amount last?

Hints

- How many more bags does Mr. Green’s class fill? - What is the total weight of those extra bags? - How many daily portions fit into the extra amount?

Solution

1. Find the number of extra bags: \(23 - 18 = 5\) bags. 2. Find the extra weight: \(5 \times 15 = 75\,\text{lb}\). 3. At about \(3\,\text{lb}\) per day, the extra amount will last approximately \(75 \div 3 \approx 25\) days.

Answer

a) Mr. Green’s class collects \(75\,\text{lb}\) more. b) The extra amount will last about \(25\) days for one deer.
5192554
An organic farm packs \(125\) bags of potatoes and \(75\) bags of onions each day. a) How many bags are packed during a \(30\)-day month? Solve in two different ways. b) The bags are loaded into shipping crates that each hold \(50\) bags. How many crates are needed for the month’s full amount?

Hints

- Find the total number of bags packed in one day. - For a second method, find each crop’s monthly total separately. - In part b, divide the full amount into equal groups of \(50\). - You can simplify the division by using place-value relationships.

Solution

1. Method 1 for part a: The farm packs \(125+75=200\) bags per day. In \(30\) days, it packs \(200\times 30=6000\) bags. 2. Method 2 for part a: It packs \(125\times 30=3750\) bags of potatoes and \(75\times 30=2250\) bags of onions. Then \(3750+2250=6000\). 3. For part b, \(6000\div 50=120\), so \(120\) crates are needed.

Answer

a) \(6000\) bags b) \(120\) crates
5193424
A youth sports club buys \(8\) soccer balls for \(\$144\). a) What is the price of one ball? b) How much would \(15\) balls cost at the same price per ball?

Hints

- Divide the total cost by \(8\) to find the unit price. - Multiply the unit price by \(15\). - You can split \(15\) into \(10 + 5\) if that makes the multiplication easier.

Solution

1. One ball costs \(\$144 \div 8 = \$18\). 2. Fifteen balls cost \(15 \times \$18 = \$270\).

Answer

a) One ball costs \(\$18\). b) Fifteen balls cost \(\$270\).
5193534
Lucas, Mia, and Tom collect \(750\) acorns altogether. Lucas and Mia collect \(480\) acorns together. Mia and Tom collect \(510\) acorns together. Who collects more acorns, Lucas or Tom? How many more?

Hints

- Which person’s amount remains when the Lucas-and-Mia total is removed from the overall total? - Subtract \(480\) from \(750\) to find Tom’s amount. - Subtract the Mia-and-Tom total from the overall total to find Lucas’s amount. - Compare Lucas’s and Tom’s amounts and find their difference.

Solution

1. Tom collected the part not included in the Lucas-and-Mia total: \(750 - 480 = 270\). 2. Lucas collected the part not included in the Mia-and-Tom total: \(750 - 510 = 240\). 3. Compare and find the difference: \(270 - 240 = 30\).

Answer

Tom collected \(30\) more acorns than Lucas.
5193804
Four friends win \(\$1348\) in a contest and split the prize equally. Leon already has \(\$57\) in savings. How much money does Leon have after adding his share of the prize to his savings?

Hints

- Divide the prize equally among the four friends. - Add Leon’s share to his existing savings. - Check the equal share by multiplying it by \(4\).

Solution

1. Each friend receives \(\$1348 \div 4 = \$337\). 2. Leon then has \(\$337 + \$57 = \$394\).

Answer

Leon has \(\$394\).
5194404
A farm stand has \(15\) crates of apples, with \(28\,\text{lb}\) in each crate. Customers buy \(135\,\text{lb}\) of apples. The remaining apples are pressed into cider. It takes \(3\,\text{lb}\) of apples to make \(1\,\text{qt}\) of cider. How many quarts of cider can be made?

Hints

- Find the total weight of all the apples. - Subtract the amount sold. - Determine how many \(3\,\text{lb}\) groups are in the remaining amount.

Solution

1. Find the total weight of the apples: \(15 \times 28 = 420\,\text{lb}\). 2. Find the weight remaining after the sales: \(420 - 135 = 285\,\text{lb}\). 3. Find the amount of cider: \(285 \div 3 = 95\,\text{qt}\).

Answer

The remaining apples can make \(95\,\text{qt}\) of cider.
5194414
A school receives \(14\) boxes of markers with \(24\) markers in each box. First, \(8\) classes each receive a set of \(12\) markers. The remaining markers are divided equally into cups with \(6\) markers in each cup. a) How many cups can be filled? b) A teacher says, “There should be enough markers for \(50\) cups.” Is the teacher correct? Explain.

Hints

- First find the total number of markers delivered. - Subtract the markers given to the classes. - Compare the number of cups you find with the teacher’s estimate.

Solution

1. The school receives \(14\times 24=336\) markers. 2. The classes receive \(8\times 12=96\) markers. 3. The number left is \(336-96=240\) markers. 4. The number of cups is \(240\div 6=40\). 5. Since \(40<50\), the teacher is incorrect.

Answer

a) \(40\) cups b) No. Only \(40\) cups can be filled.
5194434
Three children are saving for a large tent that costs \(\$600\). Mia has saved \(\$150\). Ben has saved twice as much as Mia. Noah has saved \(\$80\) less than Ben. How much have they saved altogether? Do they have enough for the tent?

Hints

- First find Ben's savings. - Use Ben's amount to find Noah's savings. - Add all three amounts and compare the total with the tent's price.

Solution

1. Find Ben's savings: \(2 \times \$150 = \$300\). 2. Find Noah's savings: \(\$300 - \$80 = \$220\). 3. Add all three amounts: \(\$150 + \$300 + \$220 = \$670\). 4. Since \(\$670 > \$600\), they have enough money.

Answer

The children have saved \(\$670\) altogether, so they have enough for the tent.
5194444
A community festival is held every \(3\) years. One festival was held in \(2002\), and the most recent festival in this period was held in \(2023\). a) How many festivals were held from \(2002\) through \(2023\), including both endpoint years? b) How many years in that same period did not have a festival?

Hints

- Count both the starting year and ending year in the full period. - Find how many \(3\)-year intervals fit between \(2002\) and \(2023\). - Remember that counting intervals is not the same as counting events. - Subtract the festival years from the total number of years.

Solution

1. Find the number of years in the inclusive period: \(2023 - 2002 + 1 = 22\) years. 2. Find the number of \(3\)-year intervals: \((2023 - 2002) \div 3 = 7\). 3. Include the festival at the start of the period: \(7 + 1 = 8\) festivals. 4. Subtract the festival years from all years in the period: \(22 - 8 = 14\) years without a festival.

Answer

a) There were \(8\) festivals. b) There were \(14\) years without a festival.
5194544
A fruit farm has a large harvest this year. It harvests \(4250\,\text{lb}\) of apples. The pear harvest is \(1340\,\text{lb}\) less than the apple harvest. The plum harvest is exactly one-fourth of the combined weight of the apples and pears. How many pounds of fruit does the farm harvest altogether?

Hints

- Find the amount of each type of fruit in the order needed. - Read carefully to identify which combined amount the plum fraction refers to. - A table may help you organize the intermediate results. - What operation finds one-fourth of a quantity?

Solution

1. Find the pear harvest: \(4250\,\text{lb} - 1340\,\text{lb} = 2910\,\text{lb}\). 2. Find the combined apple and pear harvest: \(4250\,\text{lb} + 2910\,\text{lb} = 7160\,\text{lb}\). 3. Find the plum harvest: \(\frac{1}{4} \times 7160\,\text{lb} = 1790\,\text{lb}\). 4. Add all three fruit harvests: \(7160\,\text{lb} + 1790\,\text{lb} = 8950\,\text{lb}\).

Answer

The farm harvests \(8950\,\text{lb}\) of fruit altogether.
5194554
During a school district recycling drive, students collect \(1284\,\text{lb}\) of wastepaper. The weight of the glass collected is three times the weight of the paper. The plastic collected weighs one-sixth of the combined weight of the paper and glass. How many pounds of recyclable material are collected altogether?

Hints

- What must you calculate before you can find the amount of plastic? - Which operations match “three times as much” and “one-sixth of”? - Keep the place values aligned in your calculations. - Make sure your final answer gives the total of all three materials.

Solution

1. Find the weight of the glass: \(3 \times 1284\,\text{lb} = 3852\,\text{lb}\). 2. Find the combined weight of the paper and glass: \(1284\,\text{lb} + 3852\,\text{lb} = 5136\,\text{lb}\). 3. Find the weight of the plastic: \(\frac{1}{6} \times 5136\,\text{lb} = 856\,\text{lb}\). 4. Find the total weight: \(5136\,\text{lb} + 856\,\text{lb} = 5992\,\text{lb}\).

Answer

The students collect \(5992\,\text{lb}\) of recyclable material altogether.
5194604
A farm plans its weekly hay use carefully. The pony barn uses \(1356\,\text{lb}\) of hay. The cattle barn uses five times as much hay as the pony barn. The goat barn uses one-third as much hay as the cattle barn. How many pounds of hay do the three barns use altogether in one week?

Hints

- Find the hay amount for each barn in order. - Which operation matches “five times as much,” and which matches “one-third as much”? - Did you add all three amounts at the end?

Solution

1. Find the hay used by the cattle barn: \(5 \times 1356\,\text{lb} = 6780\,\text{lb}\). 2. Find the hay used by the goat barn: \(\frac{1}{3} \times 6780\,\text{lb} = 2260\,\text{lb}\). 3. Add the three amounts: \(1356\,\text{lb} + 6780\,\text{lb} + 2260\,\text{lb} = 10{,}396\,\text{lb}\).

Answer

The three barns use \(10{,}396\,\text{lb}\) of hay altogether in one week.
5194844
A wall-mounted bookcase has \(6\) wooden shelves that are each \(3\,\text{cm}\) thick. The \(5\) spaces between the shelves are each \(32\,\text{cm}\) high. Find the total height from the bottom of the lowest shelf to the top of the highest shelf.

Hints

- Count the shelves and the spaces separately. - Find the combined thickness of the shelves. - Find the combined height of the spaces, then add the two amounts.

Solution

1. The shelves have a combined thickness of \(6\times3\,\text{cm}=18\,\text{cm}\). 2. The spaces have a combined height of \(5\times32\,\text{cm}=160\,\text{cm}\). 3. The total height is \(18\,\text{cm}+160\,\text{cm}=178\,\text{cm}\).

Answer

\(178\,\text{cm}\)
5195074
A school garden has \(124\) tulips. It has twice as many daffodils as tulips and \(50\) fewer crocuses than daffodils. How many flowers are planted altogether?

Hints

- First find the number of daffodils. - Use the daffodil count to find the number of crocuses. - Add the three flower counts.

Solution

1. Find the number of daffodils: \(124 \times 2 = 248\). 2. Find the number of crocuses: \(248 - 50 = 198\). 3. Add all three amounts: \(124 + 248 + 198 = 570\).

Answer

There are \(570\) flowers altogether.
5195084
Three classes record laps during a charity run. Class 3A completes \(210\) laps. Class 3B completes \(40\) more laps than Class 3A. Class 3C completes exactly half as many laps as Class 3B. How many laps do the three classes complete altogether?

Hints

- Find the number of laps for each class one at a time. - What operation does “more” suggest? - How do you find one-half of a number? - Add the results for all three classes at the end.

Solution

1. Find the number of laps completed by Class 3B: \(210 + 40 = 250\). 2. Find the number completed by Class 3C: \(250 \div 2 = 125\). 3. Add the laps from all three classes: \(210 + 250 + 125 = 585\).

Answer

The three classes complete \(585\) laps altogether.
5195104
At a school fair, students sell \(185\) raffle tickets in the morning. They sell \(40\) more tickets in the afternoon than in the morning. In the evening, they sell twice as many tickets as in the afternoon. How many tickets do they sell during the entire day?

Hints

- Find the sales for each part of the day in order. - Use the afternoon amount to find the evening amount. - Add the three amounts.

Solution

1. Find the afternoon sales: \(185 + 40 = 225\). 2. Find the evening sales: \(225 \times 2 = 450\). 3. Add all three amounts: \(185 + 225 + 450 = 860\).

Answer

Students sell \(860\) raffle tickets during the day.
5195254
An orchard delivers \(4\) crates of apples to a preschool. The apples in the delivery cost \(\$64\), at the same price per apple as in the orchard store. At the orchard store, \(10\) of the same apples cost \(\$4\). Each crate contains the same number of apples. How many apples are in each crate?

Hints

- Compare the delivery cost with the cost of \(10\) apples. - Use equal groups of \(\$4\) to find the total number of apples. - Divide the apples equally among the four crates.

Solution

1. The delivery price of \(\$64\) is \(64 \div 4 = 16\) groups of \(\$4\). 2. Each \(\$4\) group represents \(10\) apples, so the delivery contains \(16 \times 10 = 160\) apples. 3. Each crate contains \(160 \div 4 = 40\) apples.

Answer

Each crate contains \(40\) apples.
5195304
A produce distributor starts with \(850\,\text{kg}\) of apples. It delivers \(185\,\text{kg}\) to each of \(4\) schools. How many kilograms of apples remain after the deliveries?

Hints

- First find the total amount delivered to all four schools. - Use multiplication for the equal deliveries. - Subtract the delivered amount from the starting amount.

Solution

1. Find the total amount delivered: \(4 \times 185\,\text{kg} = 740\,\text{kg}\). 2. Subtract from the starting amount: \(850\,\text{kg} - 740\,\text{kg} = 110\,\text{kg}\).

Answer

\(110\,\text{kg}\) of apples remain.
5195314
A school has \(950\) raffle tickets to sell. Students sell \(215\) tickets in the morning and twice as many in the afternoon. A student says, “More than \(300\) tickets are left.” Is the student correct? Justify your answer.

Hints

- First find the afternoon sales. - Add the morning and afternoon sales. - Subtract from \(950\), then compare the result with \(300\).

Solution

1. Find the afternoon sales: \(215 \times 2 = 430\). 2. Find the total sold: \(215 + 430 = 645\). 3. Find the number left: \(950 - 645 = 305\). 4. Since \(305 > 300\), the statement is correct.

Answer

Yes. There are \(305\) tickets left, and \(305 > 300\).
5195654
A baker makes \(6\) trays with \(45\) rolls on each tray. The bakery sells \(142\) rolls in the morning and \(78\) in the afternoon. How many rolls are left at the end of the day?

Hints

- First find the total number baked. - Add the morning and afternoon sales. - Subtract the total sold from the total baked.

Solution

1. Find the total number baked: \(6 \times 45 = 270\). 2. Find the total number sold: \(142 + 78 = 220\). 3. Find the number left: \(270 - 220 = 50\).

Answer

There are \(50\) rolls left.
5195664
A library buys \(4\) new bookcases that each hold \(115\) books. The library places \(325\) books on the new bookcases. How many spaces remain?

Hints

- First find the capacity of all four bookcases. - Then subtract the number of books placed on them. - The difference is the number of open spaces.

Solution

1. Find the total capacity: \(4 \times 115 = 460\). 2. Subtract the books placed on the shelves: \(460 - 325 = 135\).

Answer

There are \(135\) spaces left.
5195714
A school fair sells \(1250\) tickets altogether. It sells \(180\) more adult tickets than child tickets. The principal says, “We sold exactly \(715\) adult tickets.” Is the principal correct? Support your answer with a calculation.

Hints

- Find the smaller group first. - Set aside the \(180\) extra adult tickets. - Divide the remaining tickets equally between the two groups. - Add the difference back to the adult group.

Solution

1. Remove the difference from the total: \(1250-180=1070\). 2. Divide the remaining amount equally: \(1070\div 2=535\) child tickets. 3. The number of adult tickets is \(535+180=715\). 4. Since the calculated amount is \(715\), the principal is correct.

Answer

Yes. The fair sold \(535\) child tickets and \(715\) adult tickets, for \(1250\) tickets altogether.
5195764
For a craft contest, a teacher buys \(25\) packages of construction paper and \(8\) boxes of glitter gems. Each package of construction paper costs \(\$14\). She pays with six \(\$100\) bills and receives \(\$50\) in change. How much does one box of glitter gems cost?

Hints

- Find the amount actually spent after the change is returned. - Find the total cost of the construction paper. - Divide the remaining cost equally among the eight boxes.

Solution

1. The full purchase costs \(6 \times \$100 - \$50 = \$550\). 2. The construction paper costs \(25 \times \$14 = \$350\). 3. The eight boxes of glitter gems cost \(\$550 - \$350 = \$200\). 4. One box costs \(\$200 \div 8 = \$25\).

Answer

One box of glitter gems costs \(\$25\).
5195784
Two trains travel long distances. Train A travels \(270\) miles in \(3\) hours. Train B travels \(320\) miles in \(4\) hours. Which train travels farther in one hour? Explain with calculations.

Hints

- First find how far each train travels in one hour. - How can you use the total distance and total time to find a one-hour distance? - Compare the two one-hour distances.

Solution

1. Find Train A's distance in one hour: \(270 \div 3 = 90\) miles. 2. Find Train B's distance in one hour: \(320 \div 4 = 80\) miles. 3. Since \(90 > 80\), Train A travels farther in one hour.

Answer

Train A travels farther in one hour: \(90\) miles compared with Train B's \(80\) miles.
5195864
A school library receives \(180\) new books. a) How many shelf sections are needed if each section holds \(6\) books? b) How many shelf sections are needed if each section holds \(9\) books instead? c) Explain why the second arrangement needs fewer shelf sections.

Hints

- Divide \(180\) by each number of books per section. - Compare the two quotients. - Explain how increasing the group size changes the number of groups.

Solution

1. For part a, divide: \(180 \div 6 = 30\). 2. For part b, divide: \(180 \div 9 = 20\). 3. Since each section holds more books in part b, the same total number of books is divided into fewer groups.

Answer

a) \(30\) shelf sections are needed. b) \(20\) shelf sections are needed. c) The second arrangement uses fewer sections because each section holds more books.
5196054
A farmer collected \(840\) eggs and packs \(6\) eggs in each carton. a) How many cartons can the farmer fill? b) If the farmer used cartons that held only \(3\) eggs each, would more or fewer cartons be needed? Explain.

Hints

- Break \(840\) into numbers that are easy to divide by \(6\). - Add the partial quotients for part a. - For part b, compare the two carton sizes without finding the exact second quotient.

Solution

1. For part a, divide \(840\) by \(6\). Break apart \(840 = 600 + 240\). 2. Divide each part: \(600 \div 6 = 100\) and \(240 \div 6 = 40\). Then add: \(100 + 40 = 140\). 3. For part b, cartons holding \(3\) eggs are half the size of cartons holding \(6\). Smaller groups mean more groups are needed for the same total.

Answer

a) The farmer can fill \(140\) cartons. b) More cartons would be needed because each carton holds fewer eggs.
5196064
A school prepares \(960\) raffle tickets. Class 4A receives exactly half of all the tickets to sell. The remaining tickets are divided equally between Classes 4B and 4C. How many tickets does Class 4C receive?

Hints

- First find half of \(960\). - Subtract that amount from the total. - Divide the remaining tickets equally between the two classes. - Break large numbers into place-value parts if that makes the division easier.

Solution

1. Class 4A receives half of \(960\): \(960 \div 2 = 480\). 2. The number of tickets left is \(960 - 480 = 480\). 3. Divide the remaining tickets equally between two classes: \(480 \div 2 = 240\).

Answer

Class 4C receives \(240\) tickets.
5196084
A class fund contains \(\$750\), all in \(\$5\) bills. The class spends \(\$200\) on a field trip. How many \(\$5\) bills remain?

Hints

- Find how much money remains after the purchase. - Divide the remaining amount by \(5\). - Break the remaining amount into \(500\) and another convenient part if helpful. - You can also compare the original number of bills with the number of bills spent.

Solution

1. One method is to find the remaining amount first: \(\$750 - \$200 = \$550\). 2. Divide the remaining amount by the value of each bill: \(550 \div 5 = (500 \div 5) + (50 \div 5) = 100 + 10 = 110\). 3. A second method is to find \(750 \div 5 = 150\) bills at first and \(200 \div 5 = 40\) bills spent. Then \(150 - 40 = 110\).

Answer

There are \(110\) five-dollar bills remaining.
5196514
An elementary school receives \(845\) new pencils. The office keeps \(125\) pencils in reserve. The remaining pencils are divided equally among \(8\) classes. How many pencils does each class receive?

Hints

- How many pencils remain after the reserve is set aside? - Which operation divides a quantity equally among groups? - Use place value and a related basic division fact.

Solution

1. Find the number of pencils remaining after the reserve is removed: \(845 - 125 = 720\). 2. Divide equally among eight classes: \(720 \div 8 = 90\).

Answer

Each class receives \(90\) pencils.
5196544
A school festival earns \(\$950\). The school must pay \(\$345\) for drinks and \(\$285\) for food. The remaining money is divided equally among the \(8\) participating classes. How much money does each class receive for its class fund?

Hints

- How much do the food and drinks cost altogether? - How much money remains after those costs are paid? - Once you know the remaining amount, how do you find one class's share? - A related basic division fact may help with the larger number.

Solution

1. Find the total cost of food and drinks: \(\$345 + \$285 = \$630\). 2. Find the money remaining: \(\$950 - \$630 = \$320\). 3. Divide equally among eight classes: \(\$320 \div 8 = \$40\).

Answer

Each class receives \(\$40\).
5196594
Lara says, “When I divide my number by \(6\), I get \(80\).” Jonas says, “My number is exactly \(150\) less than Lara's number.” What number did Jonas choose?

Hints

- Find Lara's number first. - Reverse division by \(6\) by multiplying by \(6\). - Then subtract \(150\) to find Jonas's number. - Check that Jonas's number is less than Lara's.

Solution

1. Lara's number is \(80 \times 6 = 480\). 2. Jonas's number is \(480 - 150 = 330\).

Answer

Jonas chose \(330\).
5196754
A community center buys new lounge furniture for \(\$4200\) and new kitchen equipment for \(\$4200\). It pays the total in \(10\) equal monthly payments. a) How much is each monthly payment? b) Suppose the furniture had cost \(\$500\) more and the kitchen equipment had cost \(\$500\) less. Would the monthly payment change? Explain.

Hints

- First find the combined cost. - Divide that total by \(10\). - For part b, decide whether adding an amount to one cost and subtracting the same amount from the other changes their sum.

Solution

1. Find the total cost: \(\$4200 + \$4200 = \$8400\). 2. Divide the total into \(10\) equal payments: \(\$8400 \div 10 = \$840\). 3. For part b, increasing one cost by \(\$500\) and decreasing the other by \(\$500\) leaves the total unchanged. The number of payments also stays the same, so the monthly payment remains \(\$840\).

Answer

a) Each monthly payment is \(\$840\). b) No. The total remains \(\$8400\), since \(\$4700 + \$3700 = \$8400\), so the monthly payment does not change.
5196954
Classes A and B are taking a camping trip. A boys’ tent holds at most \(5\) campers, and a girls’ tent holds at most \(4\) campers. The table shows the number of students. <table> <tr> <th></th> <th>Girls</th> <th>Boys</th> </tr> <tr> <td>Class A</td> <td>\(12\)</td> <td>\(13\)</td> </tr> <tr> <td>Class B</td> <td>\(14\)</td> <td>\(12\)</td> </tr> </table> Boys and girls sleep in separate tents, but students from the two classes may share tents. How many tents are needed altogether?

Hints

- Combine the two classes' counts for girls and boys separately. - A nonzero remainder means another tent is needed. - Add the numbers of girls' and boys' tents.

Solution

1. There are \(12+14=26\) girls. 2. Since \(26\div4=6\) remainder \(2\), the girls need \(7\) tents. 3. There are \(13+12=25\) boys. 4. Since \(25\div5=5\), the boys need \(5\) tents. 5. Altogether, \(7+5=12\) tents are needed.

Answer

\(12\) tents
5197084
Students at an elementary school raise money at a yard sale for new recess equipment. First-grade classes raise \(\$1420\), second-grade classes raise \(\$1250\), and the third- and fourth-grade classes together raise \(\$2530\). The school plans to spend the money equally over \(5\) months. How much can it spend each month?

Hints

- Find the total amount raised by all the grade levels. - Divide the total equally among the months. - Begin by combining all three amounts.

Solution

1. Find the total amount raised: \(\$1420 + \$1250 + \$2530 = \$5200\). 2. Divide the total equally among \(5\) months: \(\$5200 \div 5 = \$1040\).

Answer

The school can spend \(\$1040\) each month.
5197094
A wildlife park receives \(4125\,\text{kg}\) of hay during the first half of a year and \(3755\,\text{kg}\) during the second half. The hay is divided equally among \(8\) elephants. How many kilograms of hay does each elephant receive?

Hints

- Add the two donations first. - Divide the total equally among \(8\) elephants. - Check the quotient by multiplication.

Solution

1. Find the total amount of hay: \(4125\,\text{kg} + 3755\,\text{kg} = 7880\,\text{kg}\). 2. Divide equally among the elephants: \(7880\,\text{kg} \div 8 = 985\,\text{kg}\).

Answer

Each elephant receives \(985\,\text{kg}\) of hay.
5197224
A wildlife park has \(240\) visitors on Friday. On Saturday, it has \(65\) more visitors than on Friday. On Sunday, it has \(120\) fewer visitors than on Saturday. How many visitors come to the park over the three days altogether?

Hints

- Find the Saturday attendance first. - Use Saturday's attendance to find Sunday's attendance. - Add the attendance for Friday, Saturday, and Sunday.

Solution

1. Find the Saturday attendance: \(240 + 65 = 305\). 2. Find the Sunday attendance: \(305 - 120 = 185\). 3. Add the attendance for all three days: \(240 + 305 + 185 = 730\).

Answer

The wildlife park has \(730\) visitors over the three days altogether.
5197264
Lucas wants to read a \(900\)-page book during vacation. He reads \(215\) pages in the first week. In the second week, he reads \(40\) fewer pages than in the first week. In the third week, he reads \(250\) pages. How many pages must Lucas read in the fourth week to finish the book?

Hints

- First find how many pages Lucas reads in the second week. - Add the pages read during the first three weeks. - Subtract that total from \(900\).

Solution

1. Find the pages read in the second week: \(215 - 40 = 175\). 2. Find the total read in the first three weeks: \(215 + 175 + 250 = 640\). 3. Subtract from the total number of pages: \(900 - 640 = 260\).

Answer

Lucas must read \(260\) pages in the fourth week.
5197364
Three ribbons are sewn together for a festival decoration. The red ribbon is \(245\,\text{cm}\) long. The blue ribbon is \(60\,\text{cm}\) shorter than the red ribbon but \(30\,\text{cm}\) longer than the green ribbon. The seams use a total of \(25\,\text{cm}\) of ribbon. How long is the finished decoration?

Hints

- Use the red ribbon to find the blue ribbon. - Use the blue ribbon to find the green ribbon. - Add all three lengths, then subtract the amount used in the seams.

Solution

1. Find the length of the blue ribbon: \(245 - 60 = 185\,\text{cm}\). 2. Find the length of the green ribbon: \(185 - 30 = 155\,\text{cm}\). 3. Add the three ribbon lengths: \(245 + 185 + 155 = 585\,\text{cm}\). 4. Subtract the ribbon used in the seams: \(585 - 25 = 560\,\text{cm}\).

Answer

The finished decoration is \(560\,\text{cm}\) long.
5197524
A produce seller receives \(400\,\text{kg}\) of apples and pears altogether. After selling \(240\,\text{kg}\) of apples, the seller has the same weight of apples and pears left. How many kilograms of apples and pears were there at first?

Hints

- Find the total weight remaining after the apples are sold. - Divide that remaining weight equally between apples and pears. - The pear amount did not change. - Add the sold apples back to the remaining apples.

Solution

1. After the sale, \(400 - 240 = 160\), so \(160\,\text{kg}\) of fruit remain. 2. Equal amounts of apples and pears remain, so each amount is \(160 \div 2 = 80\), or \(80\,\text{kg}\). 3. No pears were sold, so the original pear amount was \(80\,\text{kg}\). 4. The original apple amount was \(80 + 240 = 320\), or \(320\,\text{kg}\).

Answer

There were \(320\,\text{kg}\) of apples and \(80\,\text{kg}\) of pears at first.
5197534
A summer camp has \(750\) campers. On the first morning, \(350\) boys go canoeing. The other boys stay at camp with all the girls. The number of boys and girls remaining at camp is then equal. How many boys and girls are enrolled altogether?

Hints

- Find how many campers remain after the canoe group leaves. - Divide the remaining campers into two equal groups. - All the girls stayed at camp. - Add the boys at camp and the boys canoeing.

Solution

1. The number of campers remaining is \(750 - 350 = 400\). 2. Equal numbers of boys and girls remain, so each group has \(400 \div 2 = 200\) campers. 3. All the girls stayed, so there are \(200\) girls altogether. 4. There are \(200 + 350 = 550\) boys altogether.

Answer

There are \(550\) boys and \(200\) girls enrolled.
5197574
Two rolls of wire are a total of \(870\,\text{m}\) long. After \(150\,\text{m}\) is cut from the first roll, the two rolls are the same length. 1. How long was each roll at first? 2. How long is each roll after the \(150\,\text{m}\) is cut?

Hints

- Find the total length left after the piece is cut. - Divide the remaining total into two equal lengths. - Decide which roll was longer at first.

Solution

1. After the cut, the total length is \(870 - 150 = 720\), or \(720\,\text{m}\). 2. The two remaining rolls are equal, so each is \(720 \div 2 = 360\), or \(360\,\text{m}\). 3. The second roll was not cut, so it was originally \(360\,\text{m}\). The first roll was originally \(360 + 150 = 510\), or \(510\,\text{m}\).

Answer

1. The first roll was \(510\,\text{m}\), and the second roll was \(360\,\text{m}\). 2. After the cut, each roll is \(360\,\text{m}\).
5197734
Two fourth-grade classes raise money for an animal shelter. One class raises \(\$36\) each school day, and the other raises \(\$44\) each school day. Their goal is \(\$1200\). a) How much do the classes raise together after \(10\) school days? b) How many school days will they need to reach their goal?

Hints

- First find how much the two classes raise together each day. - Use that daily amount to find the total after \(10\) days. - Find the number of daily amounts that make the goal by using multiplication.

Solution

1. Find the combined daily amount: \(\$36 + \$44 = \$80\). 2. Find the amount after \(10\) days: \(10 \times \$80 = \$800\). 3. Use an inverse multiplication fact to reach the goal: \(15 \times \$80 = \$1200\), so the classes need \(15\) school days.

Answer

a) After \(10\) school days, they have raised \(\$800\). b) They need \(15\) school days to reach the goal.
5197824
Twice a number is \(600\). What is three times that number?

Hints

- Decide what number multiplied by \(2\) equals \(600\). - Use the given double to find the original number. - Then multiply the original number by \(3\).

Solution

1. Find the number: \(600 \div 2 = 300\). 2. Find three times the number: \(3 \times 300 = 900\).

Answer

\(900\)
5198064
At a flea market, Jan sells comic books for \(\$3\) each and adventure books for \(\$5\) each. He sells the same number of each kind and earns \(\$112\) altogether. a) How many comic books does Jan sell? b) How much would he earn if every item he sold had instead cost \(\$3\)?

Hints

- Pair one comic book with one adventure book. - Find how many equal pairs make the total revenue. - For part b, find the total number of items and use the new unit price.

Solution

1. One comic book and one adventure book bring in \(\$3 + \$5 = \$8\). 2. Jan sells \(\$112 \div \$8 = 14\) pairs, so he sells \(14\) comic books and \(14\) adventure books. 3. He sells \(14 + 14 = 28\) items altogether. 4. At \(\$3\) per item, he would earn \(28 \times \$3 = \$84\).

Answer

a) Jan sells \(14\) comic books. b) He would earn \(\$84\).
5198144
A toy store receives building blocks packed in \(6\) red boxes and \(4\) blue boxes. Each box contains \(80\) blocks. a) How many blocks are in the shipment? Show two different methods. b) A preschool buys \(2\) of the red boxes. How many blocks from the shipment remain in the store?

Hints

- First determine the total number of boxes. - For part a, compare finding each color separately with combining the box counts first. - For part b, determine how many boxes or blocks remain after \(2\) boxes are removed.

Solution

1. Part a, Method 1: The red boxes contain \(6 \times 80 = 480\) blocks, and the blue boxes contain \(4 \times 80 = 320\) blocks. Altogether, \(480 + 320 = 800\). 2. Part a, Method 2: There are \(6 + 4 = 10\) boxes, so \(10 \times 80 = 800\) blocks. 3. Part b: The preschool buys \(2 \times 80 = 160\) blocks, so \(800 - 160 = 640\) blocks remain. Equivalently, \(10 - 2 = 8\) boxes remain, and \(8 \times 80 = 640\).

Answer

a) \(800\) blocks b) \(640\) blocks
5198384
A small bus has \(15\) passengers. A large train car has four times as many passengers. Write two different mathematical questions about the situation and answer them.

Hints

- First ask about the unknown number of train passengers. - For a second question, combine the passengers in both vehicles. - Decide whether multiplication or addition answers each question.

Solution

1. One question is, “How many passengers are in the train car?” Multiply: \(15 \times 4 = 60\). 2. Another question is, “How many passengers are in the bus and train car altogether?” Add: \(15 + 60 = 75\).

Answer

One possible pair is: 1. “How many passengers are in the train car?” There are \(60\) passengers. 2. “How many passengers are there altogether?” There are \(75\) passengers altogether.
5198494
A zoo feeds its elephants \(124\,\text{kg}\) of hay each day. How much hay do the elephants eat in one week of \(7\) days? Is a \(900\,\text{kg}\) supply enough for the week? Explain.

Hints

- How many days are in one week? - Find the total amount eaten during the week. - Compare that total with \(900\,\text{kg}\). - Decide whether your total is greater than or less than \(900\).

Solution

1. Multiply the daily amount by the number of days: \(124 \times 7 = 868\,\text{kg}\). 2. Compare the weekly amount with the supply: \(868 < 900\). 3. The supply is enough because the elephants need less than \(900\,\text{kg}\).

Answer

The elephants eat \(868\,\text{kg}\) of hay in one week. The \(900\,\text{kg}\) supply is enough.
5198504
A factory packs \(235\) marbles in each mesh bag. a) How many marbles are in \(4\) full bags? b) If the factory begins with \(1000\) marbles and fills the \(4\) bags, how many marbles remain?

Hints

- Multiply the number of marbles in one bag by \(4\). - Think about how the starting supply changes after the bags are filled. - Subtract the packed marbles from the starting amount.

Solution

1. Find the number packed: \(235 \times 4 = 940\) marbles. 2. Subtract from the starting amount: \(1000 - 940 = 60\) marbles.

Answer

a) \(940\) marbles b) \(60\) marbles
5198724
A puzzle has \(1000\) pieces. Julia places \(165\) pieces correctly on Saturday and three times as many on Sunday. How many pieces are still missing from the puzzle?

Hints

- First find the number placed on Sunday. - Add the Saturday and Sunday amounts. - Subtract the completed pieces from \(1000\).

Solution

1. Find the number placed on Sunday: \(165 \times 3 = 495\). 2. Find the total placed during the weekend: \(165 + 495 = 660\). 3. Find the number missing: \(1000 - 660 = 340\).

Answer

The puzzle is missing \(340\) pieces.
5199074
One farmer brings \(7\) sacks of potatoes weighing \(45\,\text{kg}\) each to a market. Another farmer brings \(6\) sacks of onions weighing \(55\,\text{kg}\) each. Which load is heavier, and what is the difference in weight?

Hints

- Find the total weight of each load. - Compare the two products. - Subtract the smaller weight from the larger weight.

Solution

1. Find the weight of the potatoes: \(7 \times 45\,\text{kg} = 315\,\text{kg}\). 2. Find the weight of the onions: \(6 \times 55\,\text{kg} = 330\,\text{kg}\). 3. The onion load is heavier because \(330 > 315\). 4. Find the difference: \(330\,\text{kg} - 315\,\text{kg} = 15\,\text{kg}\).

Answer

The onion load is heavier by \(15\,\text{kg}\).
5200034
The product \(60 \times 4\) is equal to \(3\) times another number. What is the other number?

Hints

- Evaluate \(60 \times 4\) first. - Then solve an unknown-factor equation. - Check that the two products are equal.

Solution

1. Find the first product: \(60 \times 4 = 240\). 2. Solve \(3 \times x = 240\): \(x = 240 \div 3 = 80\). 3. Check: \(3 \times 80 = 240\).

Answer

The number is \(80\).
5200294
A city library has \(825\) new children’s books to place in shelf sections. Each shelf section holds exactly \(5\) books. a) How many shelf sections are needed for all the books? b) Without calculating the exact number, decide whether more or fewer shelf sections would be needed if each section held \(10\) books. Explain.

Hints

- Break \(825\) into numbers that are easy to divide by \(5\). - Add the partial quotients. - For part b, compare the number of books each shelf section can hold.

Solution

1. For part a, divide: \(825 \div 5\). 2. Decompose \(825\): \(825 = 500 + 300 + 25\). 3. Divide each part: \(500 \div 5 = 100\), \(300 \div 5 = 60\), and \(25 \div 5 = 5\). Then add: \(100 + 60 + 5 = 165\). 4. For part b, each shelf section would hold more books, so fewer sections would be needed for the same total.

Answer

a) \(165\) shelf sections are needed. b) Fewer shelf sections would be needed because each section would hold more books.
5200444
A year can be divided into two halves. The first half includes January through June, and the second half includes July through December. a) How many days are in the first half of a non-leap year? b) How many days are in the second half? c) Which half is longer, and by how many days? d) How does the difference change during a leap year?

Hints

- List the number of days in each month. - Add the first six months and last six months separately. - Compare the two totals. - In a leap year, only February changes.

Solution

1. Add the days from January through June in a non-leap year: \(31 + 28 + 31 + 30 + 31 + 30 = 181\) days. 2. Add the days from July through December: \(31 + 31 + 30 + 31 + 30 + 31 = 184\) days. 3. Find the difference: \(184 - 181 = 3\) days, so the second half is longer. 4. In a leap year, February has \(29\) days, making the first half \(182\) days long. 5. The new difference is \(184 - 182 = 2\) days.

Answer

a) The first half has \(181\) days. b) The second half has \(184\) days. c) The second half is longer by \(3\) days. d) In a leap year, the second half is longer by \(2\) days.
5200704
A printing machine prints \(50\) posters in \(5\) minutes. If it works at a constant rate, how many posters does it print in one hour, or \(60\) minutes?

Hints

- How many five-minute intervals are in one hour? - Once you know the number of intervals, how can you find the total number of posters? - Will the machine print more or fewer posters in one hour than in five minutes?

Solution

1. Find how many five-minute intervals are in one hour: \(60 \div 5 = 12\). 2. Multiply by the number of posters printed in each interval: \(12 \times 50 = 600\).

Answer

The machine prints \(600\) posters in one hour.
5200714
A nursery sells young tomato plants in trays. One tray with \(6\) plants costs \(\$5\). A vegetable farmer spends \(\$125\) on these trays. How many plants does the farmer buy altogether?

Hints

- How many trays can the farmer buy for \(\$125\) if each tray costs \(\$5\)? - Each tray has \(6\) plants. How can you find the total number of plants once you know the number of trays? - Could you break \(125\) into smaller parts to make the division easier?

Solution

1. Find the number of trays the farmer buys: \(\$125 \div \$5 = 25\). 2. Find the total number of plants: \(25 \times 6 = 150\).

Answer

The farmer buys \(150\) tomato plants altogether.
5200804
A gardener has \(360\) tulip bulbs and plants \(9\) bulbs in each row. The gardener has already planted \(15\) rows. How many more rows must be planted to use all the bulbs?

Hints

- First find how many rows are needed altogether. - Identify how many rows are already planted. - Subtract to find the number of rows remaining.

Solution

1. Find the total number of rows: \(360 \div 9 = 40\) rows. 2. Subtract the rows already planted: \(40 - 15 = 25\) rows.

Answer

The gardener must plant \(25\) more rows.
5200914
Lukas wants to buy a skateboard that costs \(\$140\). He has saved \(\$85\), and his grandmother gives him \(\$20\) for his birthday. He can save \(\$5\) each week. How many more weeks must Lukas save before he can buy the skateboard?

Hints

- First find how much money Lukas has after adding the gift. - Subtract that amount from the skateboard’s price. - Determine how many weekly savings amounts fit into the amount still needed.

Solution

1. Find how much money Lukas has after the gift: \(\$85 + \$20 = \$105\). 2. Find how much more he needs: \(\$140 - \$105 = \$35\). 3. Find the number of weeks: \(\$35 \div \$5 = 7\).

Answer

Lukas must save for \(7\) more weeks.
5201014
A small truck can carry at most \(600\,\text{kg}\). It is already carrying \(4\) heavy barrels that each weigh \(80\,\text{kg}\). The driver also wants to load \(8\) boxes of tile that each weigh \(40\,\text{kg}\). Can the driver load all \(8\) boxes without exceeding the limit? Justify your answer.

Hints

- Find the total mass of the barrels. - Subtract that mass from the truck's limit. - Find the total mass of the \(8\) boxes. - Compare the box mass with the remaining capacity.

Solution

1. The barrels weigh \(4 \times 80 = 320\,\text{kg}\). 2. The remaining capacity is \(600 - 320 = 280\,\text{kg}\). 3. The boxes weigh \(8 \times 40 = 320\,\text{kg}\). 4. Since \(320 > 280\), all \(8\) boxes would exceed the remaining capacity.

Answer

No. The boxes weigh \(320\,\text{kg}\), but only \(280\,\text{kg}\) of capacity remains.
5201664
The table shows the tons of paper recycled in three cities during two consecutive years. <table> <tr><th>City</th><th>Year 1</th><th>Year 2</th></tr> <tr><td>Oakton</td><td>\(12{,}450\,\text{tons}\)</td><td>\(15{,}800\,\text{tons}\)</td></tr> <tr><td>Pineville</td><td>\(8900\,\text{tons}\)</td><td>\(12{,}150\,\text{tons}\)</td></tr> <tr><td>Maple City</td><td>\(5650\,\text{tons}\)</td><td>\(9400\,\text{tons}\)</td></tr> </table> a) Which city had the greatest increase in recycled paper from Year 1 to Year 2? b) How many more tons of paper were recycled across all three cities in Year 2 than in Year 1?

Hints

- How can you find the difference between two values? - Find the increase for each city, and then compare the increases. - Can you find the overall increase without first finding both yearly totals? - Line up place values carefully when adding large numbers.

Solution

1. Find the increase for each city. Oakton: \(15{,}800 - 12{,}450 = 3350\) tons. Pineville: \(12{,}150 - 8900 = 3250\) tons. Maple City: \(9400 - 5650 = 3750\) tons. Maple City has the greatest increase. 2. Find the total for Year 1: \(12{,}450 + 8900 + 5650 = 27{,}000\) tons. 3. Find the total for Year 2: \(15{,}800 + 12{,}150 + 9400 = 37{,}350\) tons. 4. Subtract the totals: \(37{,}350 - 27{,}000 = 10{,}350\) tons. The same result comes from adding the three city increases.

Answer

a) Maple City, with an increase of \(3750\) tons b) \(10{,}350\) tons
5201674
Three museums summarized their attendance for 2022 and 2023 in a table. <table> <tr><th>Museum</th><th>Visitors in 2022</th><th>Visitors in 2023</th></tr> <tr><td>Science Museum</td><td>\(215{,}400\)</td><td>\(248{,}900\)</td></tr> <tr><td>Art Museum</td><td>\(178{,}250\)</td><td>\(165{,}300\)</td></tr> <tr><td>Natural History Museum</td><td>\(324{,}800\)</td><td>\(352{,}150\)</td></tr> </table> a) Find the total attendance at all three museums for each year. b) Find the difference between the two yearly totals. Was there an overall increase or decrease?

Hints

- What operation does the word “total” suggest? - Notice that attendance at the Art Museum changed in a different direction from the other two museums. - Subtract the two yearly totals to find their difference. - Compare the totals to decide whether attendance increased or decreased.

Solution

1. Add the 2022 attendance figures: \(215{,}400 + 178{,}250 + 324{,}800 = 718{,}450\). 2. Add the 2023 attendance figures: \(248{,}900 + 165{,}300 + 352{,}150 = 766{,}350\). 3. Find the difference: \(766{,}350 - 718{,}450 = 47{,}900\). 4. Because the 2023 total is greater, attendance increased overall by \(47{,}900\) visitors.

Answer

a) 2022: \(718{,}450\) visitors 2023: \(766{,}350\) visitors b) An increase of \(47{,}900\) visitors
5202114
A bottle contains \(32\,\text{fl oz}\) of apple juice. Lucas first drinks \(\frac{1}{4}\) of the juice in the full bottle. Then his sister Marie drinks exactly \(\frac{1}{2}\) of the juice that remains. How many fluid ounces of juice are left in the bottle?

Hints

- How much juice does Lucas drink from the full bottle? - How much remains after Lucas drinks? - Marie drinks one-half of the amount that remains, not one-half of the original amount. - After finding Marie's amount, subtract it from the amount that was left.

Solution

1. Find the amount Lucas drinks: \(\frac{1}{4} \times 32\,\text{fl oz} = 8\,\text{fl oz}\). 2. Find the amount remaining after Lucas drinks: \(32\,\text{fl oz} - 8\,\text{fl oz} = 24\,\text{fl oz}\). 3. Marie drinks one-half of the remaining juice: \(\frac{1}{2} \times 24\,\text{fl oz} = 12\,\text{fl oz}\). 4. Find the final amount remaining: \(24\,\text{fl oz} - 12\,\text{fl oz} = 12\,\text{fl oz}\).

Answer

\(12\,\text{fl oz}\) of juice remain in the bottle.
5202214
A field day lasts exactly one hour. Students spend \(\frac{1}{4}\) of the time warming up together and \(\frac{1}{2}\) of the time in running events. The remaining time is for the awards ceremony. How many minutes are left for the awards ceremony?

Hints

- How many minutes are in one hour? - First find the minutes spent warming up and running. - Add those times to find how much of the hour is already planned. - How many minutes remain in the full hour?

Solution

1. Convert the total time to minutes: \(1\,\text{hour} = 60\,\text{minutes}\). 2. Find the warm-up time: \(\frac{1}{4} \times 60\,\text{minutes} = 15\,\text{minutes}\). 3. Find the running-event time: \(\frac{1}{2} \times 60\,\text{minutes} = 30\,\text{minutes}\). 4. Find the time already used: \(15\,\text{minutes} + 30\,\text{minutes} = 45\,\text{minutes}\). 5. Find the remaining time: \(60\,\text{minutes} - 45\,\text{minutes} = 15\,\text{minutes}\).

Answer

There are \(15\,\text{minutes}\) left for the awards ceremony.
5202274
A fruit basket contains \(30\) pieces of fruit. One-fifth of the fruit are pears, one-third are apples, and the rest are bananas. Find the number of pears and apples. How many bananas are in the basket? Are there more apples or pears?

Hints

- Find the number of pieces in each given fruit group separately. - How can you find a fraction of a known total? - After finding the pears and apples, how can you find the rest?

Solution

1. Find the number of pears: \(\frac{1}{5} \times 30 = 6\). 2. Find the number of apples: \(\frac{1}{3} \times 30 = 10\). 3. Since \(10 > 6\), there are more apples than pears. 4. Find the number of bananas: \(30 - 6 - 10 = 14\).

Answer

There are \(6\) pears, \(10\) apples, and \(14\) bananas. There are more apples than pears.
5202314
A farmer collects \(720\) eggs. One-sixth of the eggs are cracked and cannot be sold. The farmer packs the remaining eggs into cartons that hold \(12\) eggs each. How many full cartons can the farmer pack?

Hints

- First find how many eggs are cracked. - How many eggs remain to be sold? - After finding the good eggs, divide them into equal groups for the cartons.

Solution

1. Find the number of cracked eggs: \(\frac{1}{6} \times 720 = 120\). 2. Find the number of uncracked eggs: \(720 - 120 = 600\). 3. Divide by the number of eggs in each carton: \(600 \div 12 = 50\).

Answer

The farmer can pack \(50\) full cartons.
5202354
A class has \(24\) students. One-third of the students ride bicycles to school, and one-fourth walk to school. The rest ride the bus. How many students ride the bus?

Hints

- Find the bicycle and walking groups separately. - How many students are in those two groups altogether? - How can you find the number remaining for the bus? - A drawing with \(24\) dots may help you organize the groups.

Solution

1. Find the number who ride bicycles: \(\frac{1}{3} \times 24 = 8\). 2. Find the number who walk: \(\frac{1}{4} \times 24 = 6\). 3. Find the number who do not ride the bus: \(8 + 6 = 14\). 4. Subtract from the class total: \(24 - 14 = 10\).

Answer

\(10\) students ride the bus.
5202514
A bookstore has three locations. The table shows how many books each location sold in April and May. <table> <tr><th>Location</th><th>April</th><th>May</th></tr> <tr><td>Downtown</td><td>\(12{,}450\)</td><td>\(11{,}800\)</td></tr> <tr><td>Station</td><td>\(8920\)</td><td>\(10{,}250\)</td></tr> <tr><td>Westside</td><td>\(4130\)</td><td>\(4560\)</td></tr> </table> a) At which location were fewer books sold in May than in April? By how many books did sales decrease there? b) How many books were sold at all three locations over both months altogether?

Hints

- Compare the two entries in each row to find where the May value is smaller. - For part b), every sales entry in the table must be included. - You can first find a total for each month and then add the two totals.

Solution

1. Compare April and May at each location. Only Downtown decreased, from \(12{,}450\) books to \(11{,}800\) books. 2. Find the decrease: \(12{,}450 - 11{,}800 = 650\) books. 3. Add the April sales: \(12{,}450 + 8920 + 4130 = 25{,}500\). 4. Add the May sales: \(11{,}800 + 10{,}250 + 4560 = 26{,}610\). 5. Add the monthly totals: \(25{,}500 + 26{,}610 = 52{,}110\) books.

Answer

a) Downtown; sales decreased by \(650\) books. b) \(52{,}110\) books
5202604
Lucas walks from home toward a bus stop. After \(300\,\text{m}\), he realizes that he forgot his bus pass. He walks back home, gets the pass, and then walks directly to the bus stop. His fitness tracker shows that he walked \(1400\,\text{m}\) altogether. How far is the bus stop from his home?

Hints

- Which part of the route does Lucas walk twice because he forgot the pass? - Find the total extra distance caused by returning home. - Subtract that extra distance from the amount shown on the fitness tracker.

Solution

1. The forgotten pass adds a trip of \(300\,\text{m}\) away from home and \(300\,\text{m}\) back: \(2 \times 300 = 600\,\text{m}\). 2. Subtract the extra distance from the total: \(1400 - 600 = 800\,\text{m}\). 3. The direct distance from home to the bus stop is \(800\,\text{m}\).

Answer

The bus stop is \(800\,\text{m}\) from Lucas’s home.
5202894
An orchard stores \(325\,\text{kg}\) of apples and \(180\,\text{kg}\) of pears. In the morning, \(115\,\text{kg}\) of apples are delivered to a store. In the afternoon, \(95\,\text{kg}\) of newly harvested pears are added. How many kilograms of fruit are in storage at the end of the day?

Hints

- Subtract the apples that are delivered. - Add the newly harvested pears. - Add the final amounts of apples and pears.

Solution

1. Find the apples remaining: \(325 - 115 = 210\,\text{kg}\). 2. Find the new amount of pears: \(180 + 95 = 275\,\text{kg}\). 3. Add the final amounts: \(210 + 275 = 485\,\text{kg}\).

Answer

There are \(485\,\text{kg}\) of fruit in storage at the end of the day.
5203024
A gardener plants \(9\) rows with \(35\) tulip bulbs in each row. There are \(78\) bulbs left in the basket. How many tulip bulbs were in the basket at the beginning?

Hints

- First find the number of bulbs already planted. - Use multiplication for the equal rows. - Include the bulbs still in the basket.

Solution

1. Find the number planted: \(9 \times 35 = 315\). 2. Add the bulbs left in the basket: \(315 + 78 = 393\).

Answer

There were \(393\) tulip bulbs in the basket at the beginning.
5203034
Two classes collect pinecones for an art project. Class A fills \(8\) bags with \(45\) pinecones in each bag. Class B fills \(6\) bags with \(55\) pinecones in each bag. Which class collects more pinecones, and what is the difference?

Hints

- Find each class's total separately. - Compare the two products. - Subtract the smaller total from the larger total.

Solution

1. Find Class A's total: \(8 \times 45 = 360\). 2. Find Class B's total: \(6 \times 55 = 330\). 3. Since \(360 > 330\), Class A collects more. 4. Find the difference: \(360 - 330 = 30\).

Answer

Class A collects \(30\) more pinecones than Class B.
5203074
For a school fitness program, Group A receives \(7\) packages with \(12\) jump ropes in each package. Group B receives \(5\) packages with \(16\) jump ropes in each package. Which group receives more jump ropes? How many jump ropes do the two groups receive altogether?

Hints

- Find each group's total separately. - Compare the two totals. - Add them to find the combined number of jump ropes.

Solution

1. Find Group A's total: \(7 \times 12 = 84\). 2. Find Group B's total: \(5 \times 16 = 80\). 3. Since \(84 > 80\), Group A receives more. 4. Find the combined total: \(84 + 80 = 164\).

Answer

Group A receives more jump ropes. The two groups receive \(164\) jump ropes altogether.
5203084
At a field day, \(168\) students are first divided into teams of \(6\). How many teams are formed? Then calculate how many teams would be formed if each team had \(4\) students instead. Would the smaller team size create more or fewer teams?

Hints

- Divide \(168\) by each team size. - Use a decomposition that works well with each divisor. - Compare the two quotients.

Solution

1. For teams of \(6\), break \(168\) into compatible numbers: \(168 = 120 + 48\). Then \(120 \div 6 = 20\) and \(48 \div 6 = 8\), so \(20 + 8 = 28\) teams. 2. For teams of \(4\), use \(168 = 160 + 8\). Then \(160 \div 4 = 40\) and \(8 \div 4 = 2\), so \(40 + 2 = 42\) teams. 3. Since \(42 > 28\), the smaller team size creates more teams.

Answer

Teams of \(6\) produce \(28\) teams. Teams of \(4\) produce \(42\) teams, so the smaller team size creates more teams.
5203124
A toy bin contains \(150\) blue building blocks. It contains one-third as many red blocks as blue blocks. How many more blue blocks than red blocks are in the bin?

Hints

- First find the exact number of red blocks. - Dividing by \(3\) finds one-third of a quantity. - Once you know both quantities, find their difference.

Solution

1. Find the number of red blocks: \(150 \div 3 = 50\). 2. Find the difference: \(150 - 50 = 100\).

Answer

The bin contains \(100\) more blue blocks than red blocks.
5203244
A library has books on three shelves. Shelf A has \(124\) books. Shelf B has half as many books as Shelf A. Shelf C has three times as many books as Shelf B. How many more books are on Shelf C than on Shelf A?

Hints

- What operation finds half as many books? - Find the number of books on each shelf one step at a time. - The question asks for a difference, not the combined total.

Solution

1. Find the number of books on Shelf B: \(124 \div 2 = 62\). 2. Find the number of books on Shelf C: \(62 \times 3 = 186\). 3. Find the difference between Shelves C and A: \(186 - 124 = 62\).

Answer

Shelf C has \(62\) more books than Shelf A.
5203264
Students harvest \(120\,\text{kg}\) of potatoes from one garden bed. They harvest twice as much from a second bed and \(50\,\text{kg}\) less from a third bed than from the first. How many more kilograms are harvested from the second bed than from the third bed?

Hints

- Find the harvest from each bed first. - Use multiplication for the second bed and subtraction for the third. - Compare the second and third amounts.

Solution

1. Find the second-bed harvest: \(120\,\text{kg} \times 2 = 240\,\text{kg}\). 2. Find the third-bed harvest: \(120\,\text{kg} - 50\,\text{kg} = 70\,\text{kg}\). 3. Find the difference: \(240\,\text{kg} - 70\,\text{kg} = 170\,\text{kg}\).

Answer

The second bed produces \(170\,\text{kg}\) more potatoes than the third bed.
5203354
Leon and Sophie start with \(80\). Leon adds \(20\), then multiplies the result by \(4\). Sophie multiplies \(80\) by \(4\), then adds \(20\). Find both results. Who gets the greater number?

Hints

- Follow each person's operations in the stated order. - Write the two calculation chains separately. - Compare the final results.

Solution

1. Leon: \(80 + 20 = 100\), then \(100 \times 4 = 400\). 2. Sophie: \(80 \times 4 = 320\), then \(320 + 20 = 340\). 3. Since \(400 > 340\), Leon gets the greater number.

Answer

Leon gets \(400\). Sophie gets \(340\). Leon gets the greater number.
5203474
A bakery used to bake \(140\) rolls per hour. After an upgrade, it bakes \(165\) rolls per hour. How many more rolls are baked during an \(8\)-hour workday? Show two different calculation methods.

Hints

- One method starts with the difference for one hour. - Another method finds both full-day totals first. - Both methods should give the same result.

Solution

1. Method 1: Find the hourly increase and scale it to the workday. \(165 - 140 = 25\) more rolls per hour, and \(25 \times 8 = 200\) more rolls. 2. Method 2: Find each daily total and subtract. \(165 \times 8 = 1320\), \(140 \times 8 = 1120\), and \(1320 - 1120 = 200\) rolls.

Answer

The bakery produces \(200\) more rolls during an \(8\)-hour workday.
5203494
Paul buys \(4\) glitter stickers for \(35\) cents each and \(6\) animal stickers for \(45\) cents each. He pays with a \(\$10\) bill. How much change does Paul receive?

Hints

- Find the cost of each kind of sticker. - Convert the amount paid to cents. - Subtract the total cost from the amount paid.

Solution

1. The glitter stickers cost \(4 \times 35 = 140\) cents. 2. The animal stickers cost \(6 \times 45 = 270\) cents. 3. The total cost is \(140 + 270 = 410\) cents. 4. A \(\$10\) bill is \(1000\) cents, so the change is \(1000 - 410 = 590\) cents, or \(\$5.90\).

Answer

Paul receives \(\$5.90\) in change.
5203644
A tour bus has \(54\) passengers. A regular ticket costs \(\$30\), and a discount ticket costs \(\$24\). One out of every \(3\) passengers has a discount ticket. All other passengers have regular tickets. How much ticket revenue did the trip generate?

Hints

- What calculation represents “one out of every \(3\) passengers”? - How many passengers pay the regular price? - Find the revenue from each group before adding.

Solution

1. The number of discount tickets is \(54\div3=18\). 2. The number of regular tickets is \(54-18=36\). 3. Revenue from discount tickets is \(18\times\$24=\$432\). 4. Revenue from regular tickets is \(36\times\$30=\$1080\). 5. The total revenue is \(\$432+\$1080=\$1512\).

Answer

The trip generated \(\$1512\) in ticket revenue.
5203814
A small library used to have only \(90\) books. Today it has \(140\) children’s books and \(3\) times as many nonfiction books as children’s books. How many more books does the library have today than it had before?

Hints

- First find the number of nonfiction books. - Add the two current book categories. - Compare today’s total with the old total.

Solution

1. The library has \(140\times 3=420\) nonfiction books. 2. It has \(140+420=560\) books today. 3. The increase is \(560-90=470\) books.

Answer

The library has \(470\) more books today.
5203824
A sports club had \(55\) members when it was founded. Today it has \(110\) youth members and \(4\) times as many adult members as youth members. How many more members does the club need so that its current membership is exactly \(600\) greater than its founding membership?

Hints

- Find the number of adult members first. - Add to find the club’s current membership. - Determine the target membership that is \(600\) above the founding amount. - Compare the current membership with the target.

Solution

1. The number of adult members is \(110\times 4=440\). 2. The club currently has \(110+440=550\) members. 3. A membership that is \(600\) greater than the founding membership would be \(55+600=655\). 4. The club needs \(655-550=105\) more members. 5. Equivalently, the current increase is \(550-55=495\), and \(600-495=105\).

Answer

The club needs \(105\) more members.
5204034
Three fourth-grade classes collect paper for recycling. Class 4A collects \(24\,\text{kg}\). Class 4B collects three times as much as Class 4A. Class 4C collects \(15\,\text{kg}\) more than Class 4A. How many kilograms of paper do the three classes collect altogether?

Hints

- Find how much Class 4B collects. - Find how much Class 4C collects. - Add all three classes' amounts.

Solution

1. Find the amount collected by Class 4B: \(24\,\text{kg} \times 3 = 72\,\text{kg}\). 2. Find the amount collected by Class 4C: \(24\,\text{kg} + 15\,\text{kg} = 39\,\text{kg}\). 3. Add the three amounts: \(24\,\text{kg} + 72\,\text{kg} + 39\,\text{kg} = 135\,\text{kg}\).

Answer

The three classes collect \(135\,\text{kg}\) of paper altogether.
5204044
A gardener buys \(5\) bags of flower bulbs for \(\$25\). a) What is the price of one bag? b) How many bags can the gardener buy with \(\$125\) at the same price per bag?

Hints

- Divide the total cost by \(5\) to find the unit price. - Then divide the budget by the price of one bag. - Check using multiplication.

Solution

1. One bag costs \(\$25 \div 5 = \$5\). 2. With \(\$125\), the gardener can buy \(\$125 \div \$5 = 25\) bags.

Answer

a) One bag costs \(\$5\). b) The gardener can buy \(25\) bags.
5204054
A school-supply store sells notebooks in packages. Six packages cost \(\$48\). a) What is the price of one package? b) A teacher spends \(\$96\) on the same packages. How many packages does the teacher buy? c) Each package contains \(10\) notebooks. How many notebooks does the teacher buy altogether?

Hints

- Find the unit price first. - Divide the teacher’s total cost by the unit price. - For the final part, use the number of notebooks per package.

Solution

1. One package costs \(\$48 \div 6 = \$8\). 2. The teacher buys \(\$96 \div \$8 = 12\) packages. 3. The packages contain \(12 \times 10 = 120\) notebooks.

Answer

a) One package costs \(\$8\). b) The teacher buys \(12\) packages. c) The teacher buys \(120\) notebooks altogether.
5204124
Five children want to visit a zoo. A regular child ticket costs \(\$8\), and together they have saved \(\$33\). a) How much more money do they need to buy five individual tickets? b) A small-group pass for up to five people costs \(\$38\). One child says, “We need to collect less additional money for the group pass than for five individual tickets.” Is the child correct? Explain with calculations.

Hints

- Find the total price of five individual tickets. - Find how much is missing for each option. - Compare the two missing amounts.

Solution

1. Five individual tickets cost \(5 \times \$8 = \$40\). 2. The group needs \(\$40 - \$33 = \$7\) more for individual tickets. 3. The group needs \(\$38 - \$33 = \$5\) more for the small-group pass. 4. Since \(\$5 < \$7\), the child is correct.

Answer

a) The group needs \(\$7\) more for individual tickets. b) Yes. The group needs only \(\$5\) more for the small-group pass, compared with \(\$7\) for individual tickets.
5204644
An old castle was completed in \(1580\). a) A new tower was added exactly \(120\) years later. In what year was it added? b) A gatehouse was restored \(75\) years after the tower was added. In what year was the gatehouse restored? c) How many years passed from the castle’s completion to the gatehouse restoration?

Hints

- Add the first elapsed time to the castle’s completion year. - Use the tower year as the starting point for the next event. - To find the total elapsed time, compare the first and last years.

Solution

1. Find the year the tower was added: \(1580 + 120 = 1700\). 2. Find the year the gatehouse was restored: \(1700 + 75 = 1775\). 3. Find the total elapsed time: \(1775 - 1580 = 195\) years.

Answer

a) The tower was added in \(1700\). b) The gatehouse was restored in \(1775\). c) \(195\) years passed.
5205294
A warehouse shipped \(67{,}804\), \(123{,}900\), \(8560\), and \(204{,}050\) packages during four different periods. a) Find the difference between the greatest and least shipment totals. b) Add the two least shipment totals. Then subtract that sum from the greatest shipment total.

Hints

- Order the four totals before doing any calculations. - For part a, subtract the least total from the greatest. - For part b, identify and add the two least totals first. - Keep place values aligned when adding and subtracting.

Solution

1. Order the totals: \(8560 < 67{,}804 < 123{,}900 < 204{,}050\). 2. For a), subtract the least total from the greatest: \(204{,}050 - 8560 = 195{,}490\). 3. For b), add the two least totals: \(8560 + 67{,}804 = 76{,}364\). 4. Subtract that sum from the greatest total: \(204{,}050 - 76{,}364 = 127{,}686\).

Answer

a) \(195{,}490\) b) \(127{,}686\)
5205344
An orchard harvests \(4250\,\text{kg}\) of apples, pears, and plums altogether. The apples and pears together have a mass of \(3120\,\text{kg}\). The orchard harvests \(1845\,\text{kg}\) of apples. How many kilograms of each type of fruit are harvested?

Hints

- Use the combined apple-and-pear amount to find the pears. - Subtract the apple-and-pear amount from the total to find the plums. - Add all three amounts to check.

Solution

1. The apple harvest is \(1845\,\text{kg}\). 2. Find the pear harvest: \(3120\,\text{kg} - 1845\,\text{kg} = 1275\,\text{kg}\). 3. Find the plum harvest: \(4250\,\text{kg} - 3120\,\text{kg} = 1130\,\text{kg}\). 4. Check: \(1845\,\text{kg} + 1275\,\text{kg} + 1130\,\text{kg} = 4250\,\text{kg}\).

Answer

The orchard harvests \(1845\,\text{kg}\) of apples, \(1275\,\text{kg}\) of pears, and \(1130\,\text{kg}\) of plums.
5205464
An adult elephant has a mass of \(6200\,\text{kg}\), which is \(3850\,\text{kg}\) more than a rhinoceros. A hippopotamus has a mass that is \(540\,\text{kg}\) less than the rhinoceros. What are the masses of the rhinoceros and the hippopotamus? Which of the three animals has the least mass?

Hints

- The elephant's mass is greater, so subtract to find the rhinoceros's mass. - Then subtract again to find the hippopotamus's mass. - Compare all three masses.

Solution

1. Find the rhinoceros's mass: \(6200\,\text{kg} - 3850\,\text{kg} = 2350\,\text{kg}\). 2. Find the hippopotamus's mass: \(2350\,\text{kg} - 540\,\text{kg} = 1810\,\text{kg}\). 3. Compare the masses: \(1810\,\text{kg} < 2350\,\text{kg} < 6200\,\text{kg}\), so the hippopotamus has the least mass.

Answer

The rhinoceros has a mass of \(2350\,\text{kg}\), and the hippopotamus has a mass of \(1810\,\text{kg}\). The hippopotamus has the least mass.
5205644
An elementary school collects recyclable materials for a project. The table shows how many pounds of paper and plastic were collected during four months. <table> <thead> <tr> <th>Month</th> <th>Paper (lb)</th> <th>Plastic (lb)</th> </tr> </thead> <tbody> <tr> <td>March</td> <td>\(456\)</td> <td>\(212\)</td> </tr> <tr> <td>April</td> <td>\(512\)</td> <td>\(198\)</td> </tr> <tr> <td>May</td> <td>\(489\)</td> <td>\(245\)</td> </tr> <tr> <td>June</td> <td>\(534\)</td> <td>\(221\)</td> </tr> </tbody> </table> During which month was the greatest total weight of material collected? How many more pounds was that than in the month with the least total weight?

Hints

- First find the total amount collected in each month. - Compare the monthly totals to find the greatest and least. - What is the difference between those two totals?

Solution

1. Find the total for each month. March: \(456 + 212 = 668\) pounds. April: \(512 + 198 = 710\) pounds. May: \(489 + 245 = 734\) pounds. June: \(534 + 221 = 755\) pounds. 2. The greatest total is \(755\) pounds in June, and the least total is \(668\) pounds in March. 3. Find the difference: \(755 - 668 = 87\) pounds.

Answer

The greatest total was collected in June. It was \(87\) pounds more than the March total.
5205704
Write a word problem that can be solved with these two steps: 1) \(1200+500=1700\) 2) \(1700-800=900\) Write your word problem and its answer.

Hints

- Choose a situation in which something is added first and removed later. - Possible contexts include inventory, animals, or books. - Use the given numbers exactly in the story. - End with a question whose answer is \(900\).

Solution

1. A valid problem must begin with a quantity of \(1200\), increase it by \(500\), and then decrease the new quantity by \(800\). 2. The intermediate quantity is \(1700\), and the final quantity is \(900\). 3. The context, units, and wording may vary, but both operations and all three quantities must match the story.

Answer

Answers will vary. Example: A toy store has \(1200\) marbles. It receives \(500\) more and then sells \(800\). The store has \(900\) marbles left.
5205994
A fruit seller makes \(25\) identical gift baskets. Each basket contains \(6\) apples, \(4\) pears, and \(2\) bananas. a) How many pieces of fruit are used for all \(25\) baskets? b) Each apple costs \(30\) cents. What is the total cost of all the apples?

Hints

- Find the number of pieces of fruit in one basket. - Multiply by the number of baskets. - Find the total number of apples, then multiply by the price per apple.

Solution

1. Each basket contains \(6 + 4 + 2 = 12\) pieces of fruit. 2. All \(25\) baskets contain \(25 \times 12 = 300\) pieces of fruit. 3. The baskets contain \(25 \times 6 = 150\) apples. 4. The apples cost \(150 \times 30 = 4500\) cents, or \(\$45.00\).

Answer

a) The seller uses \(300\) pieces of fruit. b) All the apples cost \(\$45.00\).
5206174
A straight section of fence will be \(180\,\text{ft}\) long. Fence posts will be placed every \(10\,\text{ft}\), with a post at both the beginning and the end. How many posts are needed altogether?

Hints

- Try a shorter example and compare the number of spaces with the number of posts. - Divide the total length by the distance between posts. - Remember to include the post at the starting point.

Solution

1. Find the number of equal spaces between posts: \(180 \div 10 = 18\). 2. A row of \(18\) spaces has one more post than spaces, so \(18 + 1 = 19\) posts are needed.

Answer

\(19\) posts are needed.
5206324
A delivery van may carry at most \(500\,\text{kg}\). It is loaded with \(12\) crates of apples weighing \(25\,\text{kg}\) each and \(8\) crates of pears weighing \(20\,\text{kg}\) each. a) What is the total mass of the load? b) Is the load within the van's limit? c) Write one numerical expression that gives the total mass.

Hints

- Find the total mass of each type of fruit crate. - Add the two partial masses and compare the result with the limit. - For part c, combine the two multiplication expressions with addition.

Solution

1. Find the mass of the apple crates: \(12 \times 25\,\text{kg} = 300\,\text{kg}\). 2. Find the mass of the pear crates: \(8 \times 20\,\text{kg} = 160\,\text{kg}\). 3. Find the total mass: \(300\,\text{kg} + 160\,\text{kg} = 460\,\text{kg}\). 4. Since \(460\,\text{kg} \le 500\,\text{kg}\), the load is within the limit. 5. One expression for the total is \(12 \times 25 + 8 \times 20\).

Answer

a) The load has a total mass of \(460\,\text{kg}\). b) Yes. The load is within the \(500\)-kg limit. c) \(12 \times 25 + 8 \times 20\)
5206334
A school renovation receives \(15\) bags of cement weighing \(25\,\text{kg}\) each and \(12\) bags of plaster mix weighing \(30\,\text{kg}\) each. Which material has the greater total mass? Find the difference and write one numerical expression that gives the difference.

Hints

- Find the total mass of each material. - Subtract the smaller total from the greater total. - Combine the multiplication and subtraction into one expression.

Solution

1. Find the total mass of the cement: \(15 \times 25\,\text{kg} = 375\,\text{kg}\). 2. Find the total mass of the plaster mix: \(12 \times 30\,\text{kg} = 360\,\text{kg}\). 3. Since \(375\,\text{kg} > 360\,\text{kg}\), the cement has the greater total mass. 4. Find the difference: \(375\,\text{kg} - 360\,\text{kg} = 15\,\text{kg}\). 5. One expression for the difference is \(15 \times 25 - 12 \times 30\).

Answer

The cement has the greater total mass by \(15\,\text{kg}\). One expression for the difference is \(15 \times 25 - 12 \times 30\).
5206364
A bakery makes \(125\) rolls each day from Monday through Wednesday, \(150\) rolls each day on Thursday and Friday, and \(210\) rolls each day on Saturday and Sunday. How many rolls does the bakery make during the week?

Hints

- Count the days in each part of the week. - Find the number of rolls for each part separately. - Add the three subtotals.

Solution

1. Find the Monday-through-Wednesday total: \(3 \times 125 = 375\) rolls. 2. Find the Thursday-and-Friday total: \(2 \times 150 = 300\) rolls. 3. Find the weekend total: \(2 \times 210 = 420\) rolls. 4. Add the three totals: \(375 + 300 + 420 = 1095\) rolls.

Answer

The bakery makes \(1095\) rolls during the week.
5206374
An elementary school wants to collect \(500\) books for its new library. In the first week, it receives \(5\) boxes with \(24\) books each. In the second week, it receives \(8\) boxes with \(18\) books each. In the third week, it receives \(12\) boxes with \(15\) books each. Does the school reach its goal? How many books are still needed, or how many extra books were collected?

Hints

- Find the number of books received in each week. - Add the three weekly totals. - Compare the total with the goal of \(500\). - Subtract to find the shortage or extra amount.

Solution

1. The first week brings \(5\times 24=120\) books. 2. The second week brings \(8\times 18=144\) books. 3. The third week brings \(12\times 15=180\) books. 4. The school collects \(120+144+180=444\) books. 5. Since \(444<500\), the goal is not reached. The school still needs \(500-444=56\) books.

Answer

No. The school collects \(444\) books and still needs \(56\) more.
5206394
A small wildlife park recorded its attendance for one weekend. <table> <thead> <tr> <th>Day</th> <th>Children</th> <th>Adults</th> </tr> </thead> <tbody> <tr> <td>Friday</td> <td>\(124\)</td> <td>\(88\)</td> </tr> <tr> <td>Saturday</td> <td>\(456\)</td> <td>\(312\)</td> </tr> <tr> <td>Sunday</td> <td>\(582\)</td> <td>\(295\)</td> </tr> </tbody> </table> a) By how many people did the total attendance increase from Friday to Sunday? b) Was the number of children on Saturday more than three times the number on Friday? Justify your answer with a calculation.

Hints

- Use the row and column labels to identify each entry. - For part a), first find the total attendance for each of the two days. - What operation represents “three times” a number? - For part b), compare only the entries in the Children column.

Solution

1. Find Friday’s total attendance: \(124 + 88 = 212\). 2. Find Sunday’s total attendance: \(582 + 295 = 877\). 3. Find the increase: \(877 - 212 = 665\) people. 4. Find three times Friday’s number of children: \(124 \times 3 = 372\). 5. Since \(456 > 372\), Saturday’s number of children was more than three times Friday’s number.

Answer

a) \(665\) people b) Yes. Since \(124 \times 3 = 372\) and \(456 > 372\), the number was more than three times as great.
5206424
A hiking group starts at a mountain lodge \(4650\,\text{ft}\) above sea level. The entire trail is \(7\) miles long. The hikers climb \(2080\,\text{ft}\) to a viewpoint, then descend \(715\,\text{ft}\) to a lake. What is the lake’s elevation above sea level?

Hints

- Identify how far the hikers move upward. - A descent decreases the current elevation. - Decide whether every number in the problem is needed.

Solution

1. Find the viewpoint’s elevation: \(4650\,\text{ft} + 2080\,\text{ft} = 6730\,\text{ft}\). 2. Subtract the descent: \(6730\,\text{ft} - 715\,\text{ft} = 6015\,\text{ft}\). 3. The total trail length is not needed to determine the elevation.

Answer

The lake is \(6015\,\text{ft}\) above sea level.
5206434
In spring, the water in a reservoir is \(110\,\text{ft}\) below the top of the dam. After heavy rain, the water level rises \(40\,\text{ft}\). During a later dry period, it falls \(23\,\text{ft}\). How far below the top of the dam is the water after the dry period? Explain with calculations.

Hints

- When the water rises, does its distance below the top increase or decrease? - Treat the top of the dam as a fixed reference point. - Decide which operation matches each change in the water level.

Solution

1. When the water rises, its distance below the top becomes smaller: \(110\,\text{ft} - 40\,\text{ft} = 70\,\text{ft}\). 2. When the water falls, its distance below the top becomes larger: \(70\,\text{ft} + 23\,\text{ft} = 93\,\text{ft}\).

Answer

The water is \(93\,\text{ft}\) below the top of the dam.
5206444
A remotely operated underwater robot with a mass of \(150\,\text{kg}\) is exploring the ocean floor. It begins \(412\,\text{m}\) below the surface, descends another \(289\,\text{m}\) to collect a sample, and later rises \(154\,\text{m}\). What is the robot’s current depth below the surface?

Hints

- A descent increases the depth below the surface. - A rise decreases the depth below the surface. - Decide which information is actually needed.

Solution

1. After descending, the robot’s depth is \(412\,\text{m} + 289\,\text{m} = 701\,\text{m}\). 2. After rising, its depth is \(701\,\text{m} - 154\,\text{m} = 547\,\text{m}\). 3. The robot’s mass is not needed to calculate its depth.

Answer

The robot is \(547\,\text{m}\) below the surface.
5206624
Ethan is building a wire-frame rectangular prism. He uses small clay balls for the vertices and straws for the edges. The prism is \(12\,\text{cm}\) long, \(8\,\text{cm}\) wide, and \(5\,\text{cm}\) high. a) How many clay balls does he need? b) How many straws of each length does he need? c) What total length of straw does he need?

Hints

- Each clay ball represents one vertex. - A rectangular prism has four edges matching each dimension. - Add the lengths of all \(12\) straws.

Solution

1. A rectangular prism has \(8\) vertices, so Ethan needs \(8\) clay balls. 2. It has four edges of each dimension, so he needs four \(12\,\text{cm}\) straws, four \(8\,\text{cm}\) straws, and four \(5\,\text{cm}\) straws. 3. The total length is \(4 \times 12\,\text{cm}+4 \times 8\,\text{cm}+4 \times 5\,\text{cm}=48\,\text{cm}+32\,\text{cm}+20\,\text{cm}=100\,\text{cm}\).

Answer

a) \(8\) clay balls b) Four \(12\,\text{cm}\) straws, four \(8\,\text{cm}\) straws, and four \(5\,\text{cm}\) straws c) \(100\,\text{cm}\)
5207354
A small natural history museum offers guided tours. Admission costs \(\$9\) for each adult and \(\$5\) for each child. On Saturday, the morning tour had \(24\) adults and \(36\) children. The afternoon tour had \(38\) adults and \(22\) children. What was the museum's total admission revenue that day?

Hints

- First find the total numbers of adults and children. - How do you find the admission revenue for each group? - Add the adult and child revenue.

Solution

1. The total number of adults is \(24+38=62\). 2. Adult admission revenue is \(62 \times \$9=\$558\). 3. The total number of children is \(36+22=58\). 4. Child admission revenue is \(58 \times \$5=\$290\). 5. The total revenue is \(\$558+\$290=\$848\).

Answer

The museum's total admission revenue was \(\$848\).
5208564
Luke and Sarah are playing a number riddle game. Luke says, “My number is \(240\) less than \(710\).” Sarah says, “When I subtract \(130\) from my number, I get Luke's number.” What numbers did Luke and Sarah choose?

Hints

- Find Luke's number first. - Translate “\(240\) less than \(710\)” into subtraction. - Then use Luke's number as the result in Sarah's statement. - Undo subtracting \(130\) to recover Sarah's number.

Solution

1. Luke's number is \(710 - 240 = 470\). 2. Sarah's number minus \(130\) equals \(470\), so add \(130\): \(470 + 130 = 600\).

Answer

Luke chose \(470\), and Sarah chose \(600\).
5208954
Three numbers have a sum of \(1000\). The first number is \(350\). The second number is \(100\) less than the first. What is the third number?

Hints

- Find the second number first. - Add the first two numbers. - Find how much is still needed to reach \(1000\).

Solution

1. Find the second number: \(350 - 100 = 250\). 2. Add the first two numbers: \(350 + 250 = 600\). 3. Subtract from the total: \(1000 - 600 = 400\).

Answer

The third number is \(400\).
5208974
A public library opened in \(1974\). a) How old was the library in \(2012\)? b) The library was first expanded in \(1999\). How many years after opening was that? c) The library held a celebration in \(2024\). Which anniversary did it celebrate?

Hints

- Use the opening year as the starting point for every part. - Subtract the opening year from each later year. - Interpret each difference in the context of the question.

Solution

1. Find the library’s age in \(2012\): \(2012 - 1974 = 38\) years. 2. Find the time to the first expansion: \(1999 - 1974 = 25\) years. 3. Find the anniversary in \(2024\): \(2024 - 1974 = 50\) years.

Answer

a) The library was \(38\) years old. b) The expansion occurred \(25\) years after opening. c) The library celebrated its \(50\)th anniversary.
5209474
A small truck has a maximum gross weight of \(7700\,\text{lb}\). The empty truck weighs \(5460\,\text{lb}\), and the driver weighs \(190\,\text{lb}\). The fuel tank contains \(20\) gallons of diesel, and each gallon weighs \(7\,\text{lb}\). How many pounds of goods can still be loaded without exceeding the limit?

Hints

- Find the total weight of the fuel. - Add the truck, driver, and fuel weights. - Subtract the current weight from the maximum allowed weight.

Solution

1. Find the fuel weight: \(20 \times 7\,\text{lb} = 140\,\text{lb}\). 2. Find the current total weight: \(5460 + 190 + 140 = 5790\), so the truck, driver, and fuel weigh \(5790\,\text{lb}\). 3. Subtract from the maximum: \(7700\,\text{lb} - 5790\,\text{lb} = 1910\,\text{lb}\).

Answer

The truck can carry \(1910\,\text{lb}\) of goods.
5209614
Three friends have a combined weight of \(286\,\text{lb}\). Tim weighs \(16\,\text{lb}\) more than Max. Leo weighs the same as Max. How much does each friend weigh?

Hints

- Think about what the total would represent if all three friends weighed the same. - Set aside Tim’s extra weight before dividing. - A bar model with three equal parts and one extra piece may help. - Check that the three weights add to the given total.

Solution

1. Set aside Tim’s extra \(16\,\text{lb}\): \(286\,\text{lb} - 16\,\text{lb} = 270\,\text{lb}\). 2. Divide the remaining weight equally among the three friends: \(270\,\text{lb} \div 3 = 90\,\text{lb}\). 3. Max and Leo each weigh \(90\,\text{lb}\). 4. Tim weighs \(90\,\text{lb} + 16\,\text{lb} = 106\,\text{lb}\). 5. Check: \(90 + 90 + 106 = 286\).

Answer

Tim weighs \(106\,\text{lb}\), Max weighs \(90\,\text{lb}\), and Leo weighs \(90\,\text{lb}\).
5209864
An orchard packs apples and pears into bags. a) There are \(735\) apples, with \(7\) apples in each bag. b) There are \(852\) pears, with \(6\) pears in each bag. Find the number of bags for each fruit. Which fruit uses more bags?

Hints

- Divide the number of apples by \(7\). - Divide the number of pears by \(6\). - Break each dividend into compatible numbers, then compare the quotients.

Solution

1. For the apples, divide \(735\) by \(7\). Since \(735 = 700 + 35\), \(700 \div 7 = 100\) and \(35 \div 7 = 5\). Therefore, \(100 + 5 = 105\) bags. 2. For the pears, divide \(852\) by \(6\). Since \(852 = 600 + 240 + 12\), \(600 \div 6 = 100\), \(240 \div 6 = 40\), and \(12 \div 6 = 2\). Therefore, \(100 + 40 + 2 = 142\) bags. 3. Compare: \(142 > 105\), so the pears use more bags.

Answer

a) The apples use \(105\) bags. b) The pears use \(142\) bags. The pears use more bags.
5211044
At a juice factory, one machine fills \(45\) bottles per minute, and another fills \(55\) bottles per minute. How many minutes must both machines work together to fill \(400\) bottles?

Hints

- How many bottles do both machines fill together in one minute? - Once you know the combined rate, how many groups of that amount make \(400\)?

Solution

1. Find the combined number of bottles filled per minute: \(45 + 55 = 100\). 2. Divide the target number of bottles by the combined rate: \(400 \div 100 = 4\) minutes.

Answer

The machines must work together for \(4\) minutes.
5211314
An elementary school celebrated its \(125\)th anniversary in \(2024\). a) In what year was the school founded? b) A nearby school opened in \(1875\). How many years older is the nearby school?

Hints

- To find the founding year, work backward from \(2024\). - First find when the school was founded. - Then compare that year with \(1875\).

Solution

1. Subtract the school's age from the anniversary year: \(2024 - 125 = 1899\). The first school was founded in \(1899\). 2. Compare the founding years: \(1899 - 1875 = 24\). 3. The nearby school is \(24\) years older.

Answer

a) The school was founded in \(1899\). b) The nearby school is \(24\) years older.
5211334
A cider mill pours \(240\,\text{L}\) of apple juice equally into \(8\) barrels. a) How many liters of juice are in \(5\) of the barrels? b) An employee says, “If we fill \(10\) barrels this way, they will hold \(300\,\text{L}\) altogether.” Is the employee correct? Show your calculation.

Hints

- How many liters of juice are in one barrel? - Use place value and a related basic division fact. - For part b), find the amount in \(10\) barrels and compare it with \(300\,\text{L}\).

Solution

1. Find the amount of juice in each barrel: \(240\,\text{L} \div 8 = 30\,\text{L}\). 2. Find the amount in five barrels: \(5 \times 30\,\text{L} = 150\,\text{L}\). 3. Check the employee's statement: \(10 \times 30\,\text{L} = 300\,\text{L}\). 4. The result agrees with the statement, so the employee is correct.

Answer

a) The \(5\) barrels hold \(150\,\text{L}\) of juice. b) Yes. The employee is correct because \(10 \times 30\,\text{L} = 300\,\text{L}\).
5211344
Lucas has \(\$5.00\). He buys \(4\) pencils for \(45\) cents each and \(5\) erasers for \(60\) cents each. How many cents does he have left?

Hints

- Find the total cost of each type of item. - Express \(\$5.00\) in cents. - Subtract the total cost from the starting amount.

Solution

1. Find the cost of the pencils: \(4 \times 45 = 180\) cents. 2. Find the cost of the erasers: \(5 \times 60 = 300\) cents. 3. Find the total cost: \(180 + 300 = 480\) cents. 4. Convert the starting amount: \(\$5.00 = 500\,\text{cents}\). 5. Find the amount left: \(500 - 480 = 20\) cents.

Answer

Lucas has \(20\) cents left.
5211354
A class raises \(\$100\) for new books. The students choose \(5\) nonfiction books costing \(\$12\) each and \(4\) fiction books costing \(\$9\) each. Is there enough money for all the books? How much money will remain, or how much more is needed?

Hints

- Find the cost of each type of book. - Add the two costs. - Compare the total with \(\$100\) and find the difference.

Solution

1. Find the cost of the nonfiction books: \(5 \times \$12 = \$60\). 2. Find the cost of the fiction books: \(4 \times \$9 = \$36\). 3. Find the total cost: \(\$60 + \$36 = \$96\). 4. Since \(\$96 < \$100\), there is enough money. 5. Find the amount left: \(\$100 - \$96 = \$4\).

Answer

Yes. The class has enough money and will have \(\$4\) left.
5211394
A snack stand starts with \(95\) cans of drinks. It sells \(12\) cans on each of \(5\) weekdays and another \(24\) cans over the weekend. How many cans remain?

Hints

- First find the number sold during the five weekdays. - Add the weekend sales. - Subtract the total sold from the starting amount.

Solution

1. Find the weekday sales: \(5 \times 12 = 60\). 2. Find the total sales: \(60 + 24 = 84\). 3. Find the amount remaining: \(95 - 84 = 11\).

Answer

There are \(11\) cans left.
5211404
A bakery has \(500\,\text{kg}\) of flour. During one week, it uses \(35\,\text{kg}\) on each of \(6\) days. During the next week, it uses \(110\,\text{kg}\) on each of \(2\) days. Is there enough flour left to bake an order requiring \(45\,\text{kg}\)? Justify your answer.

Hints

- Find the flour used in each week separately. - Subtract the total used from the starting amount. - Compare the amount left with \(45\,\text{kg}\).

Solution

1. Find the first-week use: \(6 \times 35\,\text{kg} = 210\,\text{kg}\). 2. Find the second-week use: \(2 \times 110\,\text{kg} = 220\,\text{kg}\). 3. Find the total used: \(210\,\text{kg} + 220\,\text{kg} = 430\,\text{kg}\). 4. Find the flour left: \(500\,\text{kg} - 430\,\text{kg} = 70\,\text{kg}\). 5. Since \(70\,\text{kg} > 45\,\text{kg}\), there is enough flour. After the order, \(70 - 45 = 25\,\text{kg}\) would remain.

Answer

Yes. The bakery has \(70\,\text{kg}\) left, which is enough for the \(45\,\text{kg}\) order.
5211494
Ms. Miller buys \(6\) packages of colored pencils for \(\$5\) each and \(4\) drawing pads for her class. The total cost is \(\$54\). How much does one drawing pad cost?

Hints

- First find how much Ms. Miller spends on colored pencils. - Subtract that amount from the total to find the cost of all the drawing pads. - How can you find the price of one pad from the price of four pads?

Solution

1. Find the total cost of the colored pencils: \(6 \times \$5 = \$30\). 2. Find the total cost of the four drawing pads: \(\$54 - \$30 = \$24\). 3. Divide by the number of drawing pads: \(\$24 \div 4 = \$6\).

Answer

One drawing pad costs \(\$6\).
5211584
Last year, a school celebration used \(35\) blue balloons and \(18\) red balloons. This year, the school prepares three times as many blue balloons as last year and \(124\) more red balloons than last year. How many blue and red balloons does the school prepare altogether this year?

Hints

- Match each comparison with the correct balloon color. - Find the number of each color separately. - Add the two results.

Solution

1. Find the number of blue balloons: \(35 \times 3 = 105\). 2. Find the number of red balloons: \(18 + 124 = 142\). 3. Add both colors: \(105 + 142 = 247\).

Answer

The school prepares \(247\) balloons altogether.
5211624
Two crates contain \(42\,\text{kg}\) of apples altogether. After \(6\,\text{kg}\) of apples are removed from the first crate, the two crates have the same mass of apples. How many kilograms of apples were originally in each crate?

Hints

- Find the total mass remaining after \(6\,\text{kg}\) are removed. - Split that remaining mass equally between the two crates. - Add the removed mass back to the first crate.

Solution

1. Find the total mass after the apples are removed: \(42\,\text{kg} - 6\,\text{kg} = 36\,\text{kg}\). 2. Divide the remaining mass equally between the two crates: \(36\,\text{kg} \div 2 = 18\,\text{kg}\). 3. The second crate originally had \(18\,\text{kg}\). 4. Add back the removed apples to find the first crate's original amount: \(18\,\text{kg} + 6\,\text{kg} = 24\,\text{kg}\).

Answer

The first crate originally had \(24\,\text{kg}\), and the second crate originally had \(18\,\text{kg}\).
5211644
A school library has two new bookcases. The first has \(6\) shelves with \(8\) nonfiction books on each shelf. The second has \(4\) shelves with \(15\) mystery books on each shelf. How many books are on the two bookcases altogether? Which bookcase has more books?

Hints

- Find the number of books on each bookcase. - Add the two totals. - Compare the two bookcase totals.

Solution

1. Find the number on the first bookcase: \(6 \times 8 = 48\). 2. Find the number on the second bookcase: \(4 \times 15 = 60\). 3. Find the combined total: \(48 + 60 = 108\). 4. Since \(60 > 48\), the second bookcase has more books.

Answer

There are \(108\) books altogether, and the second bookcase has more books.
5211654
A school sets up \(8\) rows with \(9\) chairs in each row for a performance. Then it adds \(4\) rows with \(12\) chairs in each row. Are there enough seats for \(130\) guests? Justify your answer.

Hints

- Find the number of chairs in each group of rows. - Add the two amounts. - Compare the total with \(130\).

Solution

1. Find the chairs in the first group: \(8 \times 9 = 72\). 2. Find the chairs in the added rows: \(4 \times 12 = 48\). 3. Find the total: \(72 + 48 = 120\). 4. Since \(120 < 130\), there are not enough seats. The school needs \(130 - 120 = 10\) more chairs.

Answer

No. There are \(120\) chairs, so \(10\) more chairs are needed.
5211664
An orchard has \(13\) apple trees and \(11\) cherry trees. The fruit has already been harvested from one-third of all the trees. From how many trees has the fruit been harvested? From how many trees does the fruit still need to be harvested?

Hints

- First find the total number of trees. - What operation finds one-third of a group? - After finding the number already harvested, how can you find the rest?

Solution

1. Find the total number of trees: \(13 + 11 = 24\). 2. Find one-third of the trees: \(\frac{1}{3} \times 24 = 8\). 3. Find the number still needing harvest: \(24 - 8 = 16\).

Answer

The fruit has been harvested from \(8\) trees, and \(16\) trees still need to be harvested.
5211914
A class sells muffins during lunch for \(4\) days. It earns \(\$144\) in total and spends \(\$52\) on ingredients. How much greater is the average daily revenue than the average daily ingredient cost?

Hints

- Find the average revenue for one day. - Divide the ingredient cost equally among the \(4\) days. - Subtract the two daily averages.

Solution

1. Find the average daily revenue: \(\$144 \div 4 = \$36\). 2. Find the average daily ingredient cost: \(\$52 \div 4 = \$13\). 3. Find the difference: \(\$36 - \$13 = \$23\).

Answer

The average daily revenue is \(\$23\) greater than the average daily ingredient cost.
5211944
At a craft store, \(5\) paintbrushes and \(3\) tubes of glitter paint cost \(\$24\) altogether. Each paintbrush costs \(\$3\). Lucas has \(\$10\) and says, “I can buy \(4\) tubes of this glitter paint.” Is Lucas correct? Support your answer with a calculation.

Hints

- Find the total cost of the paintbrushes. - Use the full bill to find the cost of the three paint tubes. - Find the price of one tube, then compare the cost of four tubes with \(\$10\).

Solution

1. The five paintbrushes cost \(5 \times \$3 = \$15\). 2. The three paint tubes cost \(\$24 - \$15 = \$9\). 3. One tube costs \(\$9 \div 3 = \$3\). 4. Four tubes cost \(4 \times \$3 = \$12\). 5. Since \(\$12 > \$10\), Lucas is not correct.

Answer

No. Four tubes cost \(\$12\), but Lucas has only \(\$10\).
5211954
Six bakers work in a small bakery. Together, they bake \(150\) loaves of bread in \(5\) hours. On average, how many loaves does one baker make in one hour?

Hints

- First find how many loaves all the bakers make in one hour. - Once you know the team's hourly output, divide it equally among the bakers. - Which operation divides a total into equal shares?

Solution

1. Find how many loaves the whole team makes in one hour: \(150 \div 5 = 30\). 2. Divide that hourly total among six bakers: \(30 \div 6 = 5\).

Answer

On average, one baker makes \(5\) loaves in one hour.
5212064
A field trip needs \(336\) small juice bottles. The bottles come in cases of \(8\). A teacher says, “Exactly \(40\) cases will be enough.” Is the teacher correct? Justify your answer with a calculation.

Hints

- First calculate the exact number of cases needed. - Break \(336\) into multiples of \(8\). - Compare the quotient with \(40\).

Solution

1. Find the number of cases needed: \(336 \div 8\). 2. Break \(336\) into compatible numbers: \(336 = 320 + 16\). 3. Divide each part: \(320 \div 8 = 40\) and \(16 \div 8 = 2\). 4. Add: \(40 + 2 = 42\). Since \(42 \ne 40\), the teacher is not correct.

Answer

No. The trip needs \(42\) cases, so \(40\) cases are not enough.
5212174
A school library receives \(400\) new books. One-half are adventure stories, one-eighth are nonfiction books, and the rest are graphic novels. a) How many graphic novels are there? b) A student says, “There are three times as many graphic novels as nonfiction books.” Is the student correct? Support your answer with a calculation.

Hints

- What operation represents one-half? - How can you find one-eighth of \(400\)? - Find the other categories before determining the number of graphic novels. - To check the claim, multiply the nonfiction count by \(3\) and compare.

Solution

1. Find the adventure stories: \(\frac{1}{2} \times 400 = 200\). 2. Find the nonfiction books: \(\frac{1}{8} \times 400 = 50\). 3. Find the graphic novels: \(400 - 200 - 50 = 150\). 4. Check the claim: \(3 \times 50 = 150\). The student is correct.

Answer

a) There are \(150\) graphic novels. b) Yes. There are \(50\) nonfiction books, and \(3 \times 50 = 150\).
5212234
A garden rainwater tank holds \(3000\,\text{gal}\). It already contains \(1250\,\text{gal}\). During a storm, water flows into the tank at \(35\,\text{gal}\) per minute for \(30\) minutes. How many more gallons are needed to fill the tank?

Hints

- Find how much water enters during the entire storm. - Add that amount to the water already in the tank. - Subtract the new amount from the tank’s capacity.

Solution

1. Find the amount added during the storm: \(30 \times 35 = 1050\,\text{gal}\). 2. Find the new amount in the tank: \(1250 + 1050 = 2300\,\text{gal}\). 3. Find the amount still needed: \(3000 - 2300 = 700\,\text{gal}\).

Answer

The tank needs \(700\,\text{gal}\) more to be full.
5212294
A school garden has \(240\) tulip bulbs. a) How many bulbs go in each garden bed if the bulbs are divided equally among \(4\) beds? b) How many bulbs go in each bed if the bulbs are divided equally among \(8\) beds? What relationship do you notice between the two results?

Hints

- Divide the total number of bulbs by each number of garden beds. - Compare the two quotients. - Explain how changing the number of equal groups affects the amount in each group.

Solution

1. For part a, divide: \(240 \div 4 = 60\). 2. For part b, divide: \(240 \div 8 = 30\). 3. The number of beds doubles from \(4\) to \(8\), while the number of bulbs in each bed is halved from \(60\) to \(30\).

Answer

a) Each bed gets \(60\) bulbs. b) Each bed gets \(30\) bulbs. Doubling the number of beds halves the number of bulbs per bed.
5212394
A baker made \(420\) rolls and packs \(6\) rolls in each bag. How many bags can the baker fill? If the baker packed \(7\) rolls in each bag instead, would more or fewer bags be needed? Explain.

Hints

- Find \(420 \div 6\). - Compare the two bag sizes. - Does a larger number in each bag produce more or fewer bags for the same total?

Solution

1. Divide the total number of rolls by \(6\): \(420 \div 6 = 70\). 2. Increasing the number of rolls in each bag from \(6\) to \(7\) makes each group larger, so fewer bags are needed. 3. A calculation confirms this: \(420 \div 7 = 60\).

Answer

The baker can fill \(70\) bags of \(6\). Bags of \(7\) would require fewer bags—\(60\) bags.
5212404
Two schools order juice for field days. School A receives \(560\) bottles packed \(8\) bottles per case. School B receives \(540\) bottles packed \(6\) bottles per case. Which school receives more cases? Justify your answer with calculations.

Hints

- Calculate the number of cases for each school separately. - Divide each number of bottles by the corresponding case size. - Compare the two quotients.

Solution

1. Find the number of cases for School A: \(560 \div 8 = 70\). 2. Find the number of cases for School B: \(540 \div 6 = 90\). 3. Compare the results: \(90 > 70\), so School B receives more cases.

Answer

School B receives more cases. School A receives \(70\) cases, and School B receives \(90\) cases.
5212604
Ms. Miller saved \(\$450\) and uses all of it to buy \(3\) bicycles that cost the same amount for her grandchildren. Mr. Schmidt buys \(2\) bicycles for his grandchildren and pays \(\$320\) altogether. Who spends more per bicycle?

Hints

- First find the cost of one bicycle for Ms. Miller. - Then find the cost of one bicycle for Mr. Schmidt. - Compare the two unit prices. - Could you break each amount into hundreds and tens before dividing?

Solution

1. Find Ms. Miller's cost per bicycle: \(\$450 \div 3 = \$150\). 2. Find Mr. Schmidt's cost per bicycle: \(\$320 \div 2 = \$160\). 3. Compare the unit prices: \(\$160 > \$150\).

Answer

Mr. Schmidt spends more per bicycle. He spends \(\$160\) per bicycle, compared with Ms. Miller's \(\$150\) per bicycle.
5212614
A wooden fence has \(16\) vertical slats. The attachment points of neighboring slats are \(6\,\text{in}\) apart. The horizontal rail extends \(5\,\text{in}\) beyond the first attachment point and \(5\,\text{in}\) beyond the last. How long is the rail altogether?

Hints

- Find the number of spaces between \(16\) attachment points. - Find the distance from the first point to the last point. - Include the extension at both ends.

Solution

1. Sixteen attachment points create \(16 - 1 = 15\) spaces. 2. The distance from the first attachment point to the last is \(15 \times 6\,\text{in} = 90\,\text{in}\). 3. The two end extensions total \(2 \times 5\,\text{in} = 10\,\text{in}\). 4. The rail length is \(90\,\text{in} + 10\,\text{in} = 100\,\text{in}\).

Answer

The rail is \(100\,\text{in}\) long.
5212654
A bike trail is \(24\,\text{mi}\) long. A class plans to ride the trail over several days. On Monday, the class rides one-eighth of the entire trail. On Tuesday, the class rides one-fourth of the entire trail. How many miles does the class ride during the two days altogether? How many miles remain?

Hints

- Find the distance ridden on each day separately. - How can you find the total distance ridden during both days? - What operation finds the distance still remaining?

Solution

1. Find Monday's distance: \(\frac{1}{8} \times 24\,\text{mi} = 3\,\text{mi}\). 2. Find Tuesday's distance: \(\frac{1}{4} \times 24\,\text{mi} = 6\,\text{mi}\). 3. Add the two distances: \(3\,\text{mi} + 6\,\text{mi} = 9\,\text{mi}\). 4. Find the remaining distance: \(24\,\text{mi} - 9\,\text{mi} = 15\,\text{mi}\).

Answer

The class rides \(9\,\text{mi}\) altogether, and \(15\,\text{mi}\) remain.
5212824
A class sells \(450\) cups of juice for \(\$2\) each at a school fair. How much money does the class collect? If the juice cost \(\$300\), how much profit does the class make?

Hints

- First find the money collected from all the cups. - Multiply the number sold by the price per cup. - Profit is revenue minus cost.

Solution

1. Find the total revenue: \(450 \times \$2 = \$900\). 2. Subtract the cost: \(\$900 - \$300 = \$600\).

Answer

The class collects \(\$900\) and makes a profit of \(\$600\).
5212874
A \(945\,\text{ft}\) hiking trail is being repaired. On the first day, workers repair one-third of the trail. On the second day, they repair \(42\,\text{ft}\) less than on the first day. How many feet of the trail still need to be repaired?

Hints

- First find how many feet are repaired on the first day. - How does the second-day amount compare with the first-day amount? - How much is repaired during both days altogether? - What operation finds the part still needing repair?

Solution

1. Find the distance repaired on the first day: \(\frac{1}{3} \times 945\,\text{ft} = 315\,\text{ft}\). 2. Find the distance repaired on the second day: \(315\,\text{ft} - 42\,\text{ft} = 273\,\text{ft}\). 3. Find the total distance repaired: \(315\,\text{ft} + 273\,\text{ft} = 588\,\text{ft}\). 4. Subtract from the full trail: \(945\,\text{ft} - 588\,\text{ft} = 357\,\text{ft}\).

Answer

\(357\,\text{ft}\) of the trail still need to be repaired.
5212944
A bakery makes \(1250\) rolls in one morning. Of these, \(450\) are whole-wheat rolls. The rest are wheat rolls and rye rolls. There are \(120\) more wheat rolls than rye rolls. How many wheat rolls and how many rye rolls were made?

Hints

- Restate what is known and what must be found. - First remove the whole-wheat rolls from the total. - Imagine the two remaining types had equal amounts before accounting for the difference of \(120\).

Solution

1. The wheat and rye rolls total \(1250-450=800\). 2. Remove the difference: \(800-120=680\). 3. The smaller amount, rye rolls, is \(680\div 2=340\). 4. The number of wheat rolls is \(340+120=460\).

Answer

The bakery made \(460\) wheat rolls and \(340\) rye rolls.
5213254
A movie theater sells bags of popcorn for \(\$4\) each. On Saturday, the theater collects \(\$320\) from popcorn sales. a) How many bags of popcorn are sold on Saturday? b) On Sunday, the theater sells \(15\) more bags than on Saturday. How much money does the theater collect from popcorn sales on Sunday?

Hints

- Use the Saturday revenue and price per bag to find the number sold. - Add \(15\) to find the number sold on Sunday. - Multiply the Sunday number of bags by the price per bag.

Solution

1. Saturday sales: \(\$320 \div \$4 = 80\) bags. 2. Sunday sales: \(80 + 15 = 95\) bags. 3. Sunday revenue: \(95 \times \$4 = \$380\).

Answer

a) \(80\) bags b) \(\$380\)
5213264
At a garden center, one small flowerpot costs \(\$3\). A box of \(4\) flowerpots costs \(\$10\). a) How much would \(4\) individual flowerpots cost? b) How much money do you save by buying the box instead?

Hints

- Find the cost of four pots bought separately. - Compare that amount with the box price. - Subtract to find the savings.

Solution

1. Four individual pots cost \(4 \times 3 = 12\) dollars. 2. Compare \(\$12\) with the box price of \(\$10\). 3. The savings are \(12 - 10 = 2\) dollars.

Answer

a) \(\$12\) b) \(\$2\)
5213534
A nursery plants \(15\) window boxes with \(4\) red geraniums in each box and \(12\) window boxes with \(5\) blue petunias in each box. Compare the numbers of red and blue flowers. What do you notice? How many flowers are planted altogether?

Hints

- Find the number of each color separately. - Compare the two products. - Add the two amounts.

Solution

1. Find the number of red flowers: \(15 \times 4 = 60\). 2. Find the number of blue flowers: \(12 \times 5 = 60\). 3. Compare: \(60 = 60\), so the numbers are equal. 4. Find the total: \(60 + 60 = 120\).

Answer

There are \(60\) red flowers and \(60\) blue flowers, so the amounts are equal. There are \(120\) flowers altogether.
5213794
Lucas travels to visit his grandparents. He rides a bicycle for \(2\) hours at \(9\) miles per hour. Then he rides a bus for \(3\) hours at \(30\) miles per hour. How far does Lucas travel altogether?

Hints

- Find the distance traveled by each type of transportation. - Use distance per hour and number of hours. - Add the two distances.

Solution

1. Find the bicycle distance: \(2 \times 9 = 18\) miles. 2. Find the bus distance: \(3 \times 30 = 90\) miles. 3. Add the distances: \(18 + 90 = 108\) miles.

Answer

The entire trip is \(108\) miles long.
5215244
A community sports club recorded the numbers of child and adult members over four years. <table> <thead> <tr><th>Year</th><th>2021</th><th>2022</th><th>2023</th><th>2024</th></tr> </thead> <tbody> <tr><td>Children</td><td>\(156\)</td><td>\(188\)</td><td>\(245\)</td><td>\(290\)</td></tr> <tr><td>Adults</td><td>\(210\)</td><td>\(235\)</td><td>\(255\)</td><td>\(295\)</td></tr> </tbody> </table> a) In which year was the increase in child members from the previous year greatest? b) Find the increases in adult members for 2023 and 2024, each compared with the previous year. What do you notice when you compare the two increases? c) How many total members were added from 2021 to 2024?

Hints

- Read each row separately when finding yearly changes for children and adults. - Part b) asks for both calculations and an observation about the results. - For part c), compare the total membership at the beginning and the end. - It may help to find the combined total for each of those two years first.

Solution

1. Find each yearly increase in child members: \(188 - 156 = 32\) for 2022, \(245 - 188 = 57\) for 2023, and \(290 - 245 = 45\) for 2024. The greatest increase was \(57\), in 2023. 2. Find the adult increases: \(255 - 235 = 20\) for 2023 and \(295 - 255 = 40\) for 2024. The 2024 increase is twice the 2023 increase. 3. Find the total membership in 2021: \(156 + 210 = 366\). 4. Find the total membership in 2024: \(290 + 295 = 585\). 5. Find the overall increase: \(585 - 366 = 219\) members.

Answer

a) 2023 b) The increases were \(20\) members for 2023 and \(40\) members for 2024. The 2024 increase was twice the 2023 increase. c) \(219\) members
5216384
Leo wants to make a cube net for a cube with edge length \(4\,\text{cm}\). The net has four squares in one row, with one square attached above and one below the second square. The net's edges must be parallel to the paper's edges. He can use either sheet: - Sheet A: \(22\,\text{cm} \times 12\,\text{cm}\) - Sheet B: \(15\,\text{cm} \times 15\,\text{cm}\) Which sheet can hold the net? Justify your answer by finding the net's minimum length and width.

Hints

- Count the squares in the longest row. - Count the squares at the widest part of the net. - Compare both required dimensions with each sheet.

Solution

1. Each square has side length \(4\,\text{cm}\). 2. The row of four squares is \(4 \times 4\,\text{cm}=16\,\text{cm}\) long. 3. At its widest point, the net is three squares tall, so its width is \(3 \times 4\,\text{cm}=12\,\text{cm}\). 4. The net requires a \(16\,\text{cm} \times 12\,\text{cm}\) rectangle. 5. Sheet A is large enough because \(16\le 22\) and \(12\le 12\). Sheet B is not large enough because \(16>15\).

Answer

The net fits only on Sheet A. It requires minimum dimensions of \(16\,\text{cm} \times 12\,\text{cm}\).
5317064
The bar graph shows how many books four Grade 4 classes collected for a school book sale. a) How many books did Classes 4A, 4B, 4C, and 4D collect altogether? b) The class that collected the most books will pack them into boxes that each hold exactly \(10\) books. How many boxes will be completely filled, and how many books will be left over?
Figure for problem 531706

Hints

- Read each bar height using the vertical axis. - The scale includes intervals of \(5\), including values between multiples of \(10\). - Add all four values for part a). - Identify the tallest bar before part b). - Divide that value by \(10\) and interpret the remainder.

Solution

1. Read the values from the graph: Class 4A, \(40\); Class 4B, \(55\); Class 4C, \(30\); Class 4D, \(45\). 2. Add all four values: \(40 + 55 + 30 + 45 = 170\) books. 3. Class 4B collected the most books, with \(55\). 4. Divide by the box capacity: \(55 \div 10 = 5\) remainder \(5\). Therefore, \(5\) boxes are full and \(5\) books remain.

Answer

a) \(170\) books b) \(5\) full boxes and \(5\) books left over
5317124
Mia and Leo earn points in a game. Mia earns \(140\) points in the first round and \(260\) points in the second round. Leo starts with \(320\) points but loses \(150\) points in the second round. The calculation tree shows how to compare Mia’s total with Leo’s remaining points. a) Find the values for the empty boxes in the calculation tree. b) What is the final difference in the bottom box?
Figure for problem 531712

Hints

- On the left side of the tree, which two numbers are added? - On the right side, what number is subtracted from \(320\)? - After finding the two middle results, subtract to find the bottom result.

Solution

1. Find Mia’s total: \(140 + 260 = 400\). 2. Find Leo’s remaining points: \(320 - 150 = 170\). 3. Find the difference: \(400 - 170 = 230\).

Answer

a) The middle boxes are \(400\) on the left and \(170\) on the right. b) The final difference is \(230\) points.
5317254
Grade 4 students at an elementary school were surveyed about their favorite free-time activities. The bar graph shows the results. a) How many students were surveyed altogether? b) Which activity is most popular, and how many students chose it? c) Is this statement true: “More than one-fourth of the students prefer either making music or reading”? Justify your answer with a calculation.
Figure for problem 531725

Hints

- Read each bar height carefully. - Add all categories to find the total. - One-fourth of a total means dividing the total by \(4\). - Compare the Reading-and-Music total with that one-fourth value.

Solution

1. Read the values from the graph: Reading, \(14\); Sports, \(24\); Music, \(8\); Gaming, \(18\); Friends, \(16\). 2. Add all categories: \(14 + 24 + 8 + 18 + 16 = 80\) students. 3. The greatest value is \(24\), for Sports. 4. One-fourth of the total is \(80 \div 4 = 20\). Reading or Music accounts for \(14 + 8 = 22\) students. Since \(22 > 20\), the statement is true.

Answer

a) \(80\) students b) Sports, \(24\) students c) Yes. One-fourth of \(80\) is \(20\), and \(14 + 8 = 22\), so more than one-fourth prefer reading or music.
5350414
Four classes are collecting recyclable paper for a school challenge. The bar graph shows how many pounds each class has collected so far. a) Record the amount collected by each class. b) Before the next weigh-in, the amounts change: 1. Class C transfers \(10\,\text{lb}\) of paper to Class A. 2. Classes B and D combine their paper and divide it equally between the two classes. How many pounds of paper does each class have after both changes? c) Which class has the greatest amount at the end?
Figure for problem 535041

Hints

- Read each starting amount from the vertical scale. - Complete the changes in the order given and record each new amount. - When two amounts are combined and shared equally, add first and then divide by \(2\). - Check that the total amount of paper remains the same.

Solution

1. Read the starting amounts: Class A, \(35\,\text{lb}\); Class B, \(50\,\text{lb}\); Class C, \(45\,\text{lb}\); Class D, \(20\,\text{lb}\). 2. After Class C transfers \(10\,\text{lb}\), Class A has \(35 + 10 = 45\,\text{lb}\), and Class C has \(45 - 10 = 35\,\text{lb}\). 3. Classes B and D have \(50 + 20 = 70\,\text{lb}\) together. Dividing equally gives \(70 \div 2 = 35\,\text{lb}\) for each class. 4. The final amounts are Class A, \(45\,\text{lb}\), and Classes B, C, and D, \(35\,\text{lb}\) each. Class A has the greatest amount.

Answer

a) Class A: \(35\,\text{lb}\); Class B: \(50\,\text{lb}\); Class C: \(45\,\text{lb}\); Class D: \(20\,\text{lb}\) b) Class A: \(45\,\text{lb}\); Class B: \(35\,\text{lb}\); Class C: \(35\,\text{lb}\); Class D: \(35\,\text{lb}\) c) Class A
5351044
A school surveyed \(100\) fourth-grade students about their favorite sports. The bar graph shows the results. a) How many students took the survey altogether? Verify the total using the graph. b) Which sport was chosen by exactly \(15\) more students than basketball? c) Is the statement true or false? Explain. “Exactly one-fourth of the students chose swimming or gymnastics.”
Figure for problem 535104

Hints

- Add all the bar values to verify the total. - Add \(15\) to the basketball value for part b). - Find one-fourth of the total and compare it with the combined swimming and gymnastics votes.

Solution

1. Add all five values: \(40 + 25 + 15 + 10 + 10 = 100\) students. 2. Basketball received \(25\) votes. A value \(15\) greater is \(25 + 15 = 40\), which is the soccer value. 3. Swimming and gymnastics received \(15 + 10 = 25\) votes. One-fourth of \(100\) is \(100 \div 4 = 25\). Therefore, the statement is true.

Answer

a) \(100\) students b) Soccer c) **True**, because swimming and gymnastics received \(25\) votes altogether, and \(100 \div 4 = 25\).
5373734
Eight relay teams were planned with \(6\) students on each team. Three students are absent. How many students participate? Can the \(8\) teams still all have the same number of students without adding replacements?
Figure for problem 537373

Hints

- First find the total number of registered students. - Subtract the students who are absent. - Check whether the remaining number is divisible by \(8\).

Solution

1. Find the number of students originally registered: \(8 \times 6 = 48\). 2. Subtract the \(3\) absent students: \(48 - 3 = 45\). 3. Check whether \(45\) can be divided equally among \(8\) teams. Since \(45 \div 8 = 5\) remainder \(5\), the teams cannot all remain the same size.

Answer

\(45\) students participate. No, \(45\) cannot be divided equally among \(8\) teams.
5373804
A parking lot has \(6\) rows with \(11\) spaces in each row. Eight yellow spaces are reserved, and \(17\) blue spaces are occupied. How many spaces are neither reserved nor occupied?
Figure for problem 537380

Hints

- First find the total number of parking spaces. - Combine the reserved and occupied spaces. - Subtract that amount from the total.

Solution

1. Find the total number of spaces: \(6 \times 11 = 66\). 2. Find the number reserved or occupied: \(8 + 17 = 25\). 3. Find the remaining spaces: \(66 - 25 = 41\).

Answer

\(41\) spaces are open and not reserved.
5373834
A charging cabinet has \(54\) slots. Forty-six tablets are fully charged, and \(8\) are not. Three classes each need \(15\) charged tablets. Are there enough charged tablets? How many charged tablets will remain?
Figure for problem 537383

Hints

- Find the total number needed by the three classes. - Compare that amount with the \(46\) charged tablets. - Subtract to find how many charged tablets remain.

Solution

1. Find the number needed: \(3 \times 15 = 45\). 2. Compare with the \(46\) charged tablets. There are enough. 3. Find the number remaining: \(46 - 45 = 1\).

Answer

Yes. There are enough charged tablets, and \(1\) charged tablet will remain.
5374164
In a stadium seating diagram, each dot represents \(10\) seats. The diagram has \(6\) rows with \(8\) dots in each row. If \(85\) seats are occupied, how many seats are still available?
Figure for problem 537416

Hints

- Use the value represented by each dot. - Find the total capacity before subtracting the occupied seats.

Solution

1. The diagram contains \(6\times 8=48\) dots. 2. The total capacity is \(48\times 10=480\) seats. 3. The number available is \(480-85=395\) seats.

Answer

\(395\) seats are still available.
5374194
A reading project includes \(35\) booklets with \(24\) pages each. Six students divide all the pages equally. How many pages does each student read?
Figure for problem 537419

Hints

- Find the total number of pages first. - Divide the total equally among \(6\) students.

Solution

1. Find the total number of pages: \(35\times 24=35\times (20+4)=700+140=840\). 2. Divide the pages equally: \(840\div 6=140\).

Answer

Each student reads \(140\) pages.
5374254
A warehouse has \(56\) pallets, represented by the dots in the diagram. Each pallet holds \(12\) cartons. If \(95\) cartons are shipped, how many cartons remain?
Figure for problem 537425

Hints

- Break \(12\) into \(10+2\). - Subtract the shipped cartons after finding the full inventory.

Solution

1. Find the total number of cartons: \(56\times 12=56\times (10+2)=560+112=672\). 2. Subtract the cartons shipped: \(672-95=577\).

Answer

\(577\) cartons remain.
5374264
In a school fun run, each dot in the array represents \(\$25\) raised. The event costs \(\$420\) to organize. How much money remains after the expenses are paid?
Figure for problem 537426

Hints

- Use the rows and columns to count the dots efficiently. - Four groups of \(\$25\) make \(\$100\). - Subtract the expenses from the amount raised.

Solution

1. The array has \(6\) rows of \(10\) dots, so it shows \(6 \times 10 = 60\) dots. 2. The total amount raised is \(60 \times \$25 = \$1500\). 3. After expenses, \(\$1500 - \$420 = \$1080\) remains.

Answer

\(\$1080\) remains after expenses.
5374274
A solar array has \(45\) panels, represented by the dots in the diagram. Each panel contains \(18\) cells. After a hailstorm, \(27\) cells are damaged. How many cells still work?
Figure for problem 537427

Hints

- Calculate \(18\) as \(20-2\). - Subtract the damaged cells after finding the total.

Solution

1. Find the total number of cells: \(45\times 18=45\times (20-2)=900-90=810\). 2. Subtract the damaged cells: \(810-27=783\).

Answer

\(783\) cells still work.
5374294
In an inventory diagram, each of \(90\) dots represents \(3\) tablets. All the tablets are divided equally among \(6\) classes. How many tablets does each class receive? Show a reasonable two-step calculation.
Figure for problem 537429

Hints

- Use the value represented by each dot. - Multiply first, then divide.

Solution

1. Find the total number of tablets: \(90\times 3=270\). 2. Divide the tablets equally: \(270\div 6=45\).

Answer

\(90\times 3=270\), then \(270\div 6=45\). Each class receives \(45\) tablets.
5374404
Five bus routes normally offer \(8\) departures each. Seven departures are canceled. Can the remaining departures be distributed equally among the five routes? Explain using division and a remainder.
Figure for problem 537440

Hints

- Find the number of departures remaining first. - Check whether that total is divisible by \(5\).

Solution

1. The normal total is \(5\times 8=40\) departures. 2. After the cancellations, \(40-7=33\) departures remain. 3. Divide among the five routes: \(33\div 5=6\text{ R }3\). 4. Because the remainder is not zero, the routes cannot all have the same number of departures.

Answer

No. \(33\div 5=6\text{ R }3\), so the departures cannot be divided equally among the five routes.
5374434
A museum plans a mosaic with \(60\) tiles: \(24\) blue, \(18\) red, \(12\) green, and \(6\) yellow. Blue tiles cost \(\$2\) each, red tiles cost \(\$3\) each, green tiles cost \(\$4\) each, and yellow tiles cost \(\$5\) each. What is the total cost of the mosaic?
Figure for problem 537443

Hints

- Find the cost of each tile group separately. - Add the four partial costs at the end.

Solution

1. Blue tiles cost \(24\times \$2=\$48\). 2. Red tiles cost \(18\times \$3=\$54\). 3. Green tiles cost \(12\times \$4=\$48\). 4. Yellow tiles cost \(6\times \$5=\$30\). 5. The total cost is \(\$48+\$54+\$48+\$30=\$180\).

Answer

The mosaic costs \(\$180\).
5158394
A pool concession stand sells orders of fries at these prices: <table> <tr><th>Number of orders</th><th>Price</th></tr> <tr><td>\(1\) order</td><td>\(\$3\)</td></tr> <tr><td>\(2\) orders</td><td>\(\$5\)</td></tr> <tr><td>\(4\) orders</td><td>\(\$9\)</td></tr> </table> a) What are the different ways to buy exactly \(3\) orders, and what does each way cost? b) What is the least expensive way to buy exactly \(5\) orders? Show your work. c) How much do you save by buying the \(4\)-order deal instead of four single orders?

Hints

- Build the requested number of orders from the choices in the table. - There may be more than one way to make the same number of orders. - Find and compare the total cost of each combination. - Subtract the deal price from the cost of four single orders to find the savings.

Solution

1. For \(3\) orders, three single orders cost \(3 \times 3 = 9\) dollars. One \(2\)-order deal plus one single order costs \(5 + 3 = 8\) dollars. 2. For \(5\) orders, a \(4\)-order deal plus one single order costs \(9 + 3 = 12\) dollars. Two \(2\)-order deals plus one single order cost \(5 + 5 + 3 = 13\) dollars. One \(2\)-order deal plus three single orders costs \(5 + 3 + 3 + 3 = 14\) dollars. Five single orders cost \(5 \times 3 = 15\) dollars. Therefore, the least expensive price is \(\$12\). 3. Four single orders cost \(4 \times 3 = 12\) dollars. The \(4\)-order deal costs \(9\) dollars, so the savings are \(12 - 9 = 3\) dollars.

Answer

a) Three single orders cost \(\$9\). One \(2\)-order deal plus one single order costs \(\$8\). b) The least expensive price is \(\$12\), using the \(4\)-order deal plus one single order. c) You save \(\$3\).
5158414
A fair sells raffle tickets at these prices: <table> <tr><th>Number of tickets</th><th>Price</th></tr> <tr><td>\(1\) ticket</td><td>\(\$2\)</td></tr> <tr><td>\(4\) tickets</td><td>\(\$7\)</td></tr> <tr><td>\(6\) tickets</td><td>\(\$10\)</td></tr> <tr><td>\(10\) tickets</td><td>\(\$15\)</td></tr> </table> a) Find the least expensive way to buy exactly \(14\) tickets. b) Lucas wants exactly \(11\) tickets. What is the least amount he must pay? c) Compare two ways to buy \(12\) tickets: two \(6\)-ticket packages or three \(4\)-ticket packages. Which costs less?

Hints

- Start with the larger packages, but make sure the total number is exact. - Compare the cost of buying tickets separately with the package prices. - Test different combinations and add their prices. - For part c), calculate each total separately before comparing.

Solution

1. Fourteen single tickets would cost \(14 \times 2 = 28\) dollars. Compared with single-ticket prices, a \(4\)-ticket package saves \(\$1\), a \(6\)-ticket package saves \(\$2\), and a \(10\)-ticket package saves \(\$5\). For exactly \(14\) tickets, the greatest possible savings are \(\$6\), using \(10 + 4\) tickets. The cost is \(28 - 6 = 22\) dollars. 2. Eleven single tickets would cost \(11 \times 2 = 22\) dollars. The greatest possible savings are \(\$5\), using one \(10\)-ticket package and one single ticket. The cost is \(22 - 5 = 17\) dollars. 3. Two \(6\)-ticket packages cost \(2 \times 10 = 20\) dollars. Three \(4\)-ticket packages cost \(3 \times 7 = 21\) dollars. The first way costs \(1\) dollar less.

Answer

a) \(\$22\), using \(10 + 4\) tickets b) \(\$17\), using \(10 + 1\) tickets c) Two \(6\)-ticket packages cost less by \(\$1\).
5161744
Money puzzle: Use \(\$5\), \(\$10\), and \(\$20\) bills. You may use each type more than once. I am an amount greater than \(\$30\) and less than \(\$50\). - I can be made with exactly \(2\) bills. - I can be made with exactly \(4\) bills. - I can be made with exactly \(8\) bills. - There is no single bill worth my amount. What amount am I?

Hints

- List amounts between \(\$30\) and \(\$50\) that can be made with exactly \(2\) allowed bills. - Check whether there is a single bill with each possible value. - Test whether the remaining amount can also be made with exactly \(4\) and exactly \(8\) bills.

Solution

1. Check amounts between \(\$30\) and \(\$50\) that can be made with exactly \(2\) allowed bills. The only one is \(\$20 + \$20 = \$40\). 2. Check the four-bill condition: \(\$10 + \$10 + \$10 + \$10 = \$40\). 3. Check the eight-bill condition: eight \(\$5\) bills total \(\$40\). 4. There is no \(\$40\) bill, so every condition is satisfied.

Answer

The amount is \(\$40\).
5167674
A large archive has \(4\) sections. Each section has \(12\) aisles. Each aisle has \(15\) filing cabinets. Each cabinet has \(6\) drawers, and each drawer contains \(8\) file folders. How many file folders are in the archive altogether?

Hints

- Find the number of objects at one level before moving to the next level. - Determine the total number of cabinets before finding the drawers. - Multiply the number of drawers by the folders in each drawer. - Check that each step uses the total from the previous level.

Solution

1. Find the total number of aisles: \(4\times 12=48\). 2. Find the total number of filing cabinets: \(48\times 15=720\). 3. Find the total number of drawers: \(720\times 6=4320\). 4. Find the total number of file folders: \(4320\times 8=34{,}560\).

Answer

The archive contains \(34{,}560\) file folders.
5168364
A classroom terrarium contains \(10\) animals, all beetles or spiders. A beetle has \(6\) legs, and a spider has \(8\) legs. The animals have \(68\) legs altogether. How many beetles and how many spiders are there?

Hints

- Think about how the total number of legs changes when one beetle is replaced by one spider. - First find the number of legs if all \(10\) animals were beetles. - Compare the leg counts of one spider and one beetle. - You may organize possible combinations in a list.

Solution

1. Suppose all \(10\) animals were beetles. They would have \(10\times 6=60\) legs. 2. The actual total has \(68-60=8\) more legs. 3. Replacing one beetle with one spider adds \(8-6=2\) legs. Therefore, the number of spiders is \(8\div 2=4\). 4. The number of beetles is \(10-4=6\). 5. Check: \(4\times 8+6\times 6=32+36=68\).

Answer

There are \(6\) beetles and \(4\) spiders.
5168394
A farmer counts \(25\) chickens and pigs in a pasture. Altogether, the animals have \(74\) legs. His daughter says, “There must be \(15\) pigs and \(10\) chickens.” Check her claim, and then find the correct number of pigs and chickens.

Hints

- First find how many legs the animals in the daughter’s claim would have. - Compare the number of legs on a chicken with the number on a pig. - Think about how the total changes when one chicken is replaced by one pig.

Solution

1. Check the claim: \(15\) pigs have \(15\times 4=60\) legs, and \(10\) chickens have \(10\times 2=20\) legs. That would be \(60+20=80\) legs, so the claim is incorrect. 2. If all \(25\) animals were chickens, they would have \(25\times 2=50\) legs. 3. The actual total has \(74-50=24\) more legs. 4. Replacing one chicken with one pig adds \(4-2=2\) legs, so there are \(24\div 2=12\) pigs. 5. There are \(25-12=13\) chickens. Check: \(12\times 4+13\times 2=48+26=74\).

Answer

The daughter’s claim is incorrect. There are \(12\) pigs and \(13\) chickens.
5169184
A historic train has first-class cars with \(40\) seats and coach cars with \(70\) seats. The train must have exactly \(400\) seats, and \(3\) first-class cars are already planned. Every passenger car is \(72\,\text{ft}\) long, and the locomotive is \(50\,\text{ft}\) long. a) How many coach cars are needed? b) What is the total length of the train, including the locomotive?

Hints

- Find the number of seats already provided by the first-class cars. - Subtract to find how many seats are still needed. - Find how many groups of \(70\) seats make the remaining number of seats. - Include every passenger car and the locomotive in the total length.

Solution

1. The first-class cars provide \(3 \times 40 = 120\) seats. 2. The number of seats still needed is \(400 - 120 = 280\). 3. Since \(4 \times 70 = 280\), the train needs \(4\) coach cars. 4. The train has \(3 + 4 = 7\) passenger cars. 5. The passenger cars are \(7 \times 72\,\text{ft} = 504\,\text{ft}\) long in all. 6. Including the locomotive, the train is \(504\,\text{ft} + 50\,\text{ft} = 554\,\text{ft}\) long.

Answer

a) The train needs \(4\) coach cars. b) The train is \(554\,\text{ft}\) long.
5169874
A researcher finds a mystery number. First, the number is divided by \(100\). Then \(350\) is added, the result is tripled, and \(2000\) is subtracted. The final result is \(4000\). What is the mystery number?

Hints

- Write the operations in the order they are performed. - Start from the final result and undo each operation in reverse order. - Use the inverse operation for tripling.

Solution

1. Work backward by undoing the subtraction: \(4000 + 2000 = 6000\). 2. Undo tripling: \(6000 \div 3 = 2000\). 3. Undo adding \(350\): \(2000 - 350 = 1650\). 4. Undo division by \(100\): \(1650 \times 100 = 165{,}000\). 5. The mystery number is \(165{,}000\).

Answer

\(165{,}000\)
5172194
Two whole numbers have a sum of \(27\). The number after the smaller number is \(5\) less than the number before the larger number. Find the two numbers.

Hints

- Find the difference between the two unknown numbers. - Use the known sum and difference together. - Check both the sum and the relationship between the neighboring numbers.

Solution

1. From the smaller number to the number after it is \(1\). The gap to the number before the larger number is \(5\), and the final step to the larger number is \(1\). Therefore, the larger number is \(1+5+1=7\) greater than the smaller number. 2. Subtract the difference from the sum: \(27-7=20\). 3. Divide equally: \(20\div2=10\). The smaller number is \(10\), and the larger number is \(10+7=17\). 4. Check: \(10+17=27\), and \(11\) is \(5\) less than \(16\).

Answer

\(10\) and \(17\)
5172394
A youth group of \(42\) people is taking public transit. A single fare costs \(\$5\). A 10-ride pass costs \(\$42\) and covers \(10\) people. a) Find the total cost if everyone pays a single fare. b) What is the least number of 10-ride passes needed to cover all \(42\) people using only passes? Find the cost. c) Find the cost of buying \(4\) 10-ride passes and single fares for the remaining people. d) How much does the least expensive option save compared with part a)?

Hints

- A partial group still requires another 10-ride pass in part b. - In part c, determine how many people remain after four passes. - Compare all totals before finding the savings.

Solution

1. a) Single fares cost \(42\times\$5=\$210\). 2. b) Four passes cover \(40\) people, so \(5\) passes are needed to cover all \(42\) people. They cost \(5\times\$42=\$210\). 3. c) Four passes cost \(4\times\$42=\$168\). Two people remain, and their single fares cost \(2\times\$5=\$10\). The total is \(\$168+\$10=\$178\). 4. d) The combination in part c is least expensive. It saves \(\$210-\$178=\$32\).

Answer

a) \(\$210\) b) \(5\) passes for \(\$210\) c) \(\$178\) d) \(\$32\)
5174804
A school fair prepares \(400\) raffle tickets. During the first hour, exactly one-fourth of all the tickets are sold. During the second hour, \(35\) fewer tickets are sold than during the first hour. During the third hour, twice as many tickets are sold as during the second hour. How many tickets remain after the three hours?

Hints

- Find the number of tickets sold during each hour in order. - What operation finds one-fourth of the starting number? - After finding all three hourly amounts, how can you find the total number sold? - How can you find what remains from the starting number?

Solution

1. Find the number sold during the first hour: \(\frac{1}{4} \times 400 = 100\). 2. Find the number sold during the second hour: \(100 - 35 = 65\). 3. Find the number sold during the third hour: \(2 \times 65 = 130\). 4. Find the total number sold: \(100 + 65 + 130 = 295\). 5. Subtract from the starting number: \(400 - 295 = 105\).

Answer

There are \(105\) raffle tickets remaining.
5175324
Two packages of construction paper contain \(80\) sheets and \(48\) sheets. One-eighth of the sheets is removed from each package to make a display. How many sheets remain in the two packages altogether? Solve the problem in two different ways.

Hints

- Try finding the number of sheets remaining in each package separately. - Could you instead begin by finding the total number of sheets in both packages? - What happens if you combine the removed sheets before subtracting? - How many sheets are one-eighth of \(80\)? How many are one-eighth of \(48\)?

Solution

Method 1: 1. In the first package, \(80 - (\frac{1}{8} \times 80) = 80 - 10 = 70\) sheets remain. 2. In the second package, \(48 - (\frac{1}{8} \times 48) = 48 - 6 = 42\) sheets remain. 3. Add the remaining sheets: \(70 + 42 = 112\). Method 2: 1. Find the total number of sheets: \(80 + 48 = 128\). 2. One-eighth of all the sheets is removed: \(\frac{1}{8} \times 128 = 16\). 3. Subtract the removed sheets: \(128 - 16 = 112\).

Answer

There are \(112\) sheets of construction paper remaining altogether.
5175954
Three fourth-grade classes share the cost of a charter bus. The bus costs \(\$480\). Each student should pay the same amount. - Class 4A has \(20\) students. - Class 4B has \(22\) students. - Class 4C has \(18\) students. a) How much should each class pay? b) How much should each student pay?

Hints

- Add the class sizes to find the total number of students. - Find the amount per student by using a multiplication fact with the total number of students. - Multiply the per-student cost by each class size.

Solution

1. The total number of students is \(20 + 22 + 18 = 60\). 2. Since \(60 \times \$8 = \$480\), the cost per student is \(\$8\). 3. Class 4A pays \(20 \times \$8 = \$160\). 4. Class 4B pays \(22 \times \$8 = \$176\). 5. Class 4C pays \(18 \times \$8 = \$144\). 6. The class totals check because \(\$160 + \$176 + \$144 = \$480\).

Answer

a) Class 4A pays \(\$160\), Class 4B pays \(\$176\), and Class 4C pays \(\$144\). b) Each student pays \(\$8\).
5176104
Three friends collect equal-sized bags of acorns for a wildlife center. Leo collects \(4\) bags, Mia collects \(5\) bags, and Tom collects \(3\) bags. They receive a total reward of \(\$60\), divided according to the number of bags each person collected. a) How much money does each friend receive? b) Mia says, “I would receive less if we split the money equally among the three of us.” Is she correct? Support your answer with a calculation.

Hints

- Find the total number of bags and use multiplication to find the reward per bag. - Use each friend’s number of bags to find that person’s share. - For part b, compare Mia’s share with one-third of the total reward.

Solution

1. The friends collected \(4 + 5 + 3 = 12\) bags. 2. Since \(12 \times \$5 = \$60\), the reward is \(\$5\) per bag. 3. Leo receives \(4 \times \$5 = \$20\), Mia receives \(5 \times \$5 = \$25\), and Tom receives \(3 \times \$5 = \$15\). 4. An equal three-way split gives each friend \(\$60 \div 3 = \$20\). 5. Mia would receive \(\$25 - \$20 = \$5\) less under an equal split, so she is correct.

Answer

a) Leo receives \(\$20\), Mia receives \(\$25\), and Tom receives \(\$15\). b) Yes. An equal split would give Mia \(\$20\), which is \(\$5\) less than \(\$25\).
5176834
A bakery has a \(20\,\text{kg}\) bag of flour. The baker uses \(4\,\text{kg}\) to make rolls. a) How many kilograms of flour remain? b) Is the remaining amount more than four times the amount used? Justify your answer with a calculation. c) How many kilograms would the baker need to use altogether so that the amount remaining is exactly \(10\,\text{kg}\) greater than the amount used?

Hints

- What does “four times as much” mean? - In part b), distinguish between “more than” and “exactly.” - For part c), test possible amounts used and compare each amount with what remains.

Solution

1. Subtract to find the remaining flour: \(20 - 4 = 16\,\text{kg}\). 2. Four times the amount used is \(4 \times 4 = 16\,\text{kg}\). The remaining amount is exactly four times, not more than four times, the amount used. 3. For part c), the used and remaining amounts total \(20\,\text{kg}\), and the remaining amount must include an extra \(10\,\text{kg}\). Remove that difference: \(20\,\text{kg} - 10\,\text{kg} = 10\,\text{kg}\). Split the \(10\,\text{kg}\) equally between the used amount and the matching part of the remaining amount: \(10\,\text{kg} \div 2 = 5\,\text{kg}\). Then \(15\,\text{kg} - 5\,\text{kg} = 10\,\text{kg}\).

Answer

a) \(16\,\text{kg}\) b) No. The remaining amount is exactly four times the amount used because \(4 \times 4 = 16\). c) The baker would need to use \(5\,\text{kg}\) altogether.
5177344
Two baskets contain \(60\) apples altogether. Liam moves \(5\) apples from the first basket to the second basket. The second basket then has exactly \(10\) more apples than the first basket. a) How many apples were in each basket at the beginning? b) Explain whether the baskets began with equal or different numbers of apples.

Hints

- First find the two amounts after the move, when their total is \(60\) and their difference is \(10\). - Moving one item from one group to the other changes the difference by \(2\). - Reverse the move of \(5\) apples to find the starting amounts.

Solution

1. Moving apples between the baskets does not change the total of \(60\). 2. After the move, the two amounts differ by \(10\), so they are \(25\) and \(35\). 3. Before the move, return the \(5\) apples: the first basket had \(25+5=30\), and the second had \(35-5=30\). 4. The baskets began with equal amounts. Moving \(5\) apples out of one basket and into the other creates a difference of \(5+5=10\).

Answer

a) Each basket began with \(30\) apples. b) The amounts were equal. Moving \(5\) apples decreases one amount by \(5\) and increases the other by \(5\), creating a difference of \(10\).
5177534
Two money jars contain \(\$60\) in all. After \(\$8\) is moved from the first jar to the second jar, the second jar contains \(\$4\) more than the first jar. How much money was originally in each jar?

Hints

- The total amount stays the same when money moves between jars. - First determine the two amounts after the move, when their difference is \(\$4\). - Then reverse the transfer of \(\$8\).

Solution

1. Moving money between the jars does not change the total of \(\$60\). 2. After the move, subtract the \(\$4\) difference from the total: \(\$60 - \$4 = \$56\). 3. Split \(\$56\) equally: \(\$56 \div 2 = \$28\). After the move, the first jar has \(\$28\), and the second has \(\$28 + \$4 = \$32\). 4. Reverse the move. The first jar originally had \(\$28 + \$8 = \$36\), and the second originally had \(\$32 - \$8 = \$24\).

Answer

Originally, the first jar contained \(\$36\), and the second jar contained \(\$24\).
5178434
Mia and Ben have \(100\) stickers altogether. Mia has more stickers than Ben. If Mia gives away \(12\) stickers, they will have equal amounts. a) How many stickers did Mia have at the beginning? b) How many stickers did Ben have at the beginning? c) Instead of giving stickers away, how many stickers would Mia need to give directly to Ben so that they would have equal amounts?

Hints

- First find the total number of stickers left after Mia gives away \(12\). - The equal amount after that change is also Ben’s original amount. - In part c, moving one sticker from Mia to Ben changes their difference by \(2\).

Solution

1. After Mia gives away \(12\) stickers, the total is \(100-12=88\). 2. Equal shares of \(88\) are \(88\div 2=44\), so Ben began with \(44\) stickers. 3. Mia began with \(44+12=56\) stickers. 4. Their original difference was \(56-44=12\). Giving one sticker from Mia to Ben reduces the difference by \(2\), so Mia must give Ben \(12\div 2=6\) stickers.

Answer

a) Mia had \(56\) stickers. b) Ben had \(44\) stickers. c) Mia would need to give Ben \(6\) stickers.
5178634
A school festival orders juice. Each case of apple juice contains \(6\) bottles, and each case of orange juice contains \(10\) bottles. The order includes \(3\) cases of apple juice and \(2\) cases of orange juice. The total bill is \(\$76\), and every bottle has the same price. a) How many bottles were ordered? b) How much does one bottle cost? c) What is the total cost of the apple juice, and what is the total cost of the orange juice?

Hints

- Find the number of bottles of each kind first. - Use multiplication to find the price per bottle that makes the total bill. - Multiply the price per bottle by each kind’s bottle count.

Solution

1. The apple juice order contains \(3 \times 6 = 18\) bottles. 2. The orange juice order contains \(2 \times 10 = 20\) bottles. 3. The total number of bottles is \(18 + 20 = 38\). 4. Since \(38 \times \$2 = \$76\), one bottle costs \(\$2\). 5. The apple juice costs \(18 \times \$2 = \$36\), and the orange juice costs \(20 \times \$2 = \$40\).

Answer

a) \(38\) bottles were ordered. b) One bottle costs \(\$2\). c) The apple juice costs \(\$36\), and the orange juice costs \(\$40\).
5179514
An elementary school receives \(240\) new nonfiction books for its library. During the first \(5\) days, volunteers shelve \(24\) books each day. After that, they shelve \(30\) books each day. How many days does it take to shelve all the books?

Hints

- How many books are shelved during the first phase? - How many books remain after the first \(5\) days? - Use the faster daily rate to find the remaining number of days. - Add the days from both phases.

Solution

1. Find the number shelved during the first \(5\) days: \(5 \times 24 = 120\) books. 2. Find the number remaining: \(240 - 120 = 120\) books. 3. Find the additional days needed: \(120 \div 30 = 4\) days. 4. Add the two time periods: \(5 + 4 = 9\) days.

Answer

It takes \(9\) days to shelve all the books.
5180344
A school library begins the week with \(145\) books on a shelf. On Monday, \(18\) books are checked out and \(22\) are returned. On Tuesday, a donated box containing \(10\) mystery books, \(4\) nonfiction books, and \(1\) graphic novel is added, while \(7\) damaged books are removed. On Wednesday, \(12\) more books are returned than are checked out. On Thursday, a class checks out \(24\) books and a teacher returns \(3\). Fifteen students read in the library without checking out books. On Friday, \(31\) books are returned and \(19\) are checked out. How many books are on the shelf at the end of the week?

Hints

- Determine the net change for each day. - Decide which details affect the number of books on the shelf. - “Twelve more returned than checked out” means a net increase of \(12\). - Update the running total one day at a time.

Solution

1. After Monday, there are \(145-18+22=149\) books. 2. The donation contains \(10+4+1=15\) books. After Tuesday, there are \(149+15-7=157\) books. 3. Wednesday gives a net increase of \(12\), so there are \(157+12=169\) books. 4. The students who only read do not change the shelf total. After Thursday, there are \(169-24+3=148\) books. 5. After Friday, there are \(148+31-19=160\) books.

Answer

\(160\) books
5180654
A school cafeteria offers two snack combinations: Combination A has \(2\) pretzels and \(4\) juice boxes for \(\$8\) altogether. Combination B has \(2\) pretzels and \(2\) juice boxes for \(\$6\) altogether. How much does one juice box cost, and how much does one pretzel cost?

Hints

- Compare the two combinations. What extra items are in Combination A? - Why does Combination A cost more than Combination B? - Use the difference to find the cost of one juice box. - Then use either combination to find the cost of one pretzel.

Solution

1. Combination A has two more juice boxes than Combination B, while the number of pretzels is the same. 2. Find the difference in cost: \(\$8 - \$6 = \$2\). This is the cost of the two additional juice boxes. 3. Find the cost of one juice box: \(\$2 \div 2 = \$1\). 4. In Combination B, the two juice boxes cost \(2 \times \$1 = \$2\). 5. Find the cost of two pretzels: \(\$6 - \$2 = \$4\). 6. Find the cost of one pretzel: \(\$4 \div 2 = \$2\).

Answer

One juice box costs \(\$1\), and one pretzel costs \(\$2\).
5180914
A backpack and a pencil case cost \(\$110\) altogether. The backpack costs \(\$80\) more than the pencil case. How much does each item cost?

Hints

- A bar model can show one item as the other item plus \(\$80\). - Remove the difference from the total to leave two equal parts. - Divide the remaining amount by \(2\).

Solution

1. Subtract the price difference from the total: \(\$110 - \$80 = \$30\). This is twice the pencil case’s price. 2. The pencil case costs \(\$30 \div 2 = \$15\). 3. The backpack costs \(\$15 + \$80 = \$95\). 4. Check: \(\$15 + \$95 = \$110\).

Answer

The pencil case costs \(\$15\), and the backpack costs \(\$95\).
5180924
Two packages have a combined mass of \(750\,\text{g}\). Package A is \(150\,\text{g}\) heavier than Package B. a) What is the mass of each package? b) If \(50\,\text{g}\) of material is moved from Package A to Package B, what is the new mass of Package B?

Hints

- Use the combined mass and the difference to find the lighter package first. - Subtract the difference, then divide the remaining amount equally. - For part b, add the transferred mass to Package B.

Solution

1. Remove the difference from the combined mass: \(750\,\text{g} - 150\,\text{g} = 600\,\text{g}\). 2. Divide equally to find Package B's mass: \(600\,\text{g} \div 2 = 300\,\text{g}\). 3. Find Package A's mass: \(300\,\text{g} + 150\,\text{g} = 450\,\text{g}\). 4. After the transfer, Package B has a mass of \(300\,\text{g} + 50\,\text{g} = 350\,\text{g}\).

Answer

a) Package A has a mass of \(450\,\text{g}\), and Package B has a mass of \(300\,\text{g}\). b) Package B would have a mass of \(350\,\text{g}\).
5181464
Lukas has an \(18\,\text{ft}\) rope. He wants to cut it into jump ropes that are each \(6\,\text{ft}\) long. Each cut takes \(15\) seconds. How long will it take him to cut the rope into all the pieces?

Hints

- Find how many pieces the rope will make. - Does the final piece require another cut? - Sketch the rope and mark each place where a cut is needed.

Solution

1. Find the number of pieces: \(18 \div 6 = 3\) pieces. 2. To make \(3\) pieces from one rope, Lukas needs \(3 - 1 = 2\) cuts. 3. Find the total cutting time: \(2 \times 15 = 30\) seconds.

Answer

It will take Lukas \(30\) seconds.
5185944
Two classes are raising money for an animal shelter. Class 4A has a goal of \(\$720\) and has already raised one-eighth of that amount. Class 4B has a goal of \(\$600\) and has already raised one-fifth of that amount. Which class still needs to raise more money to reach its goal? What is the difference between the amounts the two classes still need?

Hints

- For each class, find how much has been raised and how much is still needed. - How can you compare the two remaining amounts? - A difference can be found by subtraction.

Solution

1. Class 4A has raised \(\frac{1}{8} \times \$720 = \$90\), so it still needs \(\$720 - \$90 = \$630\). 2. Class 4B has raised \(\frac{1}{5} \times \$600 = \$120\), so it still needs \(\$600 - \$120 = \$480\). 3. Since \(\$630 > \$480\), Class 4A still needs to raise more money. 4. Find the difference: \(\$630 - \$480 = \$150\).

Answer

Class 4A still needs to raise more money. The difference is \(\$150\).
5186364
Jordan and Mia collected \(240\) acorns altogether. Jordan collected \(40\) fewer acorns than Mia. a) How many acorns did each child collect? b) Suppose the total remains \(240\), but the difference between their amounts becomes greater than \(40\). Would Jordan then have more or fewer acorns than before? Explain.

Hints

- Identify which child has the larger amount. - Imagine adding \(40\) to Jordan’s amount so the two amounts become equal. - For part b, test a larger difference, such as \(60\), while keeping the total at \(240\).

Solution

1. Add the difference to the total: \(240+40=280\). 2. Divide by \(2\) to find Mia’s larger amount: \(280\div 2=140\). 3. Jordan’s amount is \(140-40=100\). 4. If the total stays fixed while the difference grows, the larger amount must increase and the smaller amount must decrease. Jordan would have fewer than \(100\) acorns.

Answer

a) Mia collected \(140\) acorns, and Jordan collected \(100\). b) Jordan would have fewer acorns. With a fixed total, increasing the difference makes the smaller share decrease.
5186434
Lucas and Sophie are saving for a trampoline that costs \(\$450\). Together, they have saved \(\$370\). Sophie has saved \(\$50\) more than Lucas. How much has each person saved? How much more do they need altogether to buy the trampoline?

Hints

- Separate the problem into finding the two savings amounts and finding the remaining amount needed. - Remove the difference before splitting the combined savings equally. - Compare their combined savings with the trampoline price.

Solution

1. Remove the \(\$50\) difference from their combined savings: \(\$370 - \$50 = \$320\). 2. Split \(\$320\) into two equal amounts: \(\$320 \div 2 = \$160\). Lucas saved \(\$160\). 3. Sophie saved \(\$160 + \$50 = \$210\). 4. They still need \(\$450 - \$370 = \$80\).

Answer

Lucas saved \(\$160\), and Sophie saved \(\$210\). Together, they still need \(\$80\).
5186994
For a school festival, Mr. Weber buys \(20\,\text{lb}\) of apples at \(\$2\) per pound. He also buys pears. He buys \(4\,\text{lb}\) fewer pears than apples, and pears cost \(\$1\) more per pound than apples. Mr. Weber has \(\$100\). Is that enough? How much money will he have left, or how much more will he need?

Hints

- First find the total cost of the apples. - Determine both the number of pounds of pears and their price per pound. - Add the two costs and compare the total with the available money.

Solution

1. Find the cost of the apples: \(20 \times \$2 = \$40\). 2. Find the amount of pears: \(20 - 4 = 16\,\text{lb}\). 3. Find the price per pound of pears: \(\$2 + \$1 = \$3\). 4. Find the cost of the pears: \(16 \times \$3 = \$48\). 5. Find the total cost: \(\$40 + \$48 = \$88\). 6. Compare the total with the budget: \(\$100 - \$88 = \$12\).

Answer

Yes, \(\$100\) is enough. Mr. Weber will have \(\$12\) left.
5187104
Two rain barrels contain \(240\,\text{L}\) of water altogether. After \(30\,\text{L}\) are poured from the first barrel into the second, the second barrel contains \(20\,\text{L}\) more than the first. How many liters were originally in each barrel?

Hints

- The total amount does not change when water is moved between barrels. - First find the amounts after the transfer using the \(20\,\text{L}\) difference. - Then reverse the \(30\,\text{L}\) transfer.

Solution

1. After the transfer, the total is still \(240\,\text{L}\), and the second barrel has \(20\,\text{L}\) more than the first. 2. Remove the extra \(20\,\text{L}\): \(240\,\text{L} - 20\,\text{L} = 220\,\text{L}\). 3. Split the remaining amount equally: \(220\,\text{L} \div 2 = 110\,\text{L}\). After the transfer, the first barrel has \(110\,\text{L}\), and the second has \(130\,\text{L}\). 4. Reverse the transfer: the first barrel originally had \(110\,\text{L} + 30\,\text{L} = 140\,\text{L}\), and the second originally had \(130\,\text{L} - 30\,\text{L} = 100\,\text{L}\).

Answer

The first barrel originally contained \(140\,\text{L}\), and the second barrel originally contained \(100\,\text{L}\).
5187204
Lucas has saved \(\$600\). He spends exactly one-half of his money on a new bicycle. A helmet costs one-fifth of his total savings. He spends the rest on a bike rack and lock set. How much more does the bicycle cost than the rack and lock set?

Hints

- Find the amount spent on each item in order. - Both fractions refer to the original \(\$600\). - Which two costs must you compare to find the difference?

Solution

1. Find the bicycle cost: \(\frac{1}{2} \times \$600 = \$300\). 2. Find the helmet cost: \(\frac{1}{5} \times \$600 = \$120\). 3. Find the total spent on the bicycle and helmet: \(\$300 + \$120 = \$420\). 4. Find the rack and lock set cost: \(\$600 - \$420 = \$180\). 5. Find the difference: \(\$300 - \$180 = \$120\).

Answer

The bicycle costs \(\$120\) more than the rack and lock set.
5187784
A dog, a cat, and a rabbit weigh \(45\) pounds altogether. The dog and cat weigh \(38\) pounds altogether. The cat and rabbit weigh \(15\) pounds altogether. How much does each animal weigh?

Hints

- Compare the total weight with the combined weight of the dog and cat. - Once one animal's weight is known, use another pair total. - Check that all three weights add to \(45\) pounds.

Solution

1. Subtract the combined weight of the dog and cat from the total to find the rabbit's weight: \(45 - 38 = 7\) pounds. 2. Subtract the rabbit's weight from the combined weight of the cat and rabbit: \(15 - 7 = 8\) pounds for the cat. 3. Subtract the cat's weight from the combined weight of the dog and cat: \(38 - 8 = 30\) pounds for the dog.

Answer

The dog weighs \(30\) pounds, the cat weighs \(8\) pounds, and the rabbit weighs \(7\) pounds.
5189254
Two wildlife parks compare their animal populations. Greenwood Park has \(156{,}780\) birds and \(42{,}350\) mammals. Lakeside Park has \(189{,}200\) birds and \(12{,}400\) mammals. During the summer: - Greenwood gains \(12{,}450\) birds, while its mammal population decreases by \(3120\). - Lakeside loses \(5600\) birds and gains \(18{,}700\) mammals. Which park has more animals after these changes, and what is the difference?

Hints

- Find the final population of each park separately. - Pay attention to which changes increase and which decrease each population. - Compare the two final totals and subtract to find the difference.

Solution

1. Greenwood’s new total is \(156{,}780+42{,}350+12{,}450-3120=208{,}460\). 2. Lakeside’s new total is \(189{,}200+12{,}400-5600+18{,}700=214{,}700\). 3. Since \(214{,}700>208{,}460\), Lakeside has more animals. 4. The difference is \(214{,}700-208{,}460=6240\) animals.

Answer

Lakeside Park has more animals. It has \(6240\) more animals than Greenwood Park.
5189854
A bakery uses \(156\,\text{lb}\) of wheat flour, \(98\,\text{lb}\) of rye flour, and \(45\,\text{lb}\) of spelt flour in one week. The next week, it uses twice as much wheat flour, three times as much rye flour, and four times as much spelt flour. How many more pounds of flour does the bakery use in the second week?

Hints

- Find the total flour used in the first week. - Use each multiplier to find the second week's amounts. - Add the second week's amounts and subtract the first week's total.

Solution

1. Find the first week's total: \(156\,\text{lb} + 98\,\text{lb} + 45\,\text{lb} = 299\,\text{lb}\). 2. Find the second week's amounts: \(156\,\text{lb} \times 2 = 312\,\text{lb}\), \(98\,\text{lb} \times 3 = 294\,\text{lb}\), and \(45\,\text{lb} \times 4 = 180\,\text{lb}\). 3. Find the second week's total: \(312\,\text{lb} + 294\,\text{lb} + 180\,\text{lb} = 786\,\text{lb}\). 4. Find the increase: \(786\,\text{lb} - 299\,\text{lb} = 487\,\text{lb}\).

Answer

The bakery uses \(487\,\text{lb}\) more flour in the second week.
5192114
A basket contains apples, pears, and plums that weigh \(950\,\text{g}\) altogether. The apples and pears weigh \(700\,\text{g}\) together. The pears and plums weigh \(550\,\text{g}\) together. How much does each type of fruit weigh?

Hints

- Identify which fruit is missing from each two-fruit total. - Subtract the apples-and-pears weight from the total to find the plums. - Once two individual weights are known, use one pair total to find the third. - A diagram or table may help organize the information.

Solution

1. The plums are the part not included in the apples-and-pears total: \(950\,\text{g} - 700\,\text{g} = 250\,\text{g}\). 2. The apples are the part not included in the pears-and-plums total: \(950\,\text{g} - 550\,\text{g} = 400\,\text{g}\). 3. Find the pears: \(700\,\text{g} - 400\,\text{g} = 300\,\text{g}\). 4. Check: \(400 + 300 + 250 = 950\).

Answer

Apples: \(400\,\text{g}\) Pears: \(300\,\text{g}\) Plums: \(250\,\text{g}\)
5192204
Three bags contain \(1\,\text{kg}\) of flour altogether. The first and second bags contain \(650\,\text{g}\) altogether. The second and third bags contain \(750\,\text{g}\) altogether. How many grams of flour are in each bag?

Hints

- Convert \(1\,\text{kg}\) to grams. - Subtract each pair total from the overall total to find the missing bag. - Use the three individual amounts to check both pair totals.

Solution

1. Convert the total mass to grams: \(1\,\text{kg} = 1000\,\text{g}\). 2. Find the amount in the third bag: \(1000 - 650 = 350\,\text{g}\). 3. Find the amount in the first bag: \(1000 - 750 = 250\,\text{g}\). 4. Find the amount in the second bag: \(1000 - 350 - 250 = 400\,\text{g}\).

Answer

The first bag contains \(250\,\text{g}\), the second contains \(400\,\text{g}\), and the third contains \(350\,\text{g}\).
5192214
A bike path is \(600\,\text{m}\) long and has three sections. The first and second sections together are \(380\,\text{m}\). The first and third sections together are \(420\,\text{m}\). Which section is longest, and how long is it?

Hints

- Subtract the first-and-second total from the full length to find the third section. - Subtract the first-and-third total from the full length to find the second section. - Find the remaining section, then compare all three lengths.

Solution

1. The third section is the part not included in the first-and-second total: \(600\,\text{m} - 380\,\text{m} = 220\,\text{m}\). 2. The second section is the part not included in the first-and-third total: \(600\,\text{m} - 420\,\text{m} = 180\,\text{m}\). 3. Find the first section: \(600 - 220 - 180 = 200\), so it is \(200\,\text{m}\). 4. Compare: \(220\,\text{m} > 200\,\text{m} > 180\,\text{m}\).

Answer

The third section is longest at \(220\,\text{m}\).
5194614
A construction company is paving a public plaza in three sections. The first section requires \(4320\) paving stones. The second section requires four times as many stones as the first. The third section requires exactly one-half as many stones as the second. Are \(25{,}000\) ordered stones enough for the entire plaza? Find how many stones will be left over or how many more are needed.

Hints

- Find the number needed for each section in order. - How many stones are needed for the entire project? - Compare the total need with the number ordered.

Solution

1. Find the stones needed for the second section: \(4 \times 4320 = 17{,}280\). 2. Find the stones needed for the third section: \(\frac{1}{2} \times 17{,}280 = 8640\). 3. Find the total number needed: \(4320 + 17{,}280 + 8640 = 30{,}240\). 4. Since \(30{,}240 > 25{,}000\), the order is not enough. 5. Find the shortage: \(30{,}240 - 25{,}000 = 5240\).

Answer

No. The order is short by \(5240\) paving stones.
5194804
A fruit warehouse distributes \(4800\,\text{lb}\) of apples: - One-half of the apples goes to grocery stores. - One-fourth is pressed for apple juice. - One-eighth is used for applesauce. - The remaining apples are donated to a farm. How many pounds of apples does the farm receive? Compare this amount with the amount used for applesauce. What do you notice? Explain.

Hints

- First find the number of pounds used for the first three purposes. - How much remains from the total of \(4800\,\text{lb}\)? - How many eighths are equal to one-half and one-fourth? - What do the three given fractions add to?

Solution

1. Find the amount sent to grocery stores: \(\frac{1}{2} \times 4800\,\text{lb} = 2400\,\text{lb}\). 2. Find the amount used for juice: \(\frac{1}{4} \times 4800\,\text{lb} = 1200\,\text{lb}\). 3. Find the amount used for applesauce: \(\frac{1}{8} \times 4800\,\text{lb} = 600\,\text{lb}\). 4. Add the assigned amounts: \(2400\,\text{lb} + 1200\,\text{lb} + 600\,\text{lb} = 4200\,\text{lb}\). 5. Find the amount donated: \(4800\,\text{lb} - 4200\,\text{lb} = 600\,\text{lb}\). 6. The donated amount equals the applesauce amount. This is because \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{7}{8}\), leaving \(\frac{1}{8}\).

Answer

The farm receives \(600\,\text{lb}\) of apples. This is the same amount used for applesauce because the assigned fractions total \(\frac{7}{8}\), leaving another \(\frac{1}{8}\).
5194904
The Parker family is saving for a trip to an amusement park. Each month, the parents put \(\$35\) into a savings jar, and each of the two children adds \(\$5\). In June and December, the grandparents add an extra \(\$50\). During the year, the family takes \(\$20\) from the jar three times for ice cream. How much money is in the jar after one year?

Hints

- Use \(12\) months for the monthly deposits. - Remember that both children contribute each month. - Add all deposits before subtracting the withdrawals.

Solution

1. The parents contribute \(12\times\$35=\$420\). 2. The two children contribute \(12\times2\times\$5=\$120\). 3. The grandparents contribute \(2\times\$50=\$100\). 4. Total deposits are \(\$420+\$120+\$100=\$640\). 5. The family removes \(3\times\$20=\$60\). 6. The final amount is \(\$640-\$60=\$580\).

Answer

\(\$580\)
5194914
Jordan wants to buy a bicycle that costs \(\$320\). Jordan has already saved \(\$45\) and receives \(\$15\) each month. Jordan also receives \(\$60\) for a birthday and \(\$30\) for a winter holiday, but spends \(\$4\) on candy each month. After one year, is there enough money for the bicycle? If not, how much more is needed? If so, how much remains?

Hints

- Include the starting savings, monthly income, and gifts. - Subtract the total monthly spending for the year. - Compare the final savings with the bicycle price.

Solution

1. Monthly income for the year is \(12\times\$15=\$180\). 2. The two gifts total \(\$60+\$30=\$90\). 3. Yearly spending is \(12\times\$4=\$48\). 4. The amount saved after one year is \(\$45+\$180+\$90-\$48=\$267\). 5. Since \(\$267<\$320\), more money is needed. The shortage is \(\$320-\$267=\$53\).

Answer

No. Jordan needs \(\$53\) more.
5194934
A wildlife park receives \(4800\,\text{lb}\) of hay for one week. The elephant habitat uses one-half of the hay. The zebra habitat uses one-fifth of the total amount. The remaining hay is for the ponies. How many more pounds of hay does the elephant habitat use than the ponies?

Hints

- First find how much hay each animal group receives. - What operations represent one-half and one-fifth of a quantity? - The final question asks for a difference, not only the remaining amount.

Solution

1. Find the hay used by the elephant habitat: \(\frac{1}{2} \times 4800\,\text{lb} = 2400\,\text{lb}\). 2. Find the hay used by the zebra habitat: \(\frac{1}{5} \times 4800\,\text{lb} = 960\,\text{lb}\). 3. Find the hay used by the first two habitats: \(2400\,\text{lb} + 960\,\text{lb} = 3360\,\text{lb}\). 4. Find the hay for the ponies: \(4800\,\text{lb} - 3360\,\text{lb} = 1440\,\text{lb}\). 5. Find the difference: \(2400\,\text{lb} - 1440\,\text{lb} = 960\,\text{lb}\).

Answer

The elephant habitat uses \(960\,\text{lb}\) more hay than the ponies.
5195814
An orchard sells crates of apples for \(\$8\) each and earns \(\$336\). It sells \(12\) more crates of red apples than green apples. How many crates of each kind does the orchard sell?

Hints

- Find the total number of crates first. - Remove the difference so the two parts can be equal. - Split the remaining crates in half, then restore the difference.

Solution

1. The orchard sells \(\$336 \div \$8 = 42\) crates altogether. 2. Remove the difference: \(42 - 12 = 30\). 3. Split the remaining \(30\) crates equally: \(30 \div 2 = 15\). This is the number of green-apple crates. 4. The number of red-apple crates is \(15 + 12 = 27\).

Answer

The orchard sells \(15\) crates of green apples and \(27\) crates of red apples.
5196204
Town A and Town B are \(782\,\text{miles}\) apart. A village lies directly between them. The village is \(36\,\text{miles}\) closer to Town B than to Town A. How far is the village from Town A?

Hints

- Draw the route as one line split into two parts. - The distance from Town A is \(36\,\text{miles}\) longer than the distance from Town B. - Think about how adding the difference to the total can create two equal copies of the longer distance.

Solution

1. The distance from Town A is the longer part. Add the difference to the total: \(782 + 36 = 818\). 2. Divide by \(2\): \(818 \div 2 = 409\,\text{miles}\) from Town A. 3. Check: The distance from Town B is \(409 - 36 = 373\,\text{miles}\), and \(409 + 373 = 782\,\text{miles}\).

Answer

The village is \(409\,\text{miles}\) from Town A.
5197804
Two fourth-grade classes earn \(\$1350\) at a flea market. Class 4A earns \(\$150\) more than Class 4B. Each class places its money into envelopes holding exactly \(\$50\) each. How many envelopes does each class fill?

Hints

- First find how much money each class earns. - Remove the difference before splitting the total equally. - Use multiplication facts with \(\$50\) to find each number of envelopes.

Solution

1. Remove the \(\$150\) difference: \(\$1350 - \$150 = \$1200\). 2. Split \(\$1200\) equally: \(\$1200 \div 2 = \$600\). Class 4B earns \(\$600\). 3. Class 4A earns \(\$600 + \$150 = \$750\). 4. Since \(15 \times \$50 = \$750\), Class 4A fills \(15\) envelopes. 5. Since \(12 \times \$50 = \$600\), Class 4B fills \(12\) envelopes.

Answer

Class 4A fills \(15\) envelopes, and Class 4B fills \(12\) envelopes.
5198274
For a school fair, the parent-teacher organization buys cases of drinks. A case of apple juice costs \(\$14\), and a case of bottled water costs \(\$6\). The organization spends \(\$952\) in all and buys \(44\) cases of apple juice. a) How many cases of bottled water are purchased? b) A member claims that the organization could have spent exactly \(\$960\) while buying the same number of apple juice cases. Use a calculation to explain why that total is impossible with the stated water price.

Hints

- First find the total cost of the known apple juice cases. - Subtract that cost from the overall total, then divide by the water price. - For part b, the amount left for water must be divisible by the price of one case.

Solution

1. The apple juice costs \(44 \times \$14 = \$616\). 2. The amount spent on water is \(\$952 - \$616 = \$336\). 3. The number of water cases is \(\$336 \div \$6 = 56\). 4. For a total of \(\$960\), the amount left for water would be \(\$960 - \$616 = \$344\). 5. Since \(344 = 6 \times 57 + 2\), \(\$344\) cannot buy a whole number of \(\$6\) cases.

Answer

a) \(56\) cases of bottled water b) The claim is impossible. After subtracting the \(\$616\) apple juice cost, \(\$344\) would remain. Since \(344 = 6 \times 57 + 2\), that amount is not divisible by \(6\).
5201784
A school garden harvests \(80\,\text{lb}\) of apples. On Monday, one-fourth of the apples is pressed for juice. On Tuesday, one-fifth of the apples that remain is made into applesauce. On Wednesday, one-half of the apples then remaining is donated for a school celebration. How many pounds of apples remain at the end?

Hints

- Break the problem into one day at a time. - After each step, record how many pounds remain. - Check whether each fraction refers to the original amount or to the amount remaining. - A table or diagram may help you track the changing amount.

Solution

1. On Monday, \(\frac{1}{4} \times 80\,\text{lb} = 20\,\text{lb}\) is used, leaving \(80\,\text{lb} - 20\,\text{lb} = 60\,\text{lb}\). 2. On Tuesday, \(\frac{1}{5} \times 60\,\text{lb} = 12\,\text{lb}\) is used, leaving \(60\,\text{lb} - 12\,\text{lb} = 48\,\text{lb}\). 3. On Wednesday, \(\frac{1}{2} \times 48\,\text{lb} = 24\,\text{lb}\) is donated, leaving \(48\,\text{lb} - 24\,\text{lb} = 24\,\text{lb}\).

Answer

\(24\,\text{lb}\) of apples remain.
5202614
A hiker plans to walk an \(18\)-mile trail from Town A to Town B. After walking for a while, the hiker realizes that a water bottle was dropped, walks back to find it, and then continues to Town B. The hiker walks \(22\,\text{miles}\) altogether. a) How many extra miles does the hiker walk compared with the direct route? b) The hiker notices the missing bottle after walking exactly \(8\,\text{miles}\) from Town A. How far from Town A is the bottle found?

Hints

- Find the difference between the actual distance and the direct distance. - A backward section must also be walked forward again. - Use the turnaround point and the backward distance to locate the bottle.

Solution

1. Find the extra distance: \(22 - 18 = 4\,\text{miles}\). 2. The hiker walks the search distance twice: once backward and once forward. Therefore, the backward distance is \(4 \div 2 = 2\,\text{miles}\). 3. The hiker turns around at mile \(8\) and walks back \(2\,\text{miles}\): \(8 - 2 = 6\). 4. The bottle is found \(6\,\text{miles}\) from Town A.

Answer

a) The hiker walks \(4\,\text{miles}\) extra. b) The bottle is found \(6\,\text{miles}\) from Town A.
5203844
A school cafeteria has three crates of apples with a total mass of \(75\,\text{kg}\). The first and second crates have the same mass. The third crate has \(15\,\text{kg}\) more than the first crate. How many kilograms of apples are in each crate?

Hints

- Imagine removing the extra \(15\,\text{kg}\) from the third crate. - Divide the remaining total equally among the three crates. - Add the extra \(15\,\text{kg}\) back to the third crate.

Solution

1. Remove the extra \(15\,\text{kg}\) from the total: \(75\,\text{kg} - 15\,\text{kg} = 60\,\text{kg}\). 2. Divide the remaining mass equally among three crates: \(60\,\text{kg} \div 3 = 20\,\text{kg}\). 3. The first and second crates each have \(20\,\text{kg}\). 4. Add the extra mass for the third crate: \(20\,\text{kg} + 15\,\text{kg} = 35\,\text{kg}\). 5. Check: \(20\,\text{kg} + 20\,\text{kg} + 35\,\text{kg} = 75\,\text{kg}\).

Answer

The first crate has \(20\,\text{kg}\), the second crate has \(20\,\text{kg}\), and the third crate has \(35\,\text{kg}\).
5205214
An orchard harvests \(52{,}480\,\text{kg}\) of apples. It harvests \(14{,}650\,\text{kg}\) fewer pears than apples and \(9320\,\text{kg}\) fewer plums than pears. The orchard sells \(95{,}000\,\text{kg}\) of fruit to grocery stores. The rest will be made into juice. How many kilograms of fruit remain for juice?

Hints

- Find the amount of each type of fruit. - Add the three harvested amounts. - Subtract the amount sold from the total harvest.

Solution

1. Find the mass of the pears: \(52{,}480\,\text{kg} - 14{,}650\,\text{kg} = 37{,}830\,\text{kg}\). 2. Start with \(37{,}830\,\text{kg}\) of pears and subtract \(9320\,\text{kg}\). The mass of the plums is \(28{,}510\,\text{kg}\). 3. Find the total harvest: \(52{,}480\,\text{kg} + 37{,}830\,\text{kg} + 28{,}510\,\text{kg} = 118{,}820\,\text{kg}\). 4. Subtract the amount sold: \(118{,}820\,\text{kg} - 95{,}000\,\text{kg} = 23{,}820\,\text{kg}\).

Answer

\(23{,}820\,\text{kg}\) of fruit remain for juice.
5205224
Three elementary schools collect paper for a recycling contest. Pine School collects \(18{,}560\,\text{kg}\). Meadow School collects \(4230\,\text{kg}\) more than Pine School. Hill School collects \(3150\,\text{kg}\) less than Meadow School. A recycling center has a container that holds \(60{,}000\,\text{kg}\). Will all the collected paper fit? If not, how many kilograms will not fit?

Hints

- Find how much each school collected. - Add the three amounts. - Compare the total with the container's capacity. - Subtract to find any excess.

Solution

1. Start with Pine School's \(18{,}560\,\text{kg}\) and add \(4230\,\text{kg}\). Meadow School collects \(22{,}790\,\text{kg}\). 2. Start with Meadow School's \(22{,}790\,\text{kg}\) and subtract \(3150\,\text{kg}\). Hill School collects \(19{,}640\,\text{kg}\). 3. Find the total: \(18{,}560\,\text{kg} + 22{,}790\,\text{kg} + 19{,}640\,\text{kg} = 60{,}990\,\text{kg}\). 4. Since \(60{,}990\,\text{kg} > 60{,}000\,\text{kg}\), all the paper will not fit. 5. Subtract the container's capacity of \(60{,}000\,\text{kg}\) from the total of \(60{,}990\,\text{kg}\). The excess is \(990\,\text{kg}\).

Answer

No. \(990\,\text{kg}\) of paper will not fit in the container.
5205304
Four fourth-grade classes collect paper for a recycling contest: - Class 4A: \(345{,}600\,\text{g}\) - Class 4B: \(120{,}850\,\text{g}\) - Class 4C: \(402{,}300\,\text{g}\) - Class 4D: \(98{,}760\,\text{g}\) a) What is the difference between the greatest and least amounts collected? b) The two classes with the least amounts combine their paper. How many more grams would they need to equal the amount collected by Class 4C?

Hints

- Identify the greatest and least amounts for part a. - For part b, identify and add the two least amounts. - Subtract their combined amount from Class 4C's amount.

Solution

1. The greatest amount is \(402{,}300\,\text{g}\), and the least amount is \(98{,}760\,\text{g}\). 2. Find the difference for part a: \(402{,}300\,\text{g} - 98{,}760\,\text{g} = 303{,}540\,\text{g}\). 3. The two least amounts are \(98{,}760\,\text{g}\) and \(120{,}850\,\text{g}\). 4. Combine them: \(98{,}760\,\text{g} + 120{,}850\,\text{g} = 219{,}610\,\text{g}\). 5. Find the amount still needed: \(402{,}300\,\text{g} - 219{,}610\,\text{g} = 182{,}690\,\text{g}\).

Answer

a) The difference is \(303{,}540\,\text{g}\). b) The two classes need \(182{,}690\,\text{g}\) more.
5205554
Three trucks carry a total of \(15{,}000\,\text{kg}\) of sand. The first and second trucks carry \(10{,}400\,\text{kg}\) altogether. The second and third trucks carry \(9100\,\text{kg}\) altogether. a) How many kilograms of sand does each truck carry? b) What is the difference between the greatest and least loads?

Hints

- Subtract the combined load of two trucks from the total to find the third truck's load. - Use each given pair to find another truck. - Compare the three loads, then subtract the least from the greatest.

Solution

1. Start with the total of \(15{,}000\,\text{kg}\) and subtract the first-and-second-truck total of \(10{,}400\,\text{kg}\). The third truck carries \(4600\,\text{kg}\). 2. Start with the total of \(15{,}000\,\text{kg}\) and subtract the second-and-third-truck total of \(9100\,\text{kg}\). The first truck carries \(5900\,\text{kg}\). 3. Start with the first-and-second-truck total of \(10{,}400\,\text{kg}\) and subtract the first truck's \(5900\,\text{kg}\). The second truck carries \(4500\,\text{kg}\). 4. The greatest load is \(5900\,\text{kg}\), and the least load is \(4500\,\text{kg}\). 5. Find the difference: \(5900\,\text{kg} - 4500\,\text{kg} = 1400\,\text{kg}\).

Answer

a) The first truck carries \(5900\,\text{kg}\), the second carries \(4500\,\text{kg}\), and the third carries \(4600\,\text{kg}\). b) The difference is \(1400\,\text{kg}\).
5205624
A nursery grows \(320\) red tulips and \(280\) yellow tulips. It grows \(150\) fewer daffodils than the total number of tulips. a) How many daffodils does the nursery grow? b) Suppose the nursery grew \(50\) more yellow tulips while the number of daffodils stayed \(150\) below the total number of tulips. By how much would the number of daffodils change? Explain without fully recalculating.

Hints

- First find the total number of tulips. - Read carefully what the daffodil count is \(150\) fewer than. - In part b, consider what happens when one quantity increases while a fixed difference is maintained.

Solution

1. Find the total number of tulips: \(320 + 280 = 600\). 2. Find the number of daffodils: \(600 - 150 = 450\). 3. In part b, the difference of \(150\) stays fixed. Increasing the tulip total by \(50\) therefore increases the daffodil count by \(50\).

Answer

a) The nursery grows \(450\) daffodils. b) The number of daffodils would increase by \(50\) because the tulip total increases by \(50\) while the difference remains fixed.
5205724
One crate has \(45\,\text{kg}\) more apples than a second crate. Then \(28\,\text{kg}\) of apples are removed from the first crate, and an unknown mass is removed from the second crate. Afterward, the first crate still has \(32\,\text{kg}\) more apples than the second crate. How many kilograms of apples were removed from the second crate?

Hints

- First find the difference after only the first crate loses \(28\,\text{kg}\). - Compare that difference with the final \(32\,\text{kg}\) difference. - Think about how removing apples from the second crate changes the difference.

Solution

1. If apples were removed only from the first crate, the difference would become \(45\,\text{kg} - 28\,\text{kg} = 17\,\text{kg}\). 2. The final difference is \(32\,\text{kg}\), which is \(32\,\text{kg} - 17\,\text{kg} = 15\,\text{kg}\) greater. 3. Removing apples from the second crate increases the first crate's lead, so \(15\,\text{kg}\) were removed from the second crate.

Answer

\(15\,\text{kg}\) of apples were removed from the second crate.
5208084
Two numbers have a sum of \(700\). The larger number is \(100\) more than the smaller number. What are the two numbers?

Hints

- Recall what “sum” and “difference” mean. - Imagine first that the two numbers were equal parts of \(700\). - Subtract the extra \(100\) from the total. - Divide the remaining amount into two equal parts.

Solution

1. Remove the extra \(100\) from the total: \(700 - 100 = 600\). 2. The remaining \(600\) represents two equal copies of the smaller number, so \(600 \div 2 = 300\). 3. The larger number is \(300 + 100 = 400\). 4. Check: \(300 + 400 = 700\) and \(400 - 300 = 100\).

Answer

The numbers are \(300\) and \(400\).
5209094
A giant tortoise was born in \(1888\) and lived for \(134\) years. An elephant in the same wildlife park was born when the tortoise was \(56\) years old. The elephant lived for \(71\) years. a) In what year did the tortoise die? b) In what year was the elephant born? c) In what year did the elephant die? Did it die before or after the tortoise?

Hints

- First find the tortoise’s birth and death years. - Use the tortoise’s age to find the elephant’s birth year. - Add the elephant’s lifespan to its birth year. - Compare the two death years.

Solution

1. Find the tortoise’s death year: \(1888 + 134 = 2022\). 2. Find the elephant’s birth year: \(1888 + 56 = 1944\). 3. Find the elephant’s death year: \(1944 + 71 = 2015\). 4. Since \(2015 < 2022\), the elephant died before the tortoise.

Answer

a) The tortoise died in \(2022\). b) The elephant was born in \(1944\). c) The elephant died in \(2015\), before the tortoise.
5209624
Three ribbons have a total length of \(150\,\text{cm}\). Ribbons A and C are the same length. Ribbon B is \(30\,\text{cm}\) shorter than Ribbon A. Find the length of each ribbon.

Hints

- Imagine making the shortest ribbon \(30\,\text{cm}\) longer. - Then all three ribbons would have the same length. - Divide the adjusted total into three equal parts. - Subtract \(30\,\text{cm}\) to recover Ribbon B’s length.

Solution

1. Imagine increasing Ribbon B by \(30\,\text{cm}\) so all three ribbons are equal. The adjusted total would be \(150\,\text{cm} + 30\,\text{cm} = 180\,\text{cm}\). 2. Divide the adjusted total equally: \(180\,\text{cm} \div 3 = 60\,\text{cm}\). 3. Ribbons A and C are each \(60\,\text{cm}\). 4. Ribbon B is \(60\,\text{cm} - 30\,\text{cm} = 30\,\text{cm}\).

Answer

Ribbon A: \(60\,\text{cm}\) Ribbon B: \(30\,\text{cm}\) Ribbon C: \(60\,\text{cm}\)
5209634
A bakery has four bags of flour with a total weight of \(220\,\text{lb}\). Bags 1 and 2 have the same weight. Bags 3 and 4 also have the same weight, and each is \(20\,\text{lb}\) heavier than Bag 1. How much does each bag weigh?

Hints

- Find the total extra weight in Bags 3 and 4. - Subtract the extra weight so that all four parts represent equal amounts. - Divide the remaining total into four equal parts. - Check that all four weights add to \(220\,\text{lb}\).

Solution

1. Bags 3 and 4 include a total of \(20\,\text{lb} + 20\,\text{lb} = 40\,\text{lb}\) of extra weight. 2. Subtract that extra weight: \(220\,\text{lb} - 40\,\text{lb} = 180\,\text{lb}\). 3. Divide the remaining weight into four equal parts: \(180\,\text{lb} \div 4 = 45\,\text{lb}\). 4. Bags 1 and 2 each weigh \(45\,\text{lb}\). 5. Bags 3 and 4 each weigh \(45\,\text{lb} + 20\,\text{lb} = 65\,\text{lb}\). 6. Check: \(45 + 45 + 65 + 65 = 220\).

Answer

Bag 1: \(45\,\text{lb}\) Bag 2: \(45\,\text{lb}\) Bag 3: \(65\,\text{lb}\) Bag 4: \(65\,\text{lb}\)
5210074
Two numbers have a sum of \(752{,}300\). The positive difference between the numbers is \(48{,}100\). What are the two numbers?

Hints

- Identify what the sum and positive difference tell you. - Combining the sum and the difference can create two copies of one number. - Check both the sum and the difference after finding the numbers.

Solution

1. Add the sum and difference: \(752{,}300 + 48{,}100 = 800{,}400\). This equals twice the greater number. 2. The greater number is \(800{,}400 \div 2 = 400{,}200\). 3. Subtract the difference to find the lesser number: \(400{,}200 - 48{,}100 = 352{,}100\). 4. Check: \(400{,}200 + 352{,}100 = 752{,}300\), and \(400{,}200 - 352{,}100 = 48{,}100\).

Answer

\(400{,}200\) and \(352{,}100\)
5211964
Two groups of children plant seedlings in a school garden. Group A has \(4\) children and plants \(48\) seedlings in \(3\) hours. Group B has \(3\) children and plants \(42\) seedlings in \(2\) hours. On average, how many more seedlings per hour does one child in Group B plant than one child in Group A?

Hints

- Find each group's average number of seedlings per child per hour separately. - For each group, account for the time first and then the number of children. - Finally, subtract the two individual hourly rates.

Solution

1. Find Group A's hourly total: \(48 \div 3 = 16\) seedlings. 2. Find one child's hourly rate in Group A: \(16 \div 4 = 4\) seedlings. 3. Find Group B's hourly total: \(42 \div 2 = 21\) seedlings. 4. Find one child's hourly rate in Group B: \(21 \div 3 = 7\) seedlings. 5. Find the difference between the individual rates: \(7 - 4 = 3\) seedlings per hour.

Answer

One child in Group B plants an average of \(3\) more seedlings per hour than one child in Group A.
5212884
A print shop must print \(720\) posters. During the first hour, one-sixth of all the posters are printed. During the second hour, \(55\) more posters are printed than during the first hour. The remaining posters are printed during the third hour. During which hour are the most posters printed? Find the number printed during each hour.

Hints

- Find the number printed during each hour in order. - How can you find the fraction printed during the first hour? - After finding the first two hourly amounts, how can you find the third? - Compare all three results.

Solution

1. Find the first-hour amount: \(\frac{1}{6} \times 720 = 120\). 2. Find the second-hour amount: \(120 + 55 = 175\). 3. Find the total for the first two hours: \(120 + 175 = 295\). 4. Find the third-hour amount: \(720 - 295 = 425\). 5. Since \(425 > 175 > 120\), the most posters are printed during the third hour.

Answer

The first hour has \(120\) posters, the second hour has \(175\), and the third hour has \(425\). The most posters are printed during the third hour.
5212954
Three elementary schools cover a combined distance of \(2450\,\text{miles}\) in a fundraising challenge. School A covers \(750\,\text{miles}\). Schools B and C cover the remaining distance, and School B covers exactly \(300\,\text{miles}\) more than School C. a) How many miles does each of Schools B and C cover? b) Each school set a goal of \(900\,\text{miles}\). Which school reaches the goal?

Hints

- First find the combined distance for Schools B and C. - Use the \(300\)-mile difference to split that total into two unequal parts. - Compare each school's distance with \(900\,\text{miles}\).

Solution

1. Find the distance covered by Schools B and C together: \(2450\,\text{miles} - 750\,\text{miles} = 1700\,\text{miles}\). 2. Remove the \(300\)-mile difference: \(1700\,\text{miles} - 300\,\text{miles} = 1400\,\text{miles}\). 3. Divide equally to find School C's distance: \(1400\,\text{miles} \div 2 = 700\,\text{miles}\). 4. Find School B's distance: \(700\,\text{miles} + 300\,\text{miles} = 1000\,\text{miles}\). 5. Compare the three distances with \(900\,\text{miles}\). Only School B reaches the goal.

Answer

a) School B covers \(1000\,\text{miles}\), and School C covers \(700\,\text{miles}\). b) Only School B reaches the \(900\)-mile goal.
5213474
A sporting-goods manufacturer makes \(6420\) basketballs and \(9150\) soccer balls. The same amount of material used for \(6\) soccer balls could make \(5\) basketballs. An employee says, “If we had used all the material only for basketballs, we could have made more than \(14{,}000\) basketballs.” Is the employee correct? Support your answer with calculations.

Hints

- First group the soccer balls in sets of \(6\). - Convert those groups to the equivalent number of basketballs. - Add that amount to the basketballs already produced. - Compare the result with \(14{,}000\).

Solution

1. The \(9150\) soccer balls form \(9150\div 6=1525\) groups of \(6\). 2. That material could make \(1525\times 5=7625\) basketballs. 3. The total possible number of basketballs is \(6420+7625=14{,}045\). 4. Since \(14{,}045>14{,}000\), the employee is correct.

Answer

Yes. The available material could have produced \(14{,}045\) basketballs.
5216394
A rectangular prism is \(6\,\text{cm}\) long, \(3\,\text{cm}\) wide, and \(2\,\text{cm}\) high. Its net can be arranged in different ways. a) In Arrangement 1, the \(6 \times 3\), \(6 \times 2\), \(6 \times 3\), and \(6 \times 2\) faces form a row with the \(6\,\text{cm}\) edges vertical. The two \(3 \times 2\) faces are attached above and below one \(6 \times 3\) face. What are the dimensions of the smallest rectangle that contains the net? b) In Arrangement 2, the \(3 \times 6\), \(3 \times 2\), \(3 \times 6\), and \(3 \times 2\) faces form a row with the \(3\,\text{cm}\) edges vertical. The two \(6 \times 2\) faces are attached above and below one \(3 \times 6\) face. What are the dimensions of the smallest rectangle that contains this net?

Hints

- Find the total length of each four-face row. - Determine how much the two attached faces add to the height. - Pay attention to which edge of each attached face touches the row.

Solution

1. In Arrangement 1, the row is \(3+2+3+2=10\,\text{cm}\) long. Its height is \(6\,\text{cm}\), and each attached face adds \(2\,\text{cm}\), so the total height is \(2+6+2=10\,\text{cm}\). The bounding rectangle is \(10\,\text{cm} \times 10\,\text{cm}\). 2. In Arrangement 2, the row is \(6+2+6+2=16\,\text{cm}\) long. Its height is \(3\,\text{cm}\), and each attached face adds \(2\,\text{cm}\), so the total height is \(2+3+2=7\,\text{cm}\). The bounding rectangle is \(16\,\text{cm} \times 7\,\text{cm}\).

Answer

a) \(10\,\text{cm} \times 10\,\text{cm}\) b) \(16\,\text{cm} \times 7\,\text{cm}\)
5383784
Anna has \(10\) stamps altogether and \(2\) more red stamps than blue stamps. Bo has as many red stamps as Anna has blue stamps, and Bo has \(9\) stamps altogether. <table><thead><tr><th>Child</th><th>Red stamps</th><th>Blue stamps</th></tr></thead><tbody><tr><td>Anna</td><td>?</td><td>?</td></tr><tr><td>Bo</td><td>?</td><td>?</td></tr></tbody></table> Complete the table.

Hints

- Start with the child whose row has the most information. - Test number pairs that satisfy all conditions, then use Anna's blue count to begin Bo's row.

Solution

1. Anna's \(10\) stamps split into \(6\) red and \(4\) blue because the red count is \(2\) greater. 2. Bo therefore has \(4\) red stamps. 3. Bo has \(9 - 4 = 5\) blue stamps.

Answer

<table><thead><tr><th>Child</th><th>Red stamps</th><th>Blue stamps</th></tr></thead><tbody><tr><td>Anna</td><td>\(6\)</td><td>\(4\)</td></tr><tr><td>Bo</td><td>\(4\)</td><td>\(5\)</td></tr></tbody></table>
5197924
A school-supply store sells notebooks for \(80\) cents and matching covers for \(35\) cents. A teacher buys the same number of notebooks and covers. She pays with three \(\$10\) bills and receives \(\$2.40\) in change. How many notebooks does she buy?

Hints

- Find the amount actually spent and convert it to cents. - Find the cost of one notebook-and-cover pair. - Use convenient multiples of the pair price to reach the total amount spent.

Solution

1. The teacher spends \(\$30.00 - \$2.40 = \$27.60\), or \(2760\) cents. 2. One notebook-and-cover pair costs \(80 + 35 = 115\) cents. 3. Twenty pairs cost \(20 \times 115 = 2300\) cents, leaving \(2760 - 2300 = 460\) cents. 4. Four more pairs cost \(4 \times 115 = 460\) cents. 5. The teacher buys \(20 + 4 = 24\) notebook-and-cover pairs, so she buys \(24\) notebooks.

Answer

The teacher buys \(24\) notebooks.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.