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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.

Two-step word problems

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5161703
A teacher has \(\$90\) for a class party. Drinks cost \(\$36\), snacks cost \(\$45\), and a fruit tray costs \(\$17\). Is the budget enough? If not, how much more is needed?

Hints

- Add the prices of all three items. - Compare the total cost with the budget. - Find the difference between the total cost and the budget.

Solution

1. Find the total cost: \(\$36 + \$45 = \$81\), and \(\$81 + \$17 = \$98\). 2. Since \(\$98 > \$90\), the budget is not enough. 3. Find the additional amount needed: \(\$98 - \$90 = \$8\).

Answer

No. The teacher needs \(\$8\) more.
5177873
A school garden has a vegetable bed that is \(15\,\text{m}\) long. Students extend it by \(6\,\text{m}\) on the left for strawberries and by \(8\,\text{m}\) on the right for herbs. How long is the garden bed after both extensions?

Hints

- Draw the original bed and the two extensions. - Add both new lengths to the original length. - Extending a length means adding, not subtracting.

Solution

1. Add the first extension: \(15 + 6 = 21\,\text{m}\). 2. Add the second extension: \(21 + 8 = 29\,\text{m}\).

Answer

The garden bed is \(29\,\text{m}\) long after both extensions.
5381393
Six more rings are added to the equipment room. Which other kind of equipment will then have the same number as the rings?
Figure for problem 538139

Hints

- Increase only the number of rings. - Find the new number of rings. - Compare that number with the other bars.

Solution

1. The number of rings changes from \(12\) to \(12 + 6 = 18\). 2. The graph shows \(18\) balls. 3. Therefore, there will be the same number of rings and balls.

Answer

There will be the same number of rings and balls.
5381943
The pie chart shows votes for ice cream flavors. 1) Which flavor has the smallest slice? 2) What is the difference between its value and the value of the most popular flavor?
Figure for problem 538194

Hints

- Find the smallest and largest slices. - Read their values. - Subtract the smaller value from the larger value.

Solution

1. Lemon has the smallest slice, with a value of \(4\). 2. Chocolate has the largest slice, with a value of \(13\). 3. The difference is \(13 - 4 = 9\).

Answer

1) Lemon has the smallest slice. 2) Its value is \(9\) less than chocolate's value.
5383453
The Lynx group has \(11\) blue cards and \(8\) green cards. The Otter group has \(7\) blue and \(12\) green. The Fox group has \(10\) of each color. The Badger group has \(13\) blue and \(6\) green. Which group has more blue cards than green cards and at least \(12\) blue cards?

Hints

- Test both conditions for each group. - Both conditions must be true for the same group.

Solution

1. Lynx has more blue than green, but only \(11\) blue cards. 2. Otter does not have more blue than green. 3. Fox has equal amounts. 4. Badger has \(13 > 6\) and \(13\) is at least \(12\).

Answer

The Badger group
5383463
Raffle tickets are counted by color. Monday has \(14\) red and \(9\) blue tickets. Tuesday has \(12\) red and \(11\) blue tickets. Wednesday has \(16\) red and \(8\) blue tickets. Thursday has \(10\) red and \(15\) blue tickets. Which two days have the same total number of tickets?

Hints

- Find the total for each day. - Compare the four totals.

Solution

1. Monday's total is \(14 + 9 = 23\). 2. Tuesday's total is \(12 + 11 = 23\). 3. Wednesday's total is \(16 + 8 = 24\). 4. Thursday's total is \(10 + 15 = 25\). 5. Monday and Tuesday have the same total.

Answer

Monday and Tuesday
5383513
On Monday and Friday, club attendance was as follows: Drumming had \(8\) and \(11\) students, Drama had \(12\) and \(10\), and Chess had \(9\) and \(9\). Which statement is true? A: Fewer students attend Drumming on Friday than on Monday. B: Two more students attend Drama on Monday than on Friday. C: One more student attends Chess on Friday than on Monday.

Hints

- Match each statement to the two numbers for that club. - Check all three statements before choosing.

Solution

1. Statement A is false because \(11 > 8\). 2. Statement B is true because \(12 - 10 = 2\). 3. Statement C is false because both values are \(9\).

Answer

Statement B is true.
5383623
A colored-pencil case has \(16\) pencils in the top section and \(13\) in the bottom section. Four pencils are added to the top, and two are removed from the bottom. How many pencils are in the case afterward?

Hints

- Find the new amount in each section. - Add the two new amounts.

Solution

1. The top section then has \(16 + 4 = 20\) pencils. 2. The bottom section then has \(13 - 2 = 11\) pencils. 3. Altogether, the case has \(20 + 11 = 31\) pencils.

Answer

There are \(31\) colored pencils in the case afterward.
5383673
Paul reads \(25\), \(17\), and \(19\) pages over three days. Rayan reads \(16\), \(24\), and \(21\) pages. Who reads more altogether, or do they tie?

Hints

- Find each student’s three-day total. - Compare the two totals.

Solution

1. Paul reads \(25+17+19=61\) pages. 2. Rayan reads \(16+24+21=61\) pages. 3. They tie.

Answer

Paul and Rayan tie with \(61\) pages each.
5383963
A music group practiced \(12\) flute sections on Monday and \(9\) on Wednesday, \(8\) guitar sections on Monday and \(13\) on Wednesday, and \(10\) drum sections on each day. Which instrument has the greatest difference between the two days?

Hints

- Find the difference between the two days for each instrument. - Compare the three differences.

Solution

1. The flute difference is \(12 - 9 = 3\). 2. The guitar difference is \(13 - 8 = 5\). 3. The drum difference is \(10 - 10 = 0\). 4. The greatest difference is \(5\), for guitar.

Answer

Guitar
5383983
A craft workshop has \(25\) pieces of cardboard and uses \(8\), has \(18\) balls of yarn and uses \(6\), and has \(30\) craft sticks and uses \(12\). Which material has the least amount remaining?

Hints

- Subtract the amount used from the amount available for each material. - Compare the three remaining amounts.

Solution

1. Cardboard remaining: \(25 - 8 = 17\). 2. Yarn remaining: \(18 - 6 = 12\). 3. Craft sticks remaining: \(30 - 12 = 18\). 4. The least remaining amount is \(12\), for yarn.

Answer

Yarn has the least amount remaining.
5384023
On Monday, Tuesday, and Wednesday, students checked out \(6\), \(8\), and \(5\) balls; \(4\), \(3\), and \(6\) jump ropes; and \(5\), \(7\), and \(4\) hula hoops. Which kind of equipment was checked out most often during the three days?

Hints

- Find the three-day total for each kind of equipment. - Compare the three totals.

Solution

1. Balls: \(6 + 8 + 5 = 19\). 2. Jump ropes: \(4 + 3 + 6 = 13\). 3. Hula hoops: \(5 + 7 + 4 = 16\). 4. The greatest total is \(19\).

Answer

Balls were checked out most often, with \(19\) checkouts.
5156623
Lucas buys \(8\) trading cards each week for \(4\) weeks. He then receives \(15\) more cards for his birthday. How many cards are in his new collection altogether?

Hints

- First find how many cards Lucas buys over the four weeks. - Equal groups suggest multiplication. - Then add the cards he receives for his birthday.

Solution

1. Find the number of cards Lucas buys: \(4 \times 8 = 32\). 2. Add the birthday cards: \(32 + 15 = 47\).

Answer

Lucas has \(47\) cards altogether.
5156873
A school event begins with \(120\) bottles of apple juice, \(95\) bottles of orange juice, and \(215\) bottles of water. By the end of the event, \(312\) bottles have been used. How many full bottles remain?

Hints

- Add the three kinds of drinks to find the starting total. - Decide whether the number of full bottles increases or decreases when bottles are used. - Subtract the number used from the starting total.

Solution

1. Add to find the total number of bottles at the start: \(120 + 95 + 215 = 430\). 2. Subtract the number used: \(430 - 312 = 118\).

Answer

There are \(118\) full bottles remaining.
5156883
A bird park has \(42\) penguins. It has \(15\) more flamingos than penguins. The park also has \(124\) parakeets. How many birds of these three kinds are in the park?

Hints

- First find the number of flamingos. - Then add the penguins, flamingos, and parakeets. - Check that all three groups are included.

Solution

1. Find the number of flamingos: \(42 + 15 = 57\). 2. Add all three groups of birds: \(42 + 57 + 124 = 223\).

Answer

The park has \(223\) birds of these three kinds.
5157413
Students are logging miles in a school bike challenge. Class 3A rides \(267\) miles, Class 3B rides \(284\) miles, and Class 3C rides \(259\) miles. a) How many miles do the three third-grade classes ride altogether? b) The fourth-grade classes ride \(945\) miles altogether. How many more miles do they ride than the third-grade classes?

Hints

- Add the three third-grade distances first. - To find how many more, subtract the smaller total from the larger total. - Check regrouping carefully.

Solution

1. Add the distances for the third-grade classes: \(267 + 284 + 259 = 810\). The third-grade classes ride \(810\) miles altogether. 2. Subtract to find the difference: \(945 - 810 = 135\).

Answer

a) The third-grade classes ride \(810\) miles altogether. b) The fourth-grade classes ride \(135\) more miles.
5161573
Paul wants to buy an electronic keyboard and a stand. The prices are: <table> <tr><td>Electronic keyboard</td><td>\(\$215\)</td></tr> <tr><td>Keyboard stand</td><td>\(\$38\)</td></tr> </table> Paul has saved \(\$195\), and his uncle gives him \(\$70\) for his birthday. How much money will Paul have left after the purchase?

Hints

- Add the prices of the keyboard and stand. - Add Paul's savings and gift. - Subtract the total cost from the total amount available.

Solution

1. Find the total cost: \(\$215 + \$38 = \$253\). 2. Find the total amount Paul has: \(\$195 + \$70 = \$265\). 3. Subtract the cost from the amount available: \(\$265 - \$253 = \$12\).

Answer

Paul will have \(\$12\) left.
5161713
A sports club has a budget of \(\$250\) for new equipment. The club plans to buy soccer balls for \(\$125\), a set of cones for \(\$48\), and practice vests for \(\$82\). Find the total cost and decide whether the \(\$250\) budget is enough.

Hints

- Add all three prices to find the total cost. - Compare the total cost with the budget. - Subtract to find how far the total is above the budget.

Solution

1. Add the cost of the soccer balls and cones: \(\$125 + \$48 = \$173\). 2. Add the cost of the practice vests: \(\$173 + \$82 = \$255\). 3. Compare the total with the budget: \(\$255 > \$250\), so the budget is not enough. 4. Find how much more is needed: \(\$255 - \$250 = \$5\).

Answer

The equipment costs \(\$255\). The budget is not enough; the club needs \(\$5\) more.
5162083
Lucas lives \(70\,\text{m}\) from school. Each school day, he walks to school in the morning and walks home in the afternoon. How many meters does Lucas walk to and from school during a \(5\)-day school week?

Hints

- Determine how many times Lucas walks the route each day. - Find his total distance for one day. - Use the daily distance to find the distance for five days.

Solution

1. Find the distance Lucas walks each day: \(70\,\text{m} \times 2 = 140\,\text{m}\). 2. Find the distance for five days: \(140\,\text{m} \times 5 = 700\,\text{m}\).

Answer

Lucas walks \(700\,\text{m}\) during the school week.
5162093
Sophie goes to music lessons twice each week. Her home is \(110\,\text{m}\) from the music school, and she walks both ways each time. How many meters does Sophie walk for music lessons over two weeks?

Hints

- Find the distance Sophie walks for one round trip. - Account for both lessons in one week. - Then find the distance for two weeks.

Solution

1. Find the distance for one round trip: \(110\,\text{m} \times 2 = 220\,\text{m}\). 2. Find the distance for two lessons in one week: \(220\,\text{m} \times 2 = 440\,\text{m}\). 3. Find the distance for two weeks: \(440\,\text{m} \times 2 = 880\,\text{m}\).

Answer

Sophie walks \(880\,\text{m}\) over two weeks.
5162103
Tim and Sarah walk to a playground on \(5\) days in one week. Tim walks \(120\,\text{m}\) to the playground and back each day. Sarah walks \(160\,\text{m}\) to the playground and back each day. How many more meters does Sarah walk than Tim during the week?

Hints

- You may find each child's weekly distance and then subtract. - A faster method is to find the difference for one day first. - The question asks for the difference, not either child's total distance.

Solution

1. Find the difference in their daily distances: \(160\,\text{m} - 120\,\text{m} = 40\,\text{m}\). 2. Find the difference over five days: \(40\,\text{m} \times 5 = 200\,\text{m}\).

Answer

Sarah walks \(200\,\text{m}\) more than Tim during the week.
5162763
During recess, Mia runs for \(4\) minutes at \(100\,\text{m}\) per minute. She then walks for \(9\) minutes at \(60\,\text{m}\) per minute. How many meters does Mia travel altogether?

Hints

- Find the running and walking distances separately. - Multiply each distance per minute by the number of minutes. - Add the two distances.

Solution

1. Find the running distance: \(4 \times 100\,\text{m} = 400\,\text{m}\). 2. Find the walking distance: \(9 \times 60\,\text{m} = 540\,\text{m}\). 3. Add the two distances: \(400\,\text{m} + 540\,\text{m} = 940\,\text{m}\).

Answer

Mia travels \(940\,\text{m}\) altogether.
5162833
The Weber family plans a three-day hike that is \(78\) miles long. They hike \(24\) miles on the first day and \(29\) miles on the second day. They hike the rest on the third day. On which day do they hike the greatest distance?

Hints

- Add the distances from the first two days. - Subtract that sum from the total distance. - Compare all three daily distances.

Solution

1. Add the distances from the first two days: \(24 + 29 = 53\) miles. 2. Subtract from the total to find the third-day distance: \(78 - 53 = 25\) miles. 3. Compare the three distances: \(24\), \(29\), and \(25\). The greatest is \(29\).

Answer

The family hikes the greatest distance on the second day: \(29\) miles.
5162843
A class takes a four-day bicycle trip. The total distance is \(215\) miles. <table> <tr> <th colspan="4">Total distance: \(215\) miles</th> </tr> <tr> <td>Day 1: \(52\) miles</td> <td>Day 2: \(64\) miles</td> <td>Day 3: \(?\) miles</td> <td>Day 4: \(48\) miles</td> </tr> </table> How many miles does the class ride on Day 3? Which day has the shortest distance?

Hints

- Add the three known daily distances. - Subtract their sum from the total distance. - Compare all four daily distances.

Solution

1. Add the known distances: \(52 + 64 + 48 = 164\) miles. 2. Subtract from the total to find the Day 3 distance: \(215 - 164 = 51\) miles. 3. Compare \(52\), \(64\), \(51\), and \(48\). The smallest distance is \(48\) miles.

Answer

The class rides \(51\) miles on Day 3. Day 4 is the shortest day at \(48\) miles.
5174233
A gardener harvests \(32\,\text{kg}\) of apples and divides them equally among \(4\) crates. a) How many kilograms of apples are in one crate? b) How many kilograms are in \(3\) crates?

Hints

- First find the weight in one crate when all four crates hold equal amounts. - Once you know the weight in one crate, how can you find the weight in three crates?

Solution

1. Find the weight in one crate: \(32\,\text{kg} \div 4 = 8\,\text{kg}\). 2. Find the weight in three crates: \(8\,\text{kg} \times 3 = 24\,\text{kg}\).

Answer

a) One crate contains \(8\,\text{kg}\) of apples. b) Three crates contain \(24\,\text{kg}\).
5174243
A farm packs \(45\,\text{kg}\) of potatoes equally into \(9\) bags. a) How many kilograms are in each bag? b) A family buys \(2\) bags, and a restaurant buys \(4\) bags. How many kilograms are sold altogether? c) How many bags remain?

Hints

- First find the weight of one bag. - Add the numbers of bags purchased. - Use the number sold to find both the weight sold and the number remaining.

Solution

1. Find the weight of one bag: \(45\,\text{kg} \div 9 = 5\,\text{kg}\). 2. Find the number of bags sold: \(2 + 4 = 6\). 3. Find the weight sold: \(6 \times 5\,\text{kg} = 30\,\text{kg}\). 4. Find the bags remaining: \(9 - 6 = 3\).

Answer

a) Each bag weighs \(5\,\text{kg}\). b) \(30\,\text{kg}\) of potatoes are sold. c) \(3\) bags remain.
5174333
At a school cafeteria, \(4\) servings of fruit salad cost \(\$12\) altogether. a) How much does one serving cost? b) How much do \(7\) servings cost? c) How much do \(10\) servings cost?

Hints

- First find the cost of one serving. - Once you know the unit price, you can find the cost of any number of servings. - Which operation finds the cost of several servings?

Solution

1. Find the cost of one serving: \(\$12 \div 4 = \$3\). 2. Find the cost of seven servings: \(7 \times \$3 = \$21\). 3. Find the cost of ten servings: \(10 \times \$3 = \$30\).

Answer

a) One serving costs \(\$3\). b) Seven servings cost \(\$21\). c) Ten servings cost \(\$30\).
5174343
A gardener plants \(27\) tulips equally in \(3\) rows. a) How many tulips are in one row? b) How many tulips would be in \(8\) rows of the same size? c) The gardener has \(45\) more tulips. How many additional rows of the same size can be planted?

Hints

- First find the number of tulips in one row. - Use that result to answer the other parts. - For part c), determine how many groups of one row fit into \(45\).

Solution

1. Find the number of tulips in one row: \(27 \div 3 = 9\). 2. Find the number of tulips in eight rows: \(8 \times 9 = 72\). 3. Find the number of additional rows: \(45 \div 9 = 5\).

Answer

a) One row contains \(9\) tulips. b) Eight rows contain \(72\) tulips. c) The remaining tulips are enough for \(5\) additional rows.
5174373
Lucas buys a notebook for \(\$7\) and \(4\) identical pencils. He pays \(\$15\) altogether. How much does one pencil cost?

Hints

- First find how much Lucas spends on the pencils altogether. - Then divide that amount equally among the four pencils. - Use the total cost and the notebook price.

Solution

1. Find the cost of all four pencils: \(\$15 - \$7 = \$8\). 2. Divide by the number of pencils: \(\$8 \div 4 = \$2\).

Answer

One pencil costs \(\$2\).
5174453
Lucas buys a poster for \(\$14\). Two identical picture frames cost \(\$6\) more altogether than the poster. How much does one picture frame cost?

Hints

- First find the total cost of the two frames. - Then divide that cost equally between the two frames. - The two frames cost more than the poster.

Solution

1. Find the cost of both frames: \(\$14 + \$6 = \$20\). 2. Divide by two: \(\$20 \div 2 = \$10\).

Answer

One picture frame costs \(\$10\).
5174463
A teacher buys \(8\) identical packages of markers. The teacher pays with \(\$50\) and receives \(\$18\) in change. How much does one package cost?

Hints

- First find how much the teacher spends altogether. - Then divide the total cost among the eight equal packages. - Change is subtracted from the amount paid.

Solution

1. Find the total amount spent: \(\$50 - \$18 = \$32\). 2. Divide by the number of packages: \(\$32 \div 8 = \$4\).

Answer

One package of markers costs \(\$4\).
5174713
A gym has \(14\) boys and \(16\) girls. They form groups of \(5\) students for a game. How many groups are formed?

Hints

- First find the total number of students. - Then divide the students into equal groups. - Solve the problem in two steps.

Solution

1. Find the total number of students: \(14 + 16 = 30\). 2. Divide into groups of five: \(30 \div 5 = 6\).

Answer

The students form \(6\) groups.
5174723
A nursery has \(160\) tulips and \(240\) daffodils. The flowers are arranged into bouquets with exactly \(8\) flowers in each bouquet. How many bouquets can be made altogether?

Hints

- Decide what you need to find first. - How many flowers are available altogether? - Which operation helps you make equal groups of flowers?

Solution

1. Find the total number of flowers: \(160 + 240 = 400\). 2. Divide by the number of flowers in each bouquet: \(400 \div 8 = 50\).

Answer

The nursery can make \(50\) bouquets.
5174813
Lucas receives \(\$15\) from his grandmother and \(\$13\) from his grandfather. He wants to buy toy cars that cost \(\$4\) each. How many toy cars can Lucas buy with all of the money?

Hints

- How much money does Lucas receive altogether? - Think about how many times the price of one car fits into the total amount. - Which operation divides a total into equal-size amounts?

Solution

1. Find the total amount of money: \(\$15 + \$13 = \$28\). 2. Divide by the cost of one toy car: \(\$28 \div \$4 = 7\).

Answer

Lucas can buy \(7\) toy cars.
5174823
Ms. Weber buys notebooks for her class. She pays with a \(\$20\) bill and a \(\$5\) bill and receives \(\$4\) in change. Each notebook costs \(\$3\). How many notebooks does Ms. Weber buy?

Hints

- First find the total amount Ms. Weber gives the cashier. - How much does she actually spend after receiving change? - Once you know the total cost, how can you find the number of notebooks?

Solution

1. Find the amount Ms. Weber gives the cashier: \(\$20 + \$5 = \$25\). 2. Subtract the change to find the amount spent: \(\$25 - \$4 = \$21\). 3. Divide by the cost of one notebook: \(\$21 \div \$3 = 7\).

Answer

Ms. Weber buys \(7\) notebooks.
5175333
A baker needs \(24\) apples to make \(4\) identical apple pies. How many apples are needed to make \(9\) of these pies?

Hints

- First find the number of apples needed for one pie. - Which operation divides a total equally? - Once you know the amount for one pie, how can you find the amount for nine pies?

Solution

1. Find the number of apples needed for one pie: \(24 \div 4 = 6\). 2. Find the number needed for nine pies: \(6 \times 9 = 54\).

Answer

The baker needs \(54\) apples.
5175393
Five classes set up booths for a school fair. At each booth, \(2\) adults and \(6\) students help. How many people help at all five booths altogether?

Hints

- Find the number of people at one booth. - Then use equal groups to find the total for five booths. - You can also find the total numbers of adults and students separately.

Solution

1. Find the number of helpers at one booth: \(2 + 6 = 8\). 2. Find the number at five booths: \(5 \times 8 = 40\).

Answer

There are \(40\) helpers altogether.
5175403
A store makes school-supply sets. Each set contains \(1\) pen and \(5\) ink refills. a) How many items are in \(9\) sets altogether? b) The store adds \(1\) highlighter to each set. How many items are now in the \(9\) sets? Explain how to use your answer from part a without recounting every item.

Hints

- First find how many items are in one original set. - Determine how many items are added across all nine sets. - Use the first answer to find the second total.

Solution

1. Each original set contains \(1 + 5 = 6\) items. 2. For part a, \(9 \times 6 = 54\) items. 3. Adding one highlighter to each of \(9\) sets adds \(9\) items. 4. For part b, \(54 + 9 = 63\) items.

Answer

a) The \(9\) sets contain \(54\) items. b) They contain \(63\) items. Add \(9\) to the first total because one item is added to each of the \(9\) sets.
5175633
A school orders \(5\) packages with \(20\) wide-ruled notebooks in each package and \(3\) packages with \(50\) graph-paper notebooks in each package. How many notebooks does the school receive altogether?

Hints

- Find the number of notebooks in each type of package. - Use multiplication for each set of equal groups. - Add the two totals.

Solution

1. Find the number of wide-ruled notebooks: \(5 \times 20 = 100\). 2. Find the number of graph-paper notebooks: \(3 \times 50 = 150\). 3. Add the two amounts: \(100 + 150 = 250\).

Answer

The school receives \(250\) notebooks altogether.
5175663
A garden has \(18\) tulips planted equally in \(3\) flower beds. a) How many tulips are in each flower bed? b) How many tulips are needed for \(8\) flower beds with the same number in each bed? c) How many of these flower beds can be filled with \(30\) tulips?

Hints

- First find the number of tulips in one flower bed. - Use that number to find the amount for \(8\) beds. - For part c), determine how many groups of that size are in \(30\).

Solution

1. Find the number in each flower bed: \(18 \div 3 = 6\) tulips. 2. For \(8\) flower beds: \(8 \times 6 = 48\) tulips. 3. With \(30\) tulips: \(30 \div 6 = 5\) flower beds.

Answer

a) \(6\) tulips b) \(48\) tulips c) \(5\) flower beds
5175673
A garden cart carries boxes of apples. Six equal-weight boxes weigh \(30\,\text{kg}\) altogether. a) How much do \(9\) boxes weigh? b) The cart can carry at most \(50\,\text{kg}\). Can it carry \(11\) boxes at once? Use a calculation to explain.

Hints

- Find the weight of one box first. - Use that weight to find the weight of \(9\) boxes. - For \(11\) boxes, find the weight of \(10\) boxes and add one more box. - Compare the weight of \(11\) boxes with the cart's limit.

Solution

1. Find the weight of one box: \(30\,\text{kg} \div 6 = 5\,\text{kg}\). 2. Nine boxes weigh \(9 \times 5\,\text{kg} = 45\,\text{kg}\). 3. Ten boxes weigh \(10 \times 5\,\text{kg} = 50\,\text{kg}\). One more box adds \(5\,\text{kg}\), so \(11\) boxes weigh \(55\,\text{kg}\). 4. Since \(55\,\text{kg} > 50\,\text{kg}\), the cart cannot carry all \(11\) boxes at once.

Answer

a) \(45\,\text{kg}\) b) No. The \(11\) boxes weigh \(55\,\text{kg}\), which is more than the \(50\,\text{kg}\) limit.
5175803
Luke uses \(8\) ounces of flour to bake \(4\) muffins. How many ounces of flour does he need for \(6\) muffins? How many ounces does he need for \(9\) muffins?

Hints

- Find the amount of flour for one muffin. - Use that amount to find the flour needed for each larger batch. - A table can help you match numbers of muffins with amounts of flour.

Solution

1. Find the amount for one muffin: \(8\,\text{oz} \div 4 = 2\,\text{oz}\). 2. For \(6\) muffins: \(6 \times 2\,\text{oz} = 12\,\text{oz}\). 3. For \(9\) muffins: \(9 \times 2\,\text{oz} = 18\,\text{oz}\).

Answer

Luke needs \(12\,\text{oz}\) of flour for \(6\) muffins and \(18\,\text{oz}\) for \(9\) muffins.
5175813
Three equal packages contain \(24\) colored pencils altogether. a) How many colored pencils are in \(7\) packages? b) How many packages are needed for \(48\) colored pencils?

Hints

- First find the number of colored pencils in one package. - Use that number to find the amount in \(7\) packages. - For part b), determine how many equal groups are in \(48\).

Solution

1. Find the number in one package: \(24 \div 3 = 8\). 2. Seven packages contain \(7 \times 8 = 56\) colored pencils. 3. For \(48\) colored pencils: \(48 \div 8 = 6\) packages.

Answer

a) \(56\) colored pencils b) \(6\) packages
5175923
A nursery delivers \(120\) seedlings for a school garden. The teachers plant \(30\) seedlings first. The remaining seedlings are shared equally among \(6\) classes for their own garden beds. How many seedlings does each class receive?

Hints

- Restate the problem in your own words. - First find how many seedlings remain after the teachers plant some. - Which operation shares a quantity equally among several groups?

Solution

1. Find the number of seedlings remaining: \(120 - 30 = 90\). 2. Divide the remaining seedlings equally among the classes: \(90 \div 6 = 15\).

Answer

Each class receives \(15\) seedlings.
5175933
A tennis club buys \(100\) new tennis balls. After the first practice, \(28\) balls are missing. The remaining balls are stored in cans that hold exactly \(4\) balls each. How many cans are needed?

Hints

- What should you do first to find how many balls remain? - How many balls fit in one can? - Think of separating the remaining balls into groups of four.

Solution

1. Find the number of tennis balls remaining: \(100 - 28 = 72\). 2. Divide by the number of balls in each can: \(72 \div 4 = 18\).

Answer

The club needs \(18\) cans.
5176183
During physical education class, a teacher divides \(40\) jump ropes equally among \(8\) groups. How many jump ropes do \(5\) groups receive altogether?

Hints

- First find how many jump ropes one group receives. - Once you know the amount for one group, how can you find the amount for five groups? - Which operation helps with equal sharing?

Solution

1. Find the number of jump ropes for one group: \(40 \div 8 = 5\). 2. Find the number for five groups: \(5 \times 5 = 25\).

Answer

Five groups receive \(25\) jump ropes altogether.
5176193
For a school event, Class 3A buys \(4\) bags containing \(32\) oranges altogether. Class 3B buys \(7\) bags of the same size. How many more oranges does Class 3B have than Class 3A?

Hints

- How many oranges are in one bag? - How many oranges does Class 3B buy altogether? - What is the difference between the two amounts? - You can also compare the numbers of bags first.

Solution

1. Find the number of oranges in one bag: \(32 \div 4 = 8\). 2. Find the number of oranges Class 3B buys: \(7 \times 8 = 56\). 3. Find the difference: \(56 - 32 = 24\).

Answer

Class 3B has \(24\) more oranges than Class 3A.
5176293
Five identical museum tickets for a class trip cost \(\$35\) altogether. How much do \(9\) of these tickets cost?

Hints

- First find the cost of one ticket. - Which operation divides the total cost equally among the tickets? - Once you know the unit price, how can you find the cost of nine tickets?

Solution

1. Find the cost of one ticket: \(\$35 \div 5 = \$7\). 2. Find the cost of nine tickets: \(9 \times \$7 = \$63\).

Answer

Nine museum tickets cost \(\$63\) altogether.
5176303
At a nursery, \(6\) identical rose bushes cost \(\$54\) altogether. Mr. Schmidt wants to buy \(12\) of these rose bushes. a) How much do the \(12\) rose bushes cost altogether? b) Can you find the answer without first finding the cost of one rose bush? Explain.

Hints

- Look closely at the relationship between \(6\) and \(12\). - What happens to the cost when the quantity is doubled? - You may use either a unit price or the relationship between the quantities.

Solution

1. One method is to find the unit price: \(\$54 \div 6 = \$9\), then calculate \(12 \times \$9 = \$108\). 2. Another method uses the relationship between the quantities. Since \(12\) is twice \(6\), the cost is twice \(\$54\). 3. Calculate the doubled cost: \(2 \times \$54 = \$108\).

Answer

a) The twelve rose bushes cost \(\$108\) altogether. b) Yes. Since twelve is twice six, double \(\$54\) to get \(\$108\).
5176713
A preschool craft group buys \(5\) packages of colorful beads for \(\$15\) altogether. Another group buys the same kind of beads and spends \(\$27\). How many packages does the second group buy?

Hints

- First find the cost of one package. - Once you know the cost per package, determine how many times it fits into the second group's total cost.

Solution

1. Find the cost of one package: \(\$15 \div 5 = \$3\). 2. Divide the second group's total cost by the cost per package: \(\$27 \div \$3 = 9\).

Answer

The second group buys \(9\) packages of beads.
5176963
A gardener buys \(5\) equal bags containing \(40\) flower bulbs altogether. The gardener needs \(72\) bulbs. How many bags are needed altogether?

Hints

- Find the number of bulbs in one bag first. - Use division to share a total equally among bags. - Then determine how many groups of that size are in \(72\).

Solution

1. Find the number of bulbs in one bag: \(40 \div 5 = 8\). 2. Find the number of bags needed: \(72 \div 8 = 9\).

Answer

The gardener needs \(9\) bags altogether.
5176973
Four equal packages of colored pencils cost \(\$32\) altogether. a) How much do \(9\) packages cost? b) A teacher has \(\$56\). How many packages can the teacher buy?

Hints

- Find the cost of one package first. - Use that cost to find the price of \(9\) packages. - Recall the multiplication facts that use the one-package price. - For part b), determine how many one-package prices fit into the budget.

Solution

1. Find the cost of one package: \(\$32 \div 4 = \$8\). 2. Nine packages cost \(9 \times \$8 = \$72\). 3. With \(\$56\), the teacher can buy \(\$56 \div \$8 = 7\) packages.

Answer

a) \(\$72\) b) \(7\) packages
5176983
Finn pays \(\$12\) for \(4\) equal packages of trading cards. He has saved \(\$21\). How many of these packages can he buy with all of his savings?

Hints

- Find the cost of one package first. - Then determine how many one-package prices fit into \(\$21\). - A table can help you match packages with prices.

Solution

1. Find the cost of one package: \(\$12 \div 4 = \$3\). 2. Find the number of packages Finn can buy: \(\$21 \div \$3 = 7\).

Answer

Finn can buy \(7\) packages of trading cards.
5177853
A school cafeteria has \(85\) apples in a crate. There are \(25\) fewer pears than apples. The number of bananas equals the total number of apples and pears. How many bananas are in the crate?

Hints

- Find the number of pears first. - Use “fewer than” to choose the first operation. - Then add the apples and pears.

Solution

1. Find the number of pears: \(85 - 25 = 60\). 2. Add the apples and pears to find the number of bananas: \(85 + 60 = 145\).

Answer

There are \(145\) bananas in the crate.
5178163
Lucas has \(38\) stickers in his first album. His second album has \(14\) more stickers than the first album. How many stickers does Lucas have in both albums altogether?

Hints

- First find the number of stickers in the second album. - Then add the amounts in the two albums. - Make sure each number in the problem is used correctly.

Solution

1. Find the number of stickers in the second album: \(38 + 14 = 52\). 2. Add the stickers in both albums: \(38 + 52 = 90\).

Answer

Lucas has \(90\) stickers in both albums altogether.
5178473
Leon has \(25\) trading cards. Ben says, “If I give you \(8\) of my cards, we will have the same number.” How many cards does Ben have before giving Leon the cards?

Hints

- First find how many cards Leon has after the gift. - At that time, Ben has the same number. - Work backward to find Ben's starting amount.

Solution

1. After receiving \(8\) cards, Leon has \(25+8=33\) cards. 2. Ben also has \(33\) cards after giving away the cards because their amounts are equal then. 3. Before giving away \(8\) cards, Ben had \(33+8=41\) cards.

Answer

Ben has \(41\) cards at the start.
5178503
Eight apple trees are planted in one straight row. The distance between each pair of neighboring trees is \(4\,\text{m}\). What is the distance from the first tree to the last tree?

Hints

- Draw the trees in a row. - Count the spaces between the trees, not the trees themselves. - Multiply the number of spaces by the distance of each space.

Solution

1. Eight trees create \(8 - 1 = 7\) spaces between neighboring trees. 2. Multiply the number of spaces by the length of each space: \(7 \times 4\,\text{m} = 28\,\text{m}\).

Answer

The distance from the first tree to the last tree is \(28\,\text{m}\).
5178513
A \(12\,\text{m}\)-long string of pennant flags has a flag at each end. There are \(5\) flags altogether, placed at equal distances. What is the distance between each pair of neighboring flags?

Hints

- Draw \(5\) flags in a row and count the spaces between them. - Remember that \(5\) objects in a row create \(4\) spaces between them. - Divide the total length equally among those spaces.

Solution

1. Five flags create \(5 - 1 = 4\) equal spaces. 2. Divide the total length among the spaces: \(12\,\text{m} \div 4 = 3\,\text{m}\).

Answer

The distance between neighboring flags is \(3\,\text{m}\).
5178723
Ms. Miller buys \(8\) packs with \(7\) glitter stickers in each pack and \(9\) packs with \(5\) star stickers in each pack. How many stickers does she buy altogether?

Hints

- Find the number of each kind of sticker separately. - Use multiplication for the equal groups. - Add the two totals.

Solution

1. Find the number of glitter stickers: \(8 \times 7 = 56\). 2. Find the number of star stickers: \(9 \times 5 = 45\). 3. Add the two amounts: \(56 + 45 = 101\).

Answer

Ms. Miller buys \(101\) stickers altogether.
5178813
A stationery store delivers \(100\) new notebooks to a school. Class 3A receives \(28\) notebooks, and Class 3B receives \(32\) notebooks. The remaining notebooks are divided equally among \(5\) study groups. How many notebooks does each study group receive?

Hints

- How many notebooks do the two classes receive altogether? - How many notebooks remain after that? - Which operation divides the remaining notebooks equally among five groups? - First find the total already distributed before sharing the rest.

Solution

1. Find the number of notebooks already distributed: \(28 + 32 = 60\). 2. Find the number of notebooks remaining: \(100 - 60 = 40\). 3. Divide the remaining notebooks equally among five groups: \(40 \div 5 = 8\).

Answer

Each study group receives \(8\) notebooks.
5179163
Emma buys \(4\) notebooks for \(\$3\) each and a watercolor set for \(\$14\). After paying, she has \(\$24\) left. How much money did Emma have before shopping?

Hints

- First find the total amount Emma spent. - Her starting amount includes both what she spent and what remained. - Break the work into smaller steps.

Solution

1. Find the cost of the notebooks: \(4 \times \$3 = \$12\). 2. Find the total amount spent: \(\$12 + \$14 = \$26\). 3. Add the amount spent and the money left: \(\$26 + \$24 = \$50\).

Answer

Emma had \(\$50\) before shopping.
5179583
Two swim teams practice three times each week. The Minnows practice for \(45\) minutes each time, and the Sharks practice for \(60\) minutes each time. Over \(4\) weeks, how many more minutes do the Sharks practice than the Minnows?

Hints

- Find the difference in time for one practice. - Determine the difference for three practices in one week. - Extend the weekly difference across four weeks.

Solution

1. Find the difference for one practice: \(60 - 45 = 15\) minutes. 2. Find the weekly difference: \(3 \times 15 = 45\) minutes. 3. Find the difference over four weeks: \(4 \times 45 = 180\) minutes.

Answer

The Sharks practice \(180\) more minutes over four weeks.
5179593
Tickets for a school play cost \(\$9\) for each adult and \(\$5\) for each child. A family pass for two adults and two children costs \(\$25\). How much does a family save by buying the family pass instead of four individual tickets?

Hints

- Find the cost of the adult and child tickets separately. - Add to find the cost of four individual tickets. - Compare that total with the family-pass price.

Solution

1. Find the cost for two adults: \(2 \times \$9 = \$18\). 2. Find the cost for two children: \(2 \times \$5 = \$10\). 3. Find the cost of the individual tickets: \(\$18 + \$10 = \$28\). 4. Find the savings: \(\$28 - \$25 = \$3\).

Answer

The family saves \(\$3\).
5180103
Three packages contain \(24\) collectible stickers altogether, with the same number in each package. Jonas wants to collect \(56\) stickers. How many packages must he buy?

Hints

- How many stickers are in one package? - Once you know the package size, how many groups of that size make \(56\)?

Solution

1. Find the number of stickers in one package: \(24 \div 3 = 8\). 2. Find the number of packages needed: \(56 \div 8 = 7\).

Answer

Jonas must buy \(7\) packages.
5180113
A sprinkler uses exactly \(45\,\text{L}\) of water in \(5\) minutes at a constant rate. a) How much water does the sprinkler use in \(8\) minutes? b) How long does it take the sprinkler to use \(90\,\text{L}\)? Can you solve part b) without first finding the amount used in one minute?

Hints

- For part a), first find the amount used in one minute. - For part b), compare \(90\) with \(45\). - What happens to the time when the amount of water doubles?

Solution

1. Find the amount used per minute: \(45\,\text{L} \div 5 = 9\,\text{L}\). 2. Find the amount used in eight minutes: \(8 \times 9\,\text{L} = 72\,\text{L}\). 3. Find the time needed to use ninety liters: \(90\,\text{L} \div 9\,\text{L} = 10\) minutes. 4. Alternatively, \(90\,\text{L}\) is twice \(45\,\text{L}\), so the time is twice five minutes, or ten minutes.

Answer

a) The sprinkler uses \(72\,\text{L}\) in \(8\) minutes. b) It takes \(10\) minutes. Since \(90\,\text{L}\) is twice \(45\,\text{L}\), the time also doubles.
5180153
A school copier makes \(10\) copies in \(20\) seconds. If it works at a constant rate, how many copies can it make in one minute, or \(60\) seconds?

Hints

- How long does one copy take? - How many groups of twenty seconds fit into one minute? - If the copier has more time, will it make more or fewer copies?

Solution

1. Find the time needed for one copy: \(20\,\text{seconds} \div 10 = 2\,\text{seconds}\). 2. Find the number of copies made in sixty seconds: \(60\,\text{seconds} \div 2\,\text{seconds} = 30\).

Answer

The copier can make \(30\) copies in one minute.
5180203
At a grocery store, \(4\) packages of juice cost \(\$12\) altogether. Ms. Weber has \(\$45\) to buy juice for a school event. What is the greatest number of packages she can buy?

Hints

- First find the cost of one package. - Once you know the unit price, determine how many times it fits into the budget. - Work through the two operations in order.

Solution

1. Find the cost of one package: \(\$12 \div 4 = \$3\). 2. Divide the budget by the cost per package: \(\$45 \div \$3 = 15\).

Answer

Ms. Weber can buy at most \(15\) packages of juice.
5180323
Five bags of potting soil weigh \(40\,\text{kg}\) altogether. How much do \(8\) identical bags weigh altogether?

Hints

- First find the weight of one bag. - Once you know the weight of one bag, how can you find the weight of eight bags? - A table may help you organize the quantities.

Solution

1. Find the weight of one bag: \(40\,\text{kg} \div 5 = 8\,\text{kg}\). 2. Find the weight of eight bags: \(8 \times 8\,\text{kg} = 64\,\text{kg}\).

Answer

Eight bags weigh \(64\,\text{kg}\) altogether.
5180373
A machine at a chocolate factory packages \(48\) chocolate bars in \(6\) minutes. A newer machine packages \(4\) more bars each minute. How many chocolate bars does the newer machine package in one minute?

Hints

- First find how many bars the original machine packages in one minute. - Once you know that rate, how can you find the newer machine's rate? - What operation does the phrase “more bars” suggest?

Solution

1. Find how many bars the first machine packages each minute: \(48 \div 6 = 8\). 2. Add the number of extra bars packaged by the newer machine: \(8 + 4 = 12\).

Answer

The newer machine packages \(12\) chocolate bars in one minute.
5180383
Two gardeners are planting flowers. Jonas plants \(56\) flowers in \(8\) minutes. Mia plants \(3\) more flowers per minute than Jonas. How many flowers does Mia plant in \(5\) minutes?

Hints

- How many flowers does Jonas plant in one minute? - How many more flowers does Mia plant each minute? - Once you know Mia's one-minute rate, how can you find her five-minute total? - Be sure to answer for five minutes, not one minute.

Solution

1. Find the number of flowers Jonas plants per minute: \(56 \div 8 = 7\). 2. Find the number of flowers Mia plants per minute: \(7 + 3 = 10\). 3. Find the number Mia plants in five minutes: \(10 \times 5 = 50\).

Answer

Mia plants \(50\) flowers in \(5\) minutes.
5180483
Students in Class 3A use \(45\,\text{L}\) of water to irrigate the school garden over \(5\) days. Class 3B uses \(4\,\text{L}\) more each day than Class 3A. How many liters of water does Class 3B use in one day?

Hints

- First find how much water Class 3A uses in one day. - Which operation divides a total amount equally among several days? - Once you know Class 3A's daily use, how can you find Class 3B's?

Solution

1. Find Class 3A's daily water use: \(45\,\text{L} \div 5 = 9\,\text{L}\). 2. Add Class 3B's additional daily use: \(9\,\text{L} + 4\,\text{L} = 13\,\text{L}\).

Answer

Class 3B uses \(13\,\text{L}\) of water in one day.
5181003
Three children have \(80\) marbles altogether. After Tom loses \(8\) marbles, all three children have the same number of marbles. a) How many marbles does each child have now? b) How many marbles did Tom have at first? c) At first, did each of the other two children have more or fewer marbles than Tom?

Hints

- First find how many marbles remain after Tom loses some. - If all three children then have equal amounts, divide the remaining marbles equally. - How many marbles must Tom have had before losing eight?

Solution

1. Find the total number of marbles remaining: \(80 - 8 = 72\). 2. Find the number each child has now: \(72 \div 3 = 24\). 3. Find Tom's original number of marbles: \(24 + 8 = 32\). 4. The other two children did not lose any marbles, so each had \(24\) at first. Since \(24 < 32\), each had fewer marbles than Tom.

Answer

a) Each child has \(24\) marbles now. b) Tom had \(32\) marbles at first. c) Each of the other children had fewer marbles than Tom.
5181293
Students plant flowers in a school garden on three Fridays. They plant \(15\) flowers on the first Friday. On the second Friday, they plant three times as many as on the first. On the third Friday, they plant \(12\) more than on the second. 1) How many flowers do they plant on the third Friday? 2) Do they plant more flowers on the third Friday than on the first two Fridays combined? Justify your answer.

Hints

- Find the number planted on each Friday in order. - “Three times as many” indicates multiplication. - Compare the third-Friday amount with the sum of the first two amounts.

Solution

1. Find the number planted on the second Friday: \(15 \times 3 = 45\). 2. Find the number planted on the third Friday: \(45 + 12 = 57\). 3. Find the total from the first two Fridays: \(15 + 45 = 60\). 4. Compare: \(57 < 60\), so the third-Friday amount is smaller.

Answer

1) They plant \(57\) flowers on the third Friday. 2) No. They plant \(60\) flowers on the first two Fridays combined, and \(57 < 60\).
5181963
Paul reads \(45\) pages of a new book in \(5\) days. His friend Mia reads \(6\) more pages each day than Paul. How many pages does Mia read in one day?

Hints

- How many pages does Paul read in one day? - Once you know Paul's daily number, how can you find Mia's? - Which detail tells you whether Mia reads more or fewer pages than Paul?

Solution

1. Find the number of pages Paul reads each day: \(45 \div 5 = 9\). 2. Add the number of extra pages Mia reads: \(9 + 6 = 15\).

Answer

Mia reads \(15\) pages in one day.
5182143
A gardener pays \(\$35\) for a new garden hose. The hose costs \(\$5\) per foot. At home, the gardener cuts off a \(2\)-foot piece to water a small garden bed. How many feet of hose remain?

Hints

- First find how many feet of hose the gardener bought. - Which operation uses the total price and the price per foot? - When a piece is cut off, does the remaining length increase or decrease?

Solution

1. Find the original length of the hose: \(\$35 \div \$5 = 7\) feet. 2. Subtract the piece that was cut off: \(7 - 2 = 5\) feet.

Answer

The gardener has \(5\) feet of hose remaining.
5182153
A class buys a long blue ribbon for a school event for \(\$54\). The ribbon costs \(\$6\) per yard. The students first cut off \(3\) yards for a large poster. Then they use exactly half of the remaining ribbon to wrap small gifts. How many yards of ribbon remain at the end?

Hints

- First find the original length of the ribbon. - How much remains after three yards are used for the poster? - Half of which amount is used for the gifts? - What operation finds one-half of a quantity?

Solution

1. Find the original length of the ribbon: \(\$54 \div \$6 = 9\) yards. 2. Find the length remaining after the first cut: \(9 - 3 = 6\) yards. 3. Find half of the remaining ribbon: \(6 \div 2 = 3\) yards. 4. Subtract the ribbon used for the gifts: \(6 - 3 = 3\) yards.

Answer

\(3\) yards of ribbon remain.
5182233
A large paint set costs \(\$18\). One paintbrush costs one-sixth as much as the paint set. How much do the paint set and paintbrush cost altogether?

Hints

- First find the cost of the paintbrush. - What operation finds one-sixth of an amount? - Remember that the question asks for the cost of both items together.

Solution

1. Find the cost of the paintbrush: \(\$18 \div 6 = \$3\). 2. Add the two prices: \(\$18 + \$3 = \$21\).

Answer

The paint set and paintbrush cost \(\$21\) altogether.
5183143
Leon has \(6\) packs with \(8\) stickers in each pack. Sophie has \(7\) packs with \(7\) stickers in each pack. Sophie says, “I have more stickers than Leon.” Is she correct? Find each total and the difference.

Hints

- Find each child's total number of stickers. - Equal groups suggest multiplication. - Compare the totals and subtract to find the difference.

Solution

1. Find Leon's total: \(6 \times 8 = 48\). 2. Find Sophie's total: \(7 \times 7 = 49\). 3. Compare: \(49 > 48\), so Sophie is correct. 4. Find the difference: \(49 - 48 = 1\).

Answer

Yes. Sophie has \(49\) stickers and Leon has \(48\), so Sophie has \(1\) more sticker.
5183153
Ms. Miller wants to bake exactly \(60\) muffins for a school fair. She first bakes \(4\) trays with \(9\) muffins on each tray. Then she bakes \(3\) trays with \(7\) muffins on each tray. Does she reach her goal? How many muffins is she short, or how many extra muffins does she have?

Hints

- Find the number of muffins in each batch. - Add the two batches. - Compare the total with the goal of \(60\).

Solution

1. Find the number from the first batch: \(4 \times 9 = 36\). 2. Find the number from the second batch: \(3 \times 7 = 21\). 3. Find the total: \(36 + 21 = 57\). 4. Since \(57 < 60\), she is short by \(60 - 57 = 3\) muffins.

Answer

No. Ms. Miller bakes \(57\) muffins, so she is \(3\) muffins short.
5185453
At a wildlife park, \(5\) small goats receive \(40\) apple slices altogether. Each goat receives the same number. How many apple slices are needed for \(8\) goats?

Hints

- Find the number of apple slices for one goat. - Use that number to find the amount for \(8\) goats. - The total should increase because more goats are being fed.

Solution

1. Find the number of apple slices for one goat: \(40 \div 5 = 8\). 2. Find the number for \(8\) goats: \(8 \times 8 = 64\).

Answer

The park needs \(64\) apple slices.
5185623
A baker needs \(12\) minutes to prepare \(3\) pizza pans. Each pan takes the same amount of time. How many minutes will the baker need to prepare \(7\) pans?

Hints

- Find how long one pan takes first. - Use that time to find how long \(7\) pans take.

Solution

1. Find the time for one pan: \(12\,\text{min} \div 3 = 4\,\text{min}\). 2. Find the time for \(7\) pans: \(7 \times 4\,\text{min} = 28\,\text{min}\).

Answer

The baker needs \(28\) minutes.
5185803
Lucas buys \(5\) identical game figures for \(\$35\) altogether. His younger brother wants to buy \(3\) of the same figures but has only \(\$15\). How much more money does his brother need?

Hints

- First find the cost of one figure. - How much do three figures cost altogether? - Compare that total with the money his brother already has.

Solution

1. Find the cost of one figure: \(\$35 \div 5 = \$7\). 2. Find the cost of three figures: \(3 \times \$7 = \$21\). 3. Find the additional amount needed: \(\$21 - \$15 = \$6\).

Answer

His brother needs \(\$6\) more.
5185983
At a flea market, a child sells packages of trading cards. Each package contains \(10\) cards and costs \(\$2\). By the end of the day, the child has collected \(\$48\). How many trading cards were sold altogether?

Hints

- Use the total money and the price per package to find the number of packages sold. - Each package contains \(10\) cards. - Use the number of packages to find the total number of cards.

Solution

1. Find the number of packages sold: \(\$48 \div \$2 = 24\) packages. 2. Find the number of cards sold: \(24 \times 10 = 240\) cards.

Answer

The child sold \(240\) trading cards altogether.
5186723
A zoo has \(4\) groups of penguins with \(8\) penguins in each group. Each penguin eats \(3\) fish per day. How many fish are needed each day for all the penguins?

Hints

- First find the total number of penguins. - Each penguin receives the same number of fish. - Use the total number of penguins to find the daily amount of fish.

Solution

1. Find the total number of penguins: \(4 \times 8 = 32\). 2. Find the number of fish needed: \(32 \times 3 = 96\).

Answer

The penguins need \(96\) fish each day.
5186783
A rain barrel holds \(400\,\text{L}\). First, \(280\,\text{L}\) of water is added. After a dry period, \(230\,\text{L}\) is needed to fill the barrel completely. How much water was removed from the barrel during the dry period?

Hints

- Find the empty space after the first filling. - Compare that amount with the amount later needed to refill the barrel. - The extra refill amount equals the water that was removed.

Solution

1. After the first filling, the empty space is \(400\,\text{L}-280\,\text{L}=120\,\text{L}\). 2. If no water had been removed, only \(120\,\text{L}\) would be needed to fill the barrel. 3. The actual refill is \(230\,\text{L}\), so the amount removed is \(230\,\text{L}-120\,\text{L}=110\,\text{L}\).

Answer

\(110\,\text{L}\)
5187473
A gardener buys \(15\) red flowers for \(\$4\) each and then has \(\$20\) remaining. The gardener wonders how many yellow flowers could have been bought with all the original money if each yellow flower cost \(\$8\). How many yellow flowers could the gardener have bought?

Hints

- First find the gardener's original amount of money. - How much do the red flowers cost altogether? - Add the money remaining after the purchase. - Once you know the original amount, divide by the price of one yellow flower.

Solution

1. Find the cost of the red flowers: \(15 \times \$4 = \$60\). 2. Find the original amount of money: \(\$60 + \$20 = \$80\). 3. Divide by the cost of one yellow flower: \(\$80 \div \$8 = 10\).

Answer

The gardener could have bought \(10\) yellow flowers.
5189273
For a class trip, Mr. Miller buys juice. Eight identical cases contain \(48\) bottles altogether. a) How many bottles are in one case? b) Mr. Miller wants each of the \(65\) children on the trip to have one bottle. Are \(10\) cases enough? Justify your answer with a calculation.

Hints

- First find how many bottles are in one case. - How many bottles are in ten cases? - Compare that amount with the number of children.

Solution

1. Find the number of bottles in one case: \(48 \div 8 = 6\). 2. Find the number of bottles in ten cases: \(10 \times 6 = 60\). 3. Compare the available bottles with the number needed: \(60 < 65\). 4. Ten cases are not enough.

Answer

a) One case contains \(6\) bottles. b) No. Ten cases contain only \(60\) bottles, but \(65\) bottles are needed.
5189333
Two movie theaters sell tickets. At the Star Theater, \(5\) tickets cost \(\$45\) altogether. At the Moon Theater, \(4\) tickets cost \(\$40\) altogether. At which theater is one ticket less expensive? Justify your answer by comparing the unit prices.

Hints

- How much does one ticket cost at each theater? - What should you do to determine which offer is better? - Compare the two unit prices.

Solution

1. Find the price of one ticket at the Star Theater: \(\$45 \div 5 = \$9\). 2. Find the price of one ticket at the Moon Theater: \(\$40 \div 4 = \$10\). 3. Since \(\$9 < \$10\), one ticket is less expensive at the Star Theater.

Answer

One ticket is less expensive at the Star Theater because it costs \(\$9\), compared with \(\$10\) at the Moon Theater.
5190013
Grandpa Henry is \(65\) years old today. His grandson Ben is \(56\) years younger. How old will Ben be when Grandpa Henry celebrates his \(75\)th birthday?

Hints

- First find Ben's age today. - Find how many years remain until Grandpa Henry turns \(75\). - Both people age by the same number of years.

Solution

1. Find Ben's current age: \(65 - 56 = 9\). 2. Find the number of years until Grandpa Henry is \(75\): \(75 - 65 = 10\). 3. Add those years to Ben's age: \(9 + 10 = 19\).

Answer

Ben will be \(19\) years old.
5190283
Ms. Berger is \(42\) years old. Her son Lucas is \(7\), and her daughter Mia is twice as old as Lucas. a) How old was Ms. Berger when Mia was born? b) How old was Ms. Berger when Lucas was born? c) At which child's birth was she older?

Hints

- First find Mia's age. - Subtract each child's age from the mother's current age. - Compare the two results.

Solution

1. Find Mia's age: \(7 \times 2 = 14\). 2. At Mia's birth, Ms. Berger was \(42 - 14 = 28\) years old. 3. At Lucas's birth, Ms. Berger was \(42 - 7 = 35\) years old. 4. Since \(35 > 28\), she was older when Lucas was born.

Answer

a) Ms. Berger was \(28\) years old when Mia was born. b) She was \(35\) years old when Lucas was born. c) She was older when Lucas was born.
5191463
A school event starts with \(450\) muffins. Students sell \(200\) muffins in the morning and \(150\) muffins in the afternoon. How many muffins remain at the end of the day?

Hints

- First find the total number of muffins sold. - Subtract the number sold from the starting amount. - You can also subtract the morning and afternoon sales one at a time.

Solution

1. Add the muffins sold: \(200 + 150 = 350\). 2. Subtract from the starting amount: \(450 - 350 = 100\).

Answer

\(100\) muffins remain at the end of the day.
5191473
Two classes collect paper for a recycling contest. Class 3A collects \(340\,\text{kg}\), and Class 3B collects \(420\,\text{kg}\). The school's goal is \(900\,\text{kg}\). How many more kilograms are needed to reach the goal?

Hints

- Add the amounts collected by both classes. - Subtract that total from the goal. - Check that the missing amount and collected amount add to \(900\,\text{kg}\).

Solution

1. Add the amounts collected by the two classes: \(340 + 420 = 760\,\text{kg}\). 2. Subtract the amount collected from the goal: \(900 - 760 = 140\,\text{kg}\).

Answer

The classes need \(140\,\text{kg}\) more to reach the goal.
5191493
A puzzle has \(950\) pieces. Marie connects \(300\) pieces on Saturday and \(400\) pieces on Sunday. How many pieces are still loose in the box?

Hints

- Add the pieces Marie completed on both days. - Subtract the completed pieces from the total. - Check that completed and loose pieces add to \(950\).

Solution

1. Add the pieces connected on both days: \(300 + 400 = 700\). 2. Subtract from the total number of pieces: \(950 - 700 = 250\).

Answer

\(250\) puzzle pieces are still loose in the box.
5191643
Two classes collect empty bottles for a recycling project. Class 3A collects \(370\) bottles. Together, Classes 3A and 3B collect \(720\) bottles. Which class collects more bottles, and how many more?

Hints

- Use the combined total and Class 3A's amount to find Class 3B's amount. - Compare the two class amounts. - Subtract to find how many more.

Solution

1. Find the number collected by Class 3B: \(720 - 370 = 350\). 2. Compare the two amounts: \(370 > 350\), so Class 3A collects more. 3. Find the difference: \(370 - 350 = 20\).

Answer

Class 3A collects more bottles—\(20\) more than Class 3B.
5191653
A book has \(560\) pages. Max has read \(240\) pages. Klara is reading the same book and has \(310\) pages left. Who has read more pages? Justify your answer with a calculation.

Hints

- First find how many pages Klara has already read. - Subtract the pages she has left from the total number of pages. - Compare Klara's result with Max's \(240\) pages.

Solution

1. Find the number of pages Klara has read: \(560 - 310 = 250\). 2. Compare the amounts read: \(250 > 240\), so Klara has read more pages.

Answer

Klara has read more. She has read \(250\) pages, while Max has read \(240\) pages.
5191663
A trail is \(720\,\text{m}\) long and has two sections: a forest section and a field section. The forest section is \(450\,\text{m}\) long. How many meters longer is the forest section than the field section?

Hints

- First find the length of the field section. - Then compare the two section lengths. - Subtract the shorter length from the longer length.

Solution

1. Find the length of the field section: \(720 - 450 = 270\,\text{m}\). 2. Find the difference between the two sections: \(450 - 270 = 180\,\text{m}\).

Answer

The forest section is \(180\,\text{m}\) longer than the field section.
5191683
A class wants to sell \(1000\) cups of juice at a school event. Students sell \(440\) cups in the morning and \(270\) cups at lunch. How many cups do they still need to sell to reach their goal?

Hints

- Add the morning and lunch sales. - Subtract the amount sold from the goal. - Check that the amount sold and the amount remaining total \(1000\).

Solution

1. Add the cups already sold: \(440 + 270 = 710\). 2. Subtract from the goal: \(1000 - 710 = 290\).

Answer

The class needs to sell \(290\) more cups to reach the goal.
5191923
Students at an elementary school want to earn \(1000\) points in a long-jump challenge. They earn \(467\) points in the morning and \(385\) points in the afternoon. How many points do they still need to reach their goal?

Hints

- First find the total number of points already earned. - Subtract that total from \(1000\). - Check that the earned points and remaining points total \(1000\).

Solution

1. Add the points earned: \(467 + 385 = 852\). 2. Subtract from the goal: \(1000 - 852 = 148\).

Answer

The students need \(148\) more points to reach their goal.
5192863
A school buys \(900\) balloons for an event. Students use \(256\) balloons in the morning and \(315\) balloons in the afternoon. How many balloons remain?

Hints

- Find the total number of balloons used. - Subtract that total from \(900\). - You can also subtract the two amounts one at a time.

Solution

1. Add the balloons used: \(256 + 315 = 571\). 2. Subtract from the starting amount: \(900 - 571 = 329\).

Answer

\(329\) balloons remain.
5192963
A hot-air balloon pilot plans a trip of \(900\) miles. The balloon travels \(342\) miles on the first day and \(285\) miles on the second day. How many miles must it travel on the third day to reach the destination?

Hints

- Find the total distance already traveled. - Subtract that total from \(900\) miles. - Check that all three daily distances add to \(900\).

Solution

1. Add the distances traveled on the first two days: \(342 + 285 = 627\) miles. 2. Subtract from the total distance: \(900 - 627 = 273\) miles.

Answer

The balloon must travel \(273\) miles on the third day.
5193063
A baker buys \(245\,\text{kg}\) of wheat flour and \(180\,\text{kg}\) of rye flour. During the day, the baker uses \(315\,\text{kg}\) of flour. How many kilograms of flour remain?

Hints

- Add the two kinds of flour first. - Subtract the amount used from the total. - Include the unit in your answer.

Solution

1. Find the total amount of flour: \(245 + 180 = 425\,\text{kg}\). 2. Subtract the amount used: \(425 - 315 = 110\,\text{kg}\).

Answer

The baker has \(110\,\text{kg}\) of flour remaining.
5193073
A school supply store begins the day with \(356\) blue notebooks, \(289\) red notebooks, and \(145\) green notebooks. At the end of the day, only \(218\) notebooks remain. How many notebooks were sold?

Hints

- Add all three notebook colors to find the starting total. - Subtract the number left at the end of the day. - Check that sold and remaining notebooks add to the starting total.

Solution

1. Find the starting total: \(356 + 289 + 145 = 790\). 2. Subtract the number remaining: \(790 - 218 = 572\).

Answer

The store sold \(572\) notebooks.
5193083
A bakery plans to make \(850\) pretzels for a community festival. It makes \(412\) pretzels on Friday and \(465\) on Saturday. How many pretzels does the bakery make altogether, and by how many does it exceed its goal?

Hints

- Add the amounts from Friday and Saturday. - Compare the total with the goal. - Subtract the goal from the total to find how far it was exceeded.

Solution

1. Add the pretzels made on both days: \(412 + 465 = 877\). 2. Subtract the goal from the total: \(877 - 850 = 27\).

Answer

The bakery makes \(877\) pretzels, exceeding its goal by \(27\) pretzels.
5193093
Lucas is saving for a mountain bike that costs \(\$450\). He saves \(\$215\) during the first year and \(\$278\) during the second year. Lucas says, “I have saved more than the bike costs.” Is Lucas correct? How much more has he saved than he needs?

Hints

- Add the two amounts Lucas saved. - Compare the total savings with the bike price. - Subtract to find the extra amount.

Solution

1. Add Lucas's savings: \(\$215 + \$278 = \$493\). 2. Compare with the bike price: \(\$493 > \$450\), so Lucas is correct. 3. Find the extra amount: \(\$493 - \$450 = \$43\).

Answer

Lucas is correct. He has saved \(\$493\), which is \(\$43\) more than the bike costs.
5193113
A produce seller has \(312\) apples. There are \(125\) fewer pears than apples, and there are also \(240\) plums. How many pieces of fruit are at the stand altogether?

Hints

- First find the number of pears. - Then add the apples, pears, and plums. - Check that all three kinds of fruit are included.

Solution

1. Find the number of pears: \(312 - 125 = 187\). 2. Add all three kinds of fruit: \(312 + 187 + 240 = 739\).

Answer

The seller has \(739\) pieces of fruit altogether.
5193353
An orchard harvests \(345\,\text{kg}\) of apples. The apples and pears together weigh \(780\,\text{kg}\). How many kilograms of pears are harvested? How many more kilograms of pears than apples are harvested?

Hints

- Use the total mass and apple mass to find the pear mass. - Then subtract the smaller harvest from the larger harvest. - Include kilograms in both answers.

Solution

1. Find the mass of the pears: \(780 - 345 = 435\,\text{kg}\). 2. Find how much greater the pear harvest is: \(435 - 345 = 90\,\text{kg}\).

Answer

The orchard harvests \(435\,\text{kg}\) of pears, which is \(90\,\text{kg}\) more than the apple harvest.
5194323
A school buys \(2\) packages containing \(250\) balloons each for a celebration. After decorating, \(112\) balloons remain. How many balloons were used?

Hints

- First find the total number of balloons available. - The balloons used and the balloons remaining make the starting total. - Subtract the remaining amount from the starting amount.

Solution

1. Find the total number purchased: \(2 \times 250 = 500\). 2. Subtract the balloons remaining: \(500 - 112 = 388\).

Answer

The school used \(388\) balloons.
5194823
A truck travels \(50\) miles in one hour. An express train travels twice that distance in one hour. How many miles does the train travel in \(5\) hours?

Hints

- First find twice the truck's one-hour distance. - Then use the train's one-hour distance to find the distance in five hours. - Solve the problem in two steps.

Solution

1. Find the train's distance in one hour: \(50 \times 2 = 100\) miles. 2. Find the distance in five hours: \(100 \times 5 = 500\) miles.

Answer

The train travels \(500\) miles in five hours.
5196703
A school library has \(345\) nonfiction books on Monday morning. A bookseller delivers \(56\) more books on Tuesday. On Wednesday, students check out \(89\) nonfiction books for a project. How many nonfiction books remain on the shelves?

Hints

- Add the books that are delivered. - Then subtract the books that are checked out. - Follow the events in time order.

Solution

1. Add the delivered books: \(345 + 56 = 401\). 2. Subtract the books checked out: \(401 - 89 = 312\).

Answer

\(312\) nonfiction books remain on the shelves.
5196723
Lucas starts with \(340\). He creates two new numbers. The first number is \(60\) greater than \(340\). The second number is \(400\) greater than the first number. What are the two new numbers?

Hints

- Which operation matches the phrase “greater than”? - Find the first new number before finding the second one. - Add hundreds to hundreds and tens to tens.

Solution

1. Find the first number: \(340 + 60 = 400\). 2. Use the first number to find the second number: \(400 + 400 = 800\).

Answer

The first number is \(400\), and the second number is \(800\).
5197193
A school library tracked the number of books borrowed in September and October. Find the increase for each category. Which category had the greatest increase? <table> <tr><th>Category</th><th>September</th><th>October</th><th>Increase</th></tr> <tr><td>Nonfiction</td><td>145</td><td>210</td><td></td></tr> <tr><td>Fiction</td><td>238</td><td>312</td><td></td></tr> <tr><td>Comics</td><td>189</td><td>254</td><td></td></tr> <tr><td>Picture books</td><td>92</td><td>167</td><td></td></tr> </table>

Hints

- Look at how each category changes from September to October. - Subtract the September number from the October number in each row. - Compare the four increases after calculating them.

Solution

1. Nonfiction increased by \(210 - 145 = 65\). 2. Fiction increased by \(312 - 238 = 74\). 3. Comics increased by \(254 - 189 = 65\). 4. Picture books increased by \(167 - 92 = 75\). 5. Since \(75\) is greatest, picture books had the greatest increase.

Answer

Nonfiction: \(65\) Fiction: \(74\) Comics: \(65\) Picture books: \(75\) Picture books had the greatest increase.
5197203
A school garden compared its fruit harvest from last year with this year. Find the increase for each fruit. How many more kilograms of fruit were harvested altogether this year? <table> <tr><th>Fruit</th><th>Last year</th><th>This year</th><th>Increase</th></tr> <tr><td>Apples</td><td>\(345\,\text{kg}\)</td><td>\(412\,\text{kg}\)</td><td></td></tr> <tr><td>Pears</td><td>\(128\,\text{kg}\)</td><td>\(205\,\text{kg}\)</td><td></td></tr> <tr><td>Plums</td><td>\(96\,\text{kg}\)</td><td>\(153\,\text{kg}\)</td><td></td></tr> </table>

Hints

- Subtract last year's amount from this year's amount for each fruit. - Align the place values carefully when subtracting. - Add the three increases to find the total increase.

Solution

1. Apples increased by \(412 - 345 = 67\,\text{kg}\). 2. Pears increased by \(205 - 128 = 77\,\text{kg}\). 3. Plums increased by \(153 - 96 = 57\,\text{kg}\). 4. Add the increases: \(67 + 77 + 57 = 201\,\text{kg}\).

Answer

Apples: \(67\,\text{kg}\) Pears: \(77\,\text{kg}\) Plums: \(57\,\text{kg}\) Altogether, the harvest increased by \(201\,\text{kg}\).
5199193
A bakery has \(8\) trays of muffins. On each tray, the muffins are arranged in \(4\) rows of \(5\). How many muffins are there altogether?

Hints

- First find how many muffins are on one tray. - Then use the number of trays to find the total. - Think of each tray as an array.

Solution

1. Find the number on one tray: \(4 \times 5 = 20\). 2. Find the number on eight trays: \(8 \times 20 = 160\).

Answer

There are \(160\) muffins altogether.
5199203
A school has \(5\) craft groups with \(6\) students in each group. Each student needs \(8\) white beads and \(2\) gold beads for a necklace. How many beads are needed altogether?

Hints

- First find the total number of students. - Find the number of beads needed by one student. - Multiply to find the total for all students.

Solution

1. Find the total number of students: \(5 \times 6 = 30\). 2. Find the number of beads each student needs: \(8 + 2 = 10\). 3. Find the total number of beads: \(30 \times 10 = 300\).

Answer

The school needs \(300\) beads altogether.
5200813
A stationery store packs pens into boxes. <table> <tr><th>Pen color</th><th>Pens per box</th><th>Total pens</th></tr> <tr><td>Blue pens</td><td>8</td><td>480</td></tr> <tr><td>Red pens</td><td>6</td><td>?</td></tr> </table> The store has the same number of boxes of each color. How many red pens are there altogether?

Hints

- Use the first row to find the number of boxes. - The number of red-pen boxes is the same. - Multiply the number of boxes by the number of red pens per box.

Solution

1. Find the number of boxes of blue pens: \(480 \div 8 = 60\). 2. There are also \(60\) boxes of red pens. 3. Find the total number of red pens: \(60 \times 6 = 360\).

Answer

There are \(360\) red pens altogether.
5203193
Students sell raffle tickets at a school event. They sell \(465\) tickets in the morning, which is \(130\) more than they sell in the afternoon. a) How many tickets do they sell in the afternoon? b) How many tickets do they sell during the entire day?

Hints

- Use “more than” to find the afternoon amount. - Then add the morning and afternoon amounts. - Add the two parts to check the daily total.

Solution

1. Find the afternoon sales: \(465 - 130 = 335\). 2. Add the morning and afternoon sales: \(465 + 335 = 800\).

Answer

a) Students sell \(335\) tickets in the afternoon. b) Students sell \(800\) tickets during the entire day.
5203763
An aquarium has \(340\) goldfish in a large tank and \(280\) silver fish in a small tank. a) How many fish are in both tanks altogether? b) The aquarium adds \(40\) more goldfish to the large tank. How many fish are there altogether now? Use your answer from part a.

Hints

- First add the fish in the two tanks. - Then add the fish that are introduced later. - Use the result from part a in part b.

Solution

1. Add the fish in both tanks: \(340 + 280 = 620\). 2. Add the new goldfish: \(620 + 40 = 660\).

Answer

a) There are \(620\) fish altogether. b) There are now \(660\) fish altogether.
5203973
Anton has \(160\) marbles, and Bea has \(125\) marbles. a) How many marbles do they have altogether? b) They give \(35\) marbles to their younger brother. How many marbles do they have left altogether?

Hints

- Add the two starting amounts. - Then subtract the marbles that are given away. - Use the answer from part a in part b.

Solution

1. Add to find the starting total: \(160 + 125 = 285\). 2. Subtract the marbles they give away: \(285 - 35 = 250\).

Answer

a) They have \(285\) marbles altogether. b) They have \(250\) marbles left.
5203983
A wildlife park has \(24\) meerkats in one habitat and \(19\) in another. a) How many meerkats are in the park altogether? b) The park sends \(7\) meerkats to another zoo. Later, \(5\) baby meerkats are born. How many meerkats are in the park now?

Hints

- Add the animals in both habitats first. - Subtract the animals that leave. - Then add the animals that are born.

Solution

1. Add the two habitats: \(24 + 19 = 43\). 2. Subtract the meerkats that leave: \(43 - 7 = 36\). 3. Add the newborn meerkats: \(36 + 5 = 41\).

Answer

a) The park has \(43\) meerkats altogether. b) The park now has \(41\) meerkats.
5204233
Maria adds \(55\) apples to a basket while her brother removes \(20\) pears. By how much does the total number of pieces of fruit change?

Hints

- Decide how adding apples changes the total. - Decide how removing pears changes the total. - Combine the increase and decrease.

Solution

1. Adding the apples changes the total by \(+55\). 2. Removing the pears changes the total by \(-20\). 3. Combine the changes: \(55 - 20 = 35\). The positive result means the total increases.

Answer

The total number of pieces of fruit increases by \(35\).
5204303
A school library has \(670\) books. Students check out \(45\) books, and then the library receives \(25\) new books. How many books are in the library now?

Hints

- Subtract the books that are checked out. - Then add the new books. - Follow the events in order.

Solution

1. Subtract the books checked out: \(670 - 45 = 625\). 2. Add the new books: \(625 + 25 = 650\).

Answer

The library now has \(650\) books.
5205613
Lucas has \(345\) trading cards, and his sister Marie has \(415\) trading cards. Their cousin Tom has \(230\) fewer cards than Lucas and Marie have altogether. How many cards does Tom have?

Hints

- First find how many cards Lucas and Marie have altogether. - Use “fewer than” to choose the next operation.

Solution

1. Add Lucas's and Marie's cards: \(345 + 415 = 760\). 2. Subtract \(230\): \(760 - 230 = 530\).

Answer

Tom has \(530\) trading cards.
5208823
A school supply store has \(340\) pencils on a shelf. A worker adds \(270\) new pencils, and later the store sells \(180\) pencils. How many pencils are on the shelf at the end of the day?

Hints

- Add the pencils that are placed on the shelf. - Then subtract the pencils that are sold. - Follow the events in order.

Solution

1. Add the new pencils: \(340 + 270 = 610\). 2. Subtract the pencils sold: \(610 - 180 = 430\).

Answer

There are \(430\) pencils on the shelf at the end of the day.
5209983
Students sell \(350\) raffle tickets on Friday. On Saturday, they sell \(125\) more tickets than on Friday. A student says, “We sold more than \(800\) tickets over the two days.” Is the student correct? Justify your answer with a calculation.

Hints

- Find the number of tickets sold on Saturday. - Add the sales from both days. - Compare the total with \(800\).

Solution

1. Find the Saturday sales: \(350 + 125 = 475\). 2. Add the sales from both days: \(350 + 475 = 825\). 3. Compare: \(825 > 800\), so the statement is correct.

Answer

Yes. Students sell \(825\) tickets over the two days, and \(825 > 800\).
5210013
Lucas has \(350\) soccer stickers. He has \(120\) fewer animal stickers than soccer stickers. How many stickers does Lucas have altogether?

Hints

- First find the number of animal stickers. - Then add the two sticker collections. - Check that the animal-sticker count is smaller than \(350\).

Solution

1. Find the number of animal stickers: \(350 - 120 = 230\). 2. Add both kinds of stickers: \(350 + 230 = 580\).

Answer

Lucas has \(580\) stickers altogether.
5210023
A bakery makes three kinds of rolls in the morning. It makes \(240\) plain rolls. It makes \(70\) more whole-wheat rolls than plain rolls and \(50\) fewer pumpkin seed rolls than plain rolls. How many rolls of these three kinds does the bakery make altogether?

Hints

- First find the number of rolls of each kind. - Both comparisons are made with the number of plain rolls. - Then add all three amounts.

Solution

1. Find the number of whole-wheat rolls: \(240 + 70 = 310\). 2. Find the number of pumpkin seed rolls: \(240 - 50 = 190\). 3. Add the three amounts: \(240 + 310 + 190 = 740\).

Answer

The bakery makes \(740\) rolls altogether.
5210083
A school library has \(450\) books. Of these, \(180\) are fiction books. The library has \(50\) fewer nonfiction books than fiction books. All the remaining books are picture books. How many picture books are in the library?

Hints

- Use the number of fiction books to find the number of nonfiction books. - Add those two categories. - Subtract their total from \(450\).

Solution

1. Find the number of nonfiction books: \(180 - 50 = 130\). 2. Find the number of fiction and nonfiction books combined: \(180 + 130 = 310\). 3. Subtract from the total: \(450 - 310 = 140\).

Answer

There are \(140\) picture books in the library.
5210093
Three third-grade classes collect paper for recycling. Together, they collect \(820\,\text{kg}\). Ms. Lee's class collects \(245\,\text{kg}\). Mr. Grant's class collects \(30\,\text{kg}\) more than Ms. Lee's class. How many kilograms of paper does Ms. Patel's class collect?

Hints

- First find how many kilograms Mr. Grant's class collects. - Add the amounts from the first two classes. - Find how much more is needed to reach \(820\,\text{kg}\).

Solution

1. Find how much Mr. Grant's class collects: \(245\,\text{kg} + 30\,\text{kg} = 275\,\text{kg}\). 2. Add the amounts collected by the first two classes: \(245\,\text{kg} + 275\,\text{kg} = 520\,\text{kg}\). 3. Subtract from the total: \(820\,\text{kg} - 520\,\text{kg} = 300\,\text{kg}\).

Answer

Ms. Patel's class collects \(300\,\text{kg}\) of paper.
5210153
A mail carrier drives \(18\) miles during the first hour of a delivery route. During the second hour, the carrier drives \(4\) miles farther than during the first hour. Then the carrier drives another \(15\) miles to finish the route. How long is the entire route?

Hints

- First find the distance driven during the second hour. - Identify the three parts of the route. - Add the three distances.

Solution

1. Find the distance driven during the second hour: \(18 + 4 = 22\) miles. 2. Add all three parts of the route: \(18 + 22 + 15 = 55\) miles.

Answer

The entire route is \(55\) miles long.
5210173
Lucas is saving for a bicycle that costs \(\$450\). He saves \(\$135\) in May. In June, he saves \(\$50\) more than he saved in May. How much more must Lucas save in July to have enough money for the bicycle?

Hints

- First find how much Lucas saves in June. - Add his savings from May and June. - Subtract that amount from the cost of the bicycle.

Solution

1. Find how much Lucas saves in June: \(\$135 + \$50 = \$185\). 2. Find how much he saves in May and June combined: \(\$135 + \$185 = \$320\). 3. Subtract from the cost of the bicycle: \(\$450 - \$320 = \$130\).

Answer

Lucas must save \(\$130\) in July.
5210183
At a field day, three third-grade classes complete a total of \(840\) laps around the track. Ms. Lee's class completes \(260\) laps. Mr. Grant's class completes \(40\) fewer laps than Ms. Lee's class. Ms. Patel's class completes the remaining laps. Which class completes the most laps? Find the number of laps completed by Ms. Patel's class and compare all three amounts.

Hints

- Find the number of laps completed by each class. - Add the laps from the first two classes. - Subtract that sum from \(840\), then compare all three amounts.

Solution

1. Find the number of laps completed by Mr. Grant's class: \(260 - 40 = 220\). 2. Add the laps completed by the first two classes: \(260 + 220 = 480\). 3. Find the remaining laps: \(840 - 480 = 360\). 4. Compare the three amounts: \(360 > 260 > 220\). Ms. Patel's class completes the most laps.

Answer

Ms. Patel's class completes the most laps, with \(360\) laps.
5211263
A baker receives \(100\) eggs. In the morning, the baker uses \(37\) eggs for cakes. The remaining eggs will be used for waffles, with exactly \(7\) eggs in each batch. How many batches of waffles can the baker make?

Hints

- Identify the important starting quantity. - How many eggs remain after the cakes are made? - Calculate the subtraction carefully. - Use a multiplication fact to find how many groups of seven fit into the remainder.

Solution

1. Find the number of eggs remaining: \(100 - 37 = 63\). 2. Divide by the number of eggs in each batch: \(63 \div 7 = 9\).

Answer

The baker can make \(9\) batches of waffles.
5211323
A toy store receives a shipment of \(84\) card games. The games are divided equally among \(7\) shelf sections. How many card games are in \(3\) of the sections altogether?

Hints

- First find how many games are in one shelf section. - Once you know how many are in one section, how can you find the number in three sections? - Decompose the dividend into compatible multiples of the divisor if helpful.

Solution

1. Find the number of card games in each shelf section: \(84 \div 7 = 12\). 2. Find the number of card games in three sections: \(12 \times 3 = 36\).

Answer

There are \(36\) card games in \(3\) shelf sections altogether.
5211413
A school library needs to shelve \(850\) books. Volunteers shelve \(260\) books on Monday. On Tuesday, they shelve \(45\) fewer books than on Monday. How many books still need to be shelved after Tuesday?

Hints

- First find how many books are shelved on Tuesday. - Add the numbers shelved on Monday and Tuesday. - Subtract that amount from \(850\).

Solution

1. Find the number of books shelved on Tuesday: \(260 - 45 = 215\). 2. Find the total shelved on both days: \(260 + 215 = 475\). 3. Subtract from the total number of books: \(850 - 475 = 375\).

Answer

There are \(375\) books left to shelve.
5211423
A walking trail is \(1000\,\text{m}\) long. The first section is \(340\,\text{m}\) long. The second section is \(120\,\text{m}\) shorter than the first section. How long is the third section?

Hints

- First find the length of the second section. - Add the lengths of the first two sections. - Find how much of the \(1000\,\text{m}\) trail remains.

Solution

1. Find the length of the second section: \(340\,\text{m} - 120\,\text{m} = 220\,\text{m}\). 2. Find the combined length of the first two sections: \(340\,\text{m} + 220\,\text{m} = 560\,\text{m}\). 3. Find the length of the third section: \(1000\,\text{m} - 560\,\text{m} = 440\,\text{m}\).

Answer

The third section is \(440\,\text{m}\) long.
5211873
A gardener has \(54\) tulip bulbs. He plants exactly \(6\) bulbs in each row. a) How many full rows can he plant? b) How many more bulbs does he need to plant \(10\) full rows?

Hints

- How many groups of \(6\) are in \(54\)? - How many bulbs are needed for \(10\) rows of \(6\)? - Compare the required number with the number the gardener has. - How many bulbs are in one additional row?

Solution

1. Divide to find the number of full rows: \(54 \div 6 = 9\). 2. Ten rows require \(10 \times 6 = 60\) bulbs. 3. The gardener needs \(60 - 54 = 6\) more bulbs.

Answer

a) \(9\) rows b) \(6\) bulbs
5212703
A school garden has a water tank holding \(850\,\text{L}\). Students use \(186\,\text{L}\) on Monday and another \(275\,\text{L}\) on Tuesday. a) How many liters of water remain in the tank? b) Did the students use more water over the two days than remains in the tank? Explain.

Hints

- First find the total amount used on both days. - Subtract the amount used from the starting amount. - Compare the amount used with the amount remaining.

Solution

1. Find the total amount of water used: \(186\,\text{L} + 275\,\text{L} = 461\,\text{L}\). 2. Find the amount remaining: \(850\,\text{L} - 461\,\text{L} = 389\,\text{L}\). 3. Compare the amounts: \(461\,\text{L} > 389\,\text{L}\), so more water was used than remains.

Answer

a) \(389\,\text{L}\) remain in the tank. b) Yes. The students used \(461\,\text{L}\), and \(461\,\text{L} > 389\,\text{L}\).
5213423
A third-grade class collects pinecones for craft booths at a school fair. The students collect \(274\) pinecones on the first day and \(358\) on the second day. Their goal is \(800\) pinecones. How many more pinecones do they need?

Hints

- Add the amounts collected on the two days. - Compare the total collected with the goal of \(800\). - Find the difference.

Solution

1. Find the number collected on both days: \(274 + 358 = 632\). 2. Subtract from the goal: \(800 - 632 = 168\).

Answer

The class needs \(168\) more pinecones.
5213433
A bakery starts the morning with \(650\) rolls. By noon, it sells \(385\) rolls. The baker then makes \(120\) more rolls. How many rolls are there now? Is this more or fewer than the bakery had at the start of the morning?

Hints

- Subtract the rolls that were sold. - Then add the newly baked rolls. - Compare the result with \(650\).

Solution

1. Find the number left after the sales: \(650 - 385 = 265\). 2. Add the newly baked rolls: \(265 + 120 = 385\). 3. Compare with the starting amount: \(385 < 650\), so there are fewer rolls now.

Answer

There are \(385\) rolls now, which is fewer than the \(650\) rolls at the start of the morning.
5213523
A store receives \(6\) large cartons of pencils. Each carton contains \(4\) packages, and each package contains \(10\) pencils. 1) How many pencils are in one large carton? 2) How many pencils are delivered altogether?

Hints

- First find the number of pencils in one carton. - Then find the total for all six cartons. - Equal groups suggest multiplication.

Solution

1. Find the number in one carton: \(4 \times 10 = 40\). 2. Find the total in six cartons: \(6 \times 40 = 240\).

Answer

1) One large carton contains \(40\) pencils. 2) The delivery contains \(240\) pencils altogether.
5214843
A bakery has \(480\) rolls in two baskets altogether. The baker removes \(65\) rolls from one basket for packaging and adds \(65\) freshly baked rolls to the other basket. a) How many rolls are now in the two baskets altogether? b) What would happen to the total if the baker removed \(70\) rolls and added \(80\) rolls instead?

Hints

- In part a, compare the number removed with the number added. - In part b, find the difference between the number added and the number removed. - Use that change to find the new total.

Solution

1. For part a, the baker removes and adds the same number of rolls. These changes cancel, so the total remains \(480\). 2. For part b, compare the number added with the number removed: \(80 - 70 = 10\). The total increases by \(10\). 3. Find the new total: \(480 + 10 = 490\).

Answer

a) There are still \(480\) rolls altogether. b) The total increases by \(10\), so there would be \(490\) rolls.
5215493
A large aquarium has \(260\) fish. A keeper moves \(45\) fish to another tank and later adds \(72\) young fish. How many fish are in the aquarium now? Is this more or fewer than the aquarium had at first?

Hints

- First subtract the fish that are moved. - Then add the young fish. - Compare the final number with \(260\).

Solution

1. Find the number after \(45\) fish are moved: \(260 - 45 = 215\). 2. Add the young fish: \(215 + 72 = 287\). 3. Compare with the starting number: \(287 > 260\), so there are more fish now.

Answer

There are \(287\) fish now, which is more than the starting number of \(260\).
5373693
An orchard has \(6\) rows with \(8\) young trees in each row. After a storm, one entire row falls; it is shown in gray. How many trees are still standing? Give two calculation methods.
Figure for problem 537369

Hints

- One complete row is affected. - You can subtract one row or multiply using the number of rows that remain.

Solution

1. Find the original total and subtract one row: \(6 \times 8 - 8 = 48 - 8 = 40\). 2. Or count the five rows still standing: \(5 \times 8 = 40\).

Answer

\(40\) trees are still standing. Two methods are \(6 \times 8 - 8 = 40\) and \(5 \times 8 = 40\).
5373713
A small movie theater has \(7\) rows with \(10\) seats in each row. The \(13\) orange seats are already reserved. How many seats are still available?
Figure for problem 537371

Hints

- Use the rows of \(10\) to find the total number of seats. - Available seats equal total seats minus reserved seats.

Solution

1. Find the total number of seats: \(7 \times 10 = 70\). 2. Subtract the reserved seats: \(70 - 13 = 57\).

Answer

There are \(57\) seats still available.
5373723
A sheet has \(6\) rows and \(9\) columns of collectible stamps. The left \(5\) columns contain blue stamps worth \(2\) cents each, and the right \(4\) columns contain green stamps worth \(3\) cents each. What is the total value of the sheet?
Figure for problem 537372

Hints

- Determine the width of each color block. - Find the number and value of each color separately. - Add the two values.

Solution

1. There are \(6 \times 5 = 30\) blue stamps and \(6 \times 4 = 24\) green stamps. 2. The blue stamps are worth \(30 \times 2 = 60\) cents. 3. The green stamps are worth \(24 \times 3 = 72\) cents. 4. The total value is \(60 + 72 = 132\) cents, or \(\$1.32\).

Answer

The sheet is worth \(132\) cents, or \(\$1.32\).
5373883
A market stand begins the day with \(72\) apples arranged in \(9\) rows of \(8\). During the morning, it sells three complete rows. During the afternoon, it sells another \(19\) apples. How many apples remain?
Figure for problem 537388

Hints

- First find the number of apples in three rows. - Add the two sales amounts. - Subtract the total sold from \(72\).

Solution

1. Find the number sold in three rows: \(3 \times 8 = 24\). 2. Find the total sold: \(24 + 19 = 43\). 3. Find the number remaining: \(72 - 43 = 29\).

Answer

\(29\) apples remain.
5381233
Teams collect nature cards during a forest scavenger hunt. How many more cards does Team Fox need, at minimum, to have more cards than Team Badger?
Figure for problem 538123

Hints

- Read the values for Team Fox and Team Badger. - Decide the smallest total that is greater than Team Badger's total. - Subtract Team Fox's current total from that target.

Solution

1. Team Fox has \(15\) cards, and Team Badger has \(20\) cards. 2. To have more than \(20\), Team Fox needs at least \(21\) cards. 3. The number still needed is \(21 - 15 = 6\) cards.

Answer

Team Fox needs at least \(6\) more cards.
5381293
During a field game, Team Green later receives \(6\) more points. Which team is then in the lead, and with how many points?
Figure for problem 538129

Hints

- Update Team Green's score first. - Keep the other teams' scores unchanged. - Compare all four scores after the change.

Solution

1. Team Green's score changes from \(30\) to \(30 + 6 = 36\). 2. Compare \(36\) with the other scores: \(25\), \(35\), and \(20\). 3. Since \(36\) is greatest, Team Green is in the lead.

Answer

Team Green is in the lead with \(36\) points.
5381303
Because of an error in a quiz, Team Star loses \(8\) points. What place is Team Star in after the deduction?
Figure for problem 538130

Hints

- Change only Team Star's score. - Compare the new score with all three other scores. - Order the four scores from greatest to least.

Solution

1. Team Star's new score is \(40 - 8 = 32\) points. 2. The other teams have \(35\), \(40\), and \(45\) points. 3. Since \(32\) is the least score, Team Star is in fourth place.

Answer

Team Star is in fourth place.
5381323
The graph shows how students travel to a sports field. Do more or fewer students travel by bike and on foot together than by car? How many more or fewer?
Figure for problem 538132

Hints

- Add the bike and walking bars first. - Compare that sum with the car bar. - Subtract to find how much greater or smaller it is.

Solution

1. By bike and on foot together, \(12 + 8 = 20\) students travel to the field. 2. By car, \(16\) students travel to the field. 3. The difference is \(20 - 16 = 4\), so the bike-and-walking total is greater.

Answer

\(4\) more students travel by bike and on foot together than by car.
5381363
The graph shows soft pretzels baked during one week. How many pretzels were baked on all days except Wednesday?
Figure for problem 538136

Hints

- Leave out the Wednesday bar. - Add the other four bar values. - Pair numbers to make the addition easier.

Solution

1. The values for Monday, Tuesday, Thursday, and Friday are \(15\), \(20\), \(25\), and \(30\). 2. Add them: \(15 + 20 + 25 + 30 = 90\).

Answer

\(90\) pretzels were baked on all days except Wednesday.
5381373
Four groups estimate the number of steps along a short path. Which group's estimate is closest to \(55\) steps?
Figure for problem 538137

Hints

- Find how far each estimate is from \(55\). - Use a positive difference whether the estimate is above or below \(55\). - Choose the least difference.

Solution

1. Find how far each estimate is from \(55\). 2. Group A: \(55 - 40 = 15\); Group B: \(55 - 50 = 5\); Group C: \(55 - 30 = 25\); Group D: \(70 - 55 = 15\). 3. The least difference is \(5\), for Group B.

Answer

Group B is closest to \(55\) steps.
5381383
Lina and Mina want to have at least \(45\) stickers altogether. How many more stickers do they need?
Figure for problem 538138

Hints

- Add Lina's and Mina's amounts first. - Compare their total with \(45\). - Subtract to find how many are still needed.

Solution

1. Lina and Mina have \(20 + 15 = 35\) stickers altogether. 2. The number still needed is \(45 - 35 = 10\) stickers.

Answer

They need \(10\) more stickers.
5381423
A class used \(60\) paper strips for a craft. The graph shows the three colors. How many strips were purple? Confirm the value using the other two bars.
Figure for problem 538142

Hints

- Add the orange and green values. - Subtract their sum from the total of \(60\). - Compare your result with the purple bar.

Solution

1. The orange and green strips total \(20 + 15 = 35\). 2. The remaining number is \(60 - 35 = 25\). 3. This matches the purple bar, which has a value of \(25\).

Answer

There were \(25\) purple paper strips.
5381433
Are the two shortest bars together greater or less than the tallest bar? By how much?
Figure for problem 538143

Hints

- Identify the two shortest bars and add their values. - Identify the tallest bar. - Compare the sum with the tallest bar by subtraction.

Solution

1. The two least values are \(10\) and \(15\), with a sum of \(10 + 15 = 25\). 2. The greatest value is \(30\). 3. Since \(30 - 25 = 5\), the sum of the two least values is less by \(5\).

Answer

The two shortest bars together are \(5\) less than the tallest bar.
5381443
Teams collect keys during a puzzle challenge. Which team is in second place, and how far behind first place is it?
Figure for problem 538144

Hints

- Find the greatest and second-greatest bar values. - The second-greatest value is second place. - Subtract the two values to find the gap.

Solution

1. The two greatest values are Team Dune with \(40\) keys and Team Meadow with \(35\) keys. 2. Team Meadow is in second place. 3. The gap is \(40 - 35 = 5\) keys.

Answer

Team Meadow is in second place, \(5\) keys behind Team Dune.
5381473
For a nature collage, the number of stones is doubled. The numbers of wood pieces and leaves stay the same. How many pieces will the collage have altogether?
Figure for problem 538147

Hints

- Double only the stone value. - Keep the wood and leaf values unchanged. - Add the three new values.

Solution

1. The number of stones changes from \(10\) to \(2 \times 10 = 20\). 2. The new total is \(15 + 20 + 20 = 55\) pieces.

Answer

The collage will have \(55\) pieces altogether.
5381483
Some birds leave the largest flock. How many birds must leave so that this flock has the same number of birds as the second-largest flock?
Figure for problem 538148

Hints

- Identify the largest and second-largest flocks. - Find the difference between their sizes. - Check that removing that number makes the flock sizes equal.

Solution

1. The largest flock is the cranes with \(40\) birds. The second-largest flock is the geese with \(35\) birds. 2. The difference is \(40 - 35 = 5\) birds.

Answer

\(5\) cranes must leave the flock.
5381523
Four collection boxes contain the amounts shown in the graph. More items will be added to Box S4 so that the boxes contain \(70\) items altogether. How many items must be added?
Figure for problem 538152

Hints

- Add the amounts in all four boxes first. - Compare that total with \(70\). - Subtract to find how many items must be added.

Solution

1. The current total is \(10 + 15 + 20 + 20 = 65\) items. 2. The number needed is \(70 - 65 = 5\) items.

Answer

\(5\) items must be added to Box S4.
5381543
Each group should have at least \(20\) points. How many points must be distributed altogether if points are added only to groups below \(20\)?
Figure for problem 538154

Hints

- Find how many points each group below \(20\) needs. - Do not add points to groups already at or above \(20\). - Add the needed amounts.

Solution

1. Group A needs \(20 - 15 = 5\) points. 2. Group C needs \(20 - 10 = 10\) points. 3. Groups B and D already meet the goal. 4. Altogether, \(5 + 10 = 15\) points must be distributed.

Answer

\(15\) points must be distributed altogether.
5381563
The graph shows red and blue beads. Check both claims: “There are \(40\) beads altogether” and “There are \(8\) more red beads than blue beads.” Do both claims match the graph?
Figure for problem 538156

Hints

- Check the total claim by addition. - Check the comparison claim by subtraction. - Decide whether each claim is true.

Solution

1. The total is \(24 + 16 = 40\), so the first claim is true. 2. The difference is \(24 - 16 = 8\), so the second claim is true. 3. Both claims match the graph.

Answer

Yes. Both claims match the graph.
5381573
Group A gives \(5\) cards to Group B. Which groups then have the same number of cards?
Figure for problem 538157

Hints

- Subtract \(5\) from Group A and add \(5\) to Group B. - Keep Group C's value unchanged. - Compare the three new values.

Solution

1. Group A will have \(30 - 5 = 25\) cards. 2. Group B will have \(20 + 5 = 25\) cards. 3. Group C already has \(25\) cards.

Answer

All three groups will have \(25\) cards each.
5381633
Nuri describes Bar D this way: “It is \(3\) greater than C and \(3\) less than B.” Is the description correct?
Figure for problem 538163

Hints

- Check each part of the description separately. - Find the difference between D and C. - Find the difference between B and D.

Solution

1. Bar C has a value of \(18\), Bar D has \(21\), and Bar B has \(24\). 2. Since \(21 - 18 = 3\) and \(24 - 21 = 3\), both parts of the description are correct.

Answer

Yes. Nuri's description is correct.
5381683
The day with the greatest value is removed from the data. What is the sum of the other three days?
Figure for problem 538168

Hints

- Identify and leave out the greatest bar. - Add the other three values. - Check that you used exactly three days.

Solution

1. Tuesday has the greatest value, \(25\). 2. The remaining values are \(15\), \(10\), and \(20\). 3. Their sum is \(15 + 10 + 20 = 45\).

Answer

The remaining three days have a sum of \(45\).
5381693
Which individual bars are taller than Bars A and B combined?
Figure for problem 538169

Hints

- Add the values of A and B. - Compare each other bar with that sum. - Remember that equal to is not the same as greater than.

Solution

1. Bars A and B have a combined value of \(10 + 15 = 25\). 2. Bar C has a value of \(20\), and Bar D has a value of \(25\). 3. Neither value is greater than \(25\).

Answer

No individual bar is taller than Bars A and B combined.
5381713
Compare Bars A and B combined with Bars C and D combined. Which pair has the greater total, and by how much?
Figure for problem 538171

Hints

- Find each pair total separately. - Compare the two totals. - Subtract to find the difference.

Solution

1. Bars A and B total \(10 + 20 = 30\). 2. Bars C and D total \(25 + 10 = 35\). 3. The difference is \(35 - 30 = 5\).

Answer

Bars C and D have the greater total, by \(5\).
5381733
Graph a) shows the points in Round \(1\), and Graph b) shows the points in Round \(2\). Which group increased its score the most?
Figure for problem 538173

Hints

- Compare the same group in both graphs. - Find each change from a) to b). - Choose the greatest positive change.

Solution

1. Group A increased by \(15 - 10 = 5\) points. 2. Group B changed by \(20 - 20 = 0\) points. 3. Group C increased by \(25 - 15 = 10\) points. 4. Group D decreased by \(25 - 20 = 5\) points. The greatest increase is \(10\), for Group C.

Answer

Group C increased its score the most, by \(10\) points.
5381743
The graphs show visitor counts in two different weeks. On which weekday did the number decrease the most from a) to b)?
Figure for problem 538174

Hints

- Compare the same weekday in both graphs. - Find each decrease from a) to b). - Choose the greatest decrease.

Solution

1. The decreases are Monday: \(30 - 25 = 5\), Tuesday: \(35 - 20 = 15\), Wednesday: \(25 - 25 = 0\), and Thursday: \(30 - 20 = 10\). 2. The greatest decrease is \(15\), on Tuesday.

Answer

The number decreased the most on Tuesday, by \(15\) visitors.
5381763
Graph a) shows bags of recyclables collected by three teams in Week \(1\), and Graph b) shows the bags collected by the same teams in Week \(2\). In which week were more bags collected altogether, and how many more?
Figure for problem 538176

Hints

- Find the total for each graph separately. - Compare the two totals. - Subtract to find how many more.

Solution

1. Week \(1\) total: \(15 + 25 + 20 = 60\) bags. 2. Week \(2\) total: \(20 + 15 + 30 = 65\) bags. 3. The difference is \(65 - 60 = 5\) bags.

Answer

Week \(2\) had \(5\) more bags collected.
5381773
Do the two graphs have the same total? Justify your answer using the values shown.
Figure for problem 538177

Hints

- Add all four values in Graph a). - Add all four values in Graph b). - Compare the totals.

Solution

1. Graph a) has a total of \(10 + 20 + 15 + 25 = 70\). 2. Graph b) has a total of \(15 + 10 + 25 + 20 = 70\). 3. The totals are equal.

Answer

Yes. Both graphs have a total of \(70\).
5381783
Compare Graphs a) and b). 1) Which regions increased? 2) Which regions decreased?
Figure for problem 538178

Hints

- Compare the same region in both graphs. - Mark each change as an increase or a decrease. - Check all four regions.

Solution

1. South increased from \(25\) to \(30\), and West increased from \(20\) to \(25\). 2. North decreased from \(30\) to \(25\), and East decreased from \(35\) to \(30\).

Answer

1) South and West increased. 2) North and East decreased.
5381803
How does the gap between A and B change from Graph a) to Graph b)?
Figure for problem 538180

Hints

- Find the A–B gap in each graph. - Compare the two gaps. - State whether the gap became larger or smaller.

Solution

1. In a), the gap is \(40 - 30 = 10\) points. 2. In b), the gap is \(45 - 40 = 5\) points. 3. The gap becomes \(10 - 5 = 5\) points smaller.

Answer

The gap becomes \(5\) points smaller.
5381813
Compare both graphs. 1) In Graph a), do more, fewer, or the same number of students travel by bike and on foot together than by bus? 2) In Graph b), do more, fewer, or the same number of students travel by bike and on foot together than by bus?
Figure for problem 538181

Hints

- Add the bike and walking values in each graph. - Compare each sum with the bus value in the same graph. - State the result for both a) and b).

Solution

1. In a), bike and walking total \(15 + 10 = 25\), the same as the bus value of \(25\). 2. In b), bike and walking total \(20 + 15 = 35\), while the bus value is \(20\). 3. In b), the combined total is \(35 - 20 = 15\) greater.

Answer

1) In Graph a), the numbers are equal. 2) In Graph b), \(15\) more students travel by bike and on foot together than by bus.
5381833
The total increases from a) to b). Which product contributes the most to this increase?
Figure for problem 538183

Hints

- Compare each product in the two graphs. - Find each increase or decrease. - Choose the greatest positive change.

Solution

1. Apple crates increase by \(25 - 20 = 5\). 2. Pear crates increase by \(25 - 15 = 10\). 3. Nut crates decrease by \(25 - 20 = 5\). 4. The greatest increase is \(10\), for pears.

Answer

Pears contribute the most, with an increase of \(10\) crates.
5381853
For each group, add its value from a) to its value from b). Which groups have the same sum?
Figure for problem 538185

Hints

- Add the two values for one group at a time. - Record all four sums. - Compare the sums.

Solution

1. Group A: \(5 + 20 = 25\). 2. Group B: \(10 + 15 = 25\). 3. Group C: \(15 + 10 = 25\). 4. Group D: \(20 + 5 = 25\). All four sums are equal.

Answer

All four groups have the same sum: \(25\) each.
5381863
In which graph are the greatest and least values farther apart?
Figure for problem 538186

Hints

- Find the greatest and least value in each graph. - Subtract to find each difference. - Compare the two differences.

Solution

1. In a), the difference between the greatest and least values is \(25 - 10 = 15\). 2. In b), the difference between the greatest and least values is \(20 - 15 = 5\). 3. Since \(15 > 5\), the values are farther apart in a).

Answer

The greatest and least values are farther apart in Graph a).
5381893
How many objects do the two graphs show altogether?
Figure for problem 538189

Hints

- Find the total in each graph first. - Add the two graph totals. - Check that all six bars were included.

Solution

1. Graph a) has \(15 + 20 + 10 = 45\) objects. 2. Graph b) has \(20 + 15 + 15 = 50\) objects. 3. Together, the graphs show \(45 + 50 = 95\) objects.

Answer

The two graphs show \(95\) objects altogether.
5381903
Which group is below \(20\) in a) but reaches at least \(20\) in b)?
Figure for problem 538190

Hints

- Check the value in a) first. - Then check the same group in b). - Apply both conditions to every group.

Solution

1. Group A increases from \(15\) to \(20\), so it meets both conditions. 2. Group C remains below \(20\). 3. Groups B and D were already above \(20\) in a).

Answer

Only Group A meets both conditions.
5381913
Which group has the greatest combined total from the two graphs?
Figure for problem 538191

Hints

- Add each group's two values. - Record the four totals. - Choose the greatest total.

Solution

1. Group A: \(10 + 20 = 30\). 2. Group B: \(15 + 10 = 25\). 3. Group C: \(20 + 15 = 35\). 4. Group D: \(25 + 20 = 45\). The greatest total is \(45\).

Answer

Group D has the greatest combined total, \(45\).
5381923
In Graph b), only Bar C will be increased until the two graphs have the same total. By how much must C increase?
Figure for problem 538192

Hints

- Find the total of each graph. - Find the difference between the totals. - Only Bar C changes.

Solution

1. Graph a) has a total of \(20 + 15 + 25 = 60\). 2. Graph b) has a total of \(15 + 20 + 20 = 55\). 3. The difference is \(60 - 55 = 5\), so C must increase by \(5\).

Answer

Bar C in b) must increase by \(5\).
5381993
How many students did not choose “Outdoors”?
Figure for problem 538199

Hints

- Add all slice values to find the total. - Identify the Outdoors value. - Subtract it from the total.

Solution

1. The chart total is \(14 + 10 + 8 = 32\) students. 2. The Outdoors slice represents \(10\) students. 3. The number who did not choose Outdoors is \(32 - 10 = 22\).

Answer

\(22\) students did not choose Outdoors.
5382013
Which statements are true? A: Lake and zoo have equal-sized slices. B: Castle has a larger slice than forest. C: Forest and castle have a combined value of \(20\). D: Zoo is twice as large as castle.
Figure for problem 538201

Hints

- Check each statement separately. - Use addition for a combined value and multiplication for “twice as large.” - Compare every statement with the labeled slice values.

Solution

1. A is true because Lake and Zoo each have a value of \(10\). 2. B is false because \(6 < 14\). 3. C is true because \(14 + 6 = 20\). 4. D is false because \(2 \times 6 = 12\), not \(10\).

Answer

Statements A and C are true.
5382023
A total of \(30\) color cards were counted. The number for yellow is missing from the pie chart. How many yellow cards are there?
Figure for problem 538202

Hints

- Add the three visible values. - Subtract their sum from the total of \(30\). - Check that all four values add to \(30\).

Solution

1. The visible values total \(8 + 7 + 9 = 24\). 2. The missing value is \(30 - 24 = 6\).

Answer

There are \(6\) yellow cards.
5382033
Four more students choose singing. Which activity will then have the greatest value?
Figure for problem 538203

Hints

- Increase only the singing value. - Keep the other two values unchanged. - Compare the three new values.

Solution

1. Singing increases from \(8\) to \(8 + 4 = 12\). 2. Dancing stays at \(11\), and painting stays at \(9\). 3. Since \(12\) is greatest, singing will have the greatest value.

Answer

Singing will have the greatest value, with \(12\) students.
5382063
The pie chart shows kinds of trees that were counted. 1) How many trees were counted altogether? 2) Which kind was most common? 3) Which kind was least common?
Figure for problem 538206

Hints

- Add all four slice values. - Find the greatest slice value. - Find the least slice value.

Solution

1. The total is \(9 + 13 + 6 + 12 = 40\) trees. 2. The greatest value, \(13\), belongs to maple. 3. The least value, \(6\), belongs to birch.

Answer

1) \(40\) trees were counted altogether. 2) Maple was most common. 3) Birch was least common.
5382093
Which pie chart represents more students altogether, and how many more?
Figure for problem 538209

Hints

- Add all slice values in each graph. - Compare the two totals. - Subtract to find how many more.

Solution

1. Graph a) represents \(8 + 7 + 5 = 20\) students. 2. Graph b) represents \(10 + 9 + 6 = 25\) students. 3. The difference is \(25 - 20 = 5\) students.

Answer

Graph b) represents \(5\) more students.
5382113
The total stays the same. Which section changes the most from a) to b), and in which direction?
Figure for problem 538211

Hints

- Compare each section in the two charts. - Find the size of each change. - Identify the greatest change and state its direction.

Solution

1. Reading increases from \(8\) to \(11\), a change of \(3\). 2. Sports decreases from \(9\) to \(5\), a change of \(4\). 3. Music increases from \(7\) to \(8\), a change of \(1\), and Games stays at \(6\). 4. The greatest change is \(4\), for Sports, and it is a decrease.

Answer

Sports changes the most. It decreases by \(4\).
5382123
In the pie chart, Noon's value is twice Morning's value, and Evening's value is three times Morning's value. Check both descriptions using the numbers shown, and find the total.
Figure for problem 538212

Hints

- Check the “twice” relationship with multiplication. - Check the “three times” relationship with multiplication. - Add all three values for the total.

Solution

1. Noon's value is twice Morning's value because \(2 \times 6 = 12\). 2. Evening's value is three times Morning's value because \(3 \times 6 = 18\). 3. The total is \(6 + 12 + 18 = 36\).

Answer

Both descriptions are correct. The total is \(36\).
5382943
Art studio: <table><tr><th>Main material</th><th>Number of artworks</th></tr><tr><td>Clay</td><td>\(13\)</td></tr><tr><td>Wood</td><td>\(19\)</td></tr><tr><td>Fabric</td><td>\(11\)</td></tr><tr><td>Paper</td><td>\(17\)</td></tr></table> For the final bar graph, \(4\) artworks are moved from the Wood category to the Paper category. What are the new bar lengths?

Hints

- One category loses \(4\) artworks and another category gains \(4\). - Check that the total number of artworks stays the same.

Solution

1. For Wood, subtract \(4\): \(19 - 4 = 15\). 2. For Paper, add \(4\): \(17 + 4 = 21\). 3. The Clay and Fabric values stay the same.

Answer

Clay: \(13\) Wood: \(15\) Fabric: \(11\) Paper: \(21\)
5383193
Three teams make paper airplanes. <table><thead><tr><th>Team</th><th>Round 1</th><th>Round 2</th></tr></thead><tbody><tr><td>Red</td><td>\(12\)</td><td>\(8\)</td></tr><tr><td>Blue</td><td>\(15\)</td><td>\(11\)</td></tr><tr><td>Green</td><td>\(9\)</td><td>\(14\)</td></tr></tbody></table> How many paper airplanes did the teams make in both rounds altogether?

Hints

- Use all six numbers in the table. - Find the total for each round first. - Add the two round totals.

Solution

1. In Round 1, the teams made \(12 + 15 + 9 = 36\) airplanes. 2. In Round 2, the teams made \(8 + 11 + 14 = 33\) airplanes. 3. Altogether, they made \(36 + 33 = 69\) airplanes.

Answer

The teams made \(69\) paper airplanes altogether.
5383603
A classroom book box is counted before and after a book drive. <table><thead><tr><th>Day</th><th>Before</th></tr></thead><tbody><tr><td>Monday</td><td>\(11\)</td></tr><tr><td>Tuesday</td><td>\(15\)</td></tr><tr><td>Wednesday</td><td>\(13\)</td></tr></tbody></table> <table><thead><tr><th>Day</th><th>After</th></tr></thead><tbody><tr><td>Monday</td><td>\(14\)</td></tr><tr><td>Tuesday</td><td>\(14\)</td></tr><tr><td>Wednesday</td><td>\(18\)</td></tr></tbody></table> On which day did the number increase the most?

Hints

- Compare the Before and After values for each day. - Find each increase, then compare the increases.

Solution

1. Monday's increase is \(14 - 11 = 3\). 2. Tuesday's value decreased from \(15\) to \(14\). 3. Wednesday's increase is \(18 - 13 = 5\). 4. The greatest increase is \(5\), on Wednesday.

Answer

Wednesday
5383643
A theater booth counts tickets sold. <table><thead><tr><th>Time</th><th>Child tickets</th><th>Adult tickets</th></tr></thead><tbody><tr><td>Morning</td><td>\(24\)</td><td>\(17\)</td></tr><tr><td>Afternoon</td><td>\(19\)</td><td>\(21\)</td></tr></tbody></table> Were more child tickets or adult tickets sold? How many more?

Hints

- Add the two values in the Child tickets column. - Add the two values in the Adult tickets column. - Compare the totals and find their difference.

Solution

1. The number of child tickets is \(24 + 19 = 43\). 2. The number of adult tickets is \(17 + 21 = 38\). 3. The difference is \(43 - 38 = 5\).

Answer

\(5\) more child tickets were sold.
5383923
A bin has small and large cubes in three colors. There are \(4\) small and \(3\) large red cubes, \(5\) small and \(2\) large blue cubes, and \(3\) small and \(6\) large green cubes. a) How many cubes are there of each color? b) How many cubes are there in all?

Hints

- Add the small and large cubes for each color. - Then add the three color totals.

Solution

1. Red: \(4 + 3 = 7\). 2. Blue: \(5 + 2 = 7\). 3. Green: \(3 + 6 = 9\). 4. The total is \(7 + 7 + 9 = 23\).

Answer

a) Red: \(7\); blue: \(7\); green: \(9\) b) \(23\) cubes
5384033
Fill in the two missing numbers. <table><thead><tr><th>Group</th><th>Red cards</th><th>Blue cards</th><th>Row total</th></tr></thead><tbody><tr><td>Group A</td><td>\(8\)</td><td>?</td><td>\(15\)</td></tr><tr><td>Group B</td><td>?</td><td>\(9\)</td><td>\(15\)</td></tr><tr><td>Column total</td><td>\(14\)</td><td>\(16\)</td><td>\(30\)</td></tr></tbody></table>

Hints

- Start with a row or column that has only one missing value. - Check each result against the other totals.

Solution

1. Group A has \(15 - 8 = 7\) blue cards. 2. Group B has \(14 - 8 = 6\) red cards. 3. These values also satisfy the remaining row and column totals.

Answer

Group A: \(7\) blue cards Group B: \(6\) red cards
5384073
A score table is changed after a review: Team North receives \(2\) more points in Round 2. Team South loses \(1\) point from Round 1. <table><thead><tr><th>Team</th><th>Round 1</th><th>Round 2</th></tr></thead><tbody><tr><td>North</td><td>\(14\)</td><td>\(11\)</td></tr><tr><td>South</td><td>\(12\)</td><td>\(13\)</td></tr></tbody></table> Which team has more points after the changes?

Hints

- Update the two affected table values first. - Find each team's new total, then compare.

Solution

1. Team North's new total is \(14 + (11 + 2) = 27\). 2. Team South's new total is \((12 - 1) + 13 = 24\). 3. Since \(27 > 24\), Team North has more points.

Answer

Team North has more points, with a total of \(27\).
5179173
A class buys \(6\) cases of drinks for \(\$9\) each and spends another \(\$38\) on food for a class party. After the purchases, \(\$18\) remains in the class fund. a) How much money was in the class fund before the purchases? b) Would the class have had enough money if the food had cost \(\$20\) more? Explain.

Hints

- Find the cost of the drinks and food together. - Add the money left to find the starting amount. - For part b, increase the total cost by \(\$20\) and compare.

Solution

1. Find the cost of the drinks: \(6 \times \$9 = \$54\). 2. Find the original total spent: \(\$54 + \$38 = \$92\). 3. Find the starting amount: \(\$92 + \$18 = \$110\). 4. If the food cost \(\$20\) more, the new total cost would be \(\$92 + \$20 = \$112\). 5. Since \(\$112 > \$110\), the class would not have enough money.

Answer

a) The class fund originally contained \(\$110\). b) No. The new cost would be \(\$112\), which is \(\$2\) more than the \(\$110\) available.
5185903
A box in the school library contains some nonfiction books at the start of the day. A class checks out \(15\) books and returns \(11\) of them at lunchtime. Later, three students each take \(4\) books home. One of those students returns \(2\) books that evening, while the other two keep their books for the weekend. At the end of the day, \(28\) books remain in the box. How many books were in the box at the start of the day?

Hints

- Find the net number of books removed by each group. - Three students each take the same number of books. - Add the net removals to the ending amount to work backward.

Solution

1. The first class removes a net total of \(15-11=4\) books. 2. The three students take \(3\times4=12\) books and return \(2\), for a net removal of \(12-2=10\) books. 3. Altogether, \(4+10=14\) books are missing from the starting amount. 4. Work backward from the \(28\) books remaining: \(28+14=42\).

Answer

\(42\) books
5204503
A school library removes \(18\) old books on Monday and buys \(45\) new books on Tuesday. On Wednesday, a community group donates a box of books. After all three changes, the library has \(110\) more books than it had before Monday. How many books were in the donated box?

Hints

- Find the net change after removing and buying books. - Compare that change with the final increase of \(110\). - The difference is the number of donated books.

Solution

1. Find the net change after Monday and Tuesday: \(45 - 18 = 27\). The library has \(27\) more books at that point. 2. The final increase is \(110\) books. Subtract the known increase: \(110 - 27 = 83\). 3. The donated box contains \(83\) books.

Answer

The donated box contained \(83\) books.
5213483
A bakery sells \(145\) loaves of bread in its store. It delivers \(25\) fewer loaves to a restaurant than it sells in the store. After the sales and delivery, the bakery has exactly as many loaves left as it sold and delivered altogether. How many loaves did the bakery have at the beginning of the morning?

Hints

- First find the number of loaves delivered to the restaurant. - Find the total number sold and delivered. - The number left is equal to that total. - Combine the loaves given out and the loaves left.

Solution

1. Find the number delivered to the restaurant: \(145 - 25 = 120\). 2. Find the total number sold and delivered: \(145 + 120 = 265\). 3. The bakery also has \(265\) loaves left. 4. Add the loaves given out and the loaves left: \(265 + 265 = 530\).

Answer

The bakery had \(530\) loaves at the beginning of the morning.
5381243
Three boxes contain marbles. Move marbles from Box A to Box B until the two boxes contain the same number. How many marbles must be moved?
Figure for problem 538124

Hints

- Find the difference between Box A and Box B. - Think about how moving one marble changes both boxes. - Check that the two new amounts are equal.

Solution

1. Box A contains \(24\) marbles, and Box B contains \(16\) marbles. 2. Their difference is \(24 - 16 = 8\). 3. Each marble moved decreases the difference by \(2\), because Box A loses one while Box B gains one. 4. Therefore, \(8 \div 2 = 4\) marbles must be moved.

Answer

\(4\) marbles must be moved from Box A to Box B.
5381273
Which two statements match the graph? A: Dogs and rabbits received the same number of votes. B: Horses received the most votes. C: Cats received \(4\) more votes than dogs. D: There were fewer than \(40\) votes altogether.
Figure for problem 538127

Hints

- Check each statement separately against the graph. - Pay attention to words such as “same,” “more,” and “most.” - For the total, add all four bar values.

Solution

1. Dogs and rabbits each received \(12\) votes, so A is true. 2. Cats received \(16 - 12 = 4\) more votes than dogs, so C is true. 3. B is false because cats received the most votes. 4. D is false because \(12 + 16 + 8 + 12 = 48\), which is not fewer than \(40\).

Answer

Statements A and C are true.
5381283
The graph shows glass containers collected for recycling. Which statement cannot be true? A: In February, \(15\) more containers were collected than in January. B: In March, half as many containers were collected as in February. C: In April, \(5\) more containers were collected than in March.
Figure for problem 538128

Hints

- Check each statement against the graph. - For “half as many,” divide the larger value by \(2\). - For “more than,” find the difference.

Solution

1. A is true because \(40 - 25 = 15\). 2. B is false because half of \(40\) is \(20\), not \(30\). 3. C is true because \(35 - 30 = 5\).

Answer

Statement B cannot be true.
5381413
Which two animal teams differ by exactly \(8\) collected items? Name every pair that works.
Figure for problem 538141

Hints

- Write down the four values. - Find the difference for each possible pair in an organized way. - Make sure you list every pair with a difference of \(8\).

Solution

1. Hedgehog and Mouse differ by \(20 - 12 = 8\). 2. Frog and Owl differ by \(16 - 8 = 8\). 3. Checking the other pairs shows that none of their differences is \(8\).

Answer

Hedgehog and Mouse; Frog and Owl.
5381513
Which two groups have the closest scores? What is the difference between their scores?
Figure for problem 538151

Hints

- Write down all four scores. - Compare the differences between possible pairs. - Choose the least positive difference.

Solution

1. Compare the score differences for all pairs. 2. The least difference is between Group B with \(30\) points and Group C with \(25\) points. 3. Their difference is \(30 - 25 = 5\) points.

Answer

Groups B and C have the closest scores. Their scores differ by \(5\) points.
5381603
Divide the four groups into two pairs so that the pairs have equal point totals. Give the pairing.
Figure for problem 538160

Hints

- Write down the four point values. - Test ways to split the groups into two pairs. - Add both pair totals to check that they are equal.

Solution

1. Groups A and B have \(10 + 25 = 35\) points. 2. Groups C and D have \(15 + 20 = 35\) points. 3. The two pair totals are equal.

Answer

Pair A with B and pair C with D. Each pair has \(35\) points.
5381663
Only Group C may receive more points. What is the least number of points it needs to be alone in first place?
Figure for problem 538166

Hints

- Find the current greatest score. - Determine the least score that is greater than it. - Subtract Group C's current score from that target.

Solution

1. Group C has \(20\) points, and the current greatest score is \(25\). 2. To be alone in first place, Group C must have at least \(26\) points. 3. It needs \(26 - 20 = 6\) more points.

Answer

Group C needs at least \(6\) more points.
5381703
On which two days were exactly \(36\) admission tickets sold altogether? Name every possible pair.
Figure for problem 538170

Hints

- Write down all five day values. - Test pairs in an organized way. - List every pair with a sum of \(36\).

Solution

1. Monday and Friday give \(12 + 24 = 36\). 2. Tuesday and Thursday give \(15 + 21 = 36\). 3. Checking the other day pairs shows that none has a sum of \(36\).

Answer

Monday and Friday; Tuesday and Thursday.
5381823
In Graph b), the four values from Graph a) were rearranged. 1) Which value moved from A to C? 2) Which value moved from B to D? 3) Which value moved from C to A? 4) Which value moved from D to B?
Figure for problem 538182

Hints

- Match equal values across the two graphs. - Follow each starting bar named in the question. - Record all four value moves.

Solution

1. A in a) and C in b) both have \(12\). 2. B in a) and D in b) both have \(18\). 3. C in a) and A in b) both have \(24\). 4. D in a) and B in b) both have \(30\).

Answer

1) A to C: \(12\) 2) B to D: \(18\) 3) C to A: \(24\) 4) D to B: \(30\)
5382043
Some students switch their vote from cocoa to tea. How many must switch so that cocoa and tea have equal numbers of votes?
Figure for problem 538204

Hints

- Find the difference between cocoa and tea. - Think about how one switch changes both values. - Check that the two new values are equal.

Solution

1. Cocoa has \(14\) votes, and tea has \(8\) votes, a difference of \(14 - 8 = 6\). 2. Each student who switches decreases cocoa by \(1\) and increases tea by \(1\), reducing the difference by \(2\). 3. Therefore, \(6 \div 2 = 3\) students must switch.

Answer

\(3\) students must switch from cocoa to tea.
5384043
Fill in the three missing numbers. <table><thead><tr><th>Color</th><th>Small stars</th><th>Large stars</th><th>Combined</th></tr></thead><tbody><tr><td>Silver</td><td>\(9\)</td><td>?</td><td>\(14\)</td></tr><tr><td>Gold</td><td>?</td><td>?</td><td>\(15\)</td></tr><tr><td>Total</td><td>\(16\)</td><td>\(13\)</td><td>\(29\)</td></tr></tbody></table>

Hints

- Begin with a row or column that has only one missing value. - Enter each result before using another total.

Solution

1. The number of large silver stars is \(14 - 9 = 5\). 2. The number of small gold stars is \(16 - 9 = 7\). 3. The number of large gold stars is \(15 - 7 = 8\).

Answer

Large silver stars: \(5\) Small gold stars: \(7\) Large gold stars: \(8\)
5384053
Exactly one value inside the table is incorrect. The totals along the edges are correct. <table><thead><tr><th>Team</th><th>Round 1</th><th>Round 2</th><th>Combined</th></tr></thead><tbody><tr><td>Team A</td><td>\(6\)</td><td>\(8\)</td><td>\(14\)</td></tr><tr><td>Team B</td><td>\(7\)</td><td>\(10\)</td><td>\(16\)</td></tr><tr><td>Total</td><td>\(13\)</td><td>\(17\)</td><td>\(30\)</td></tr></tbody></table> Find and correct the incorrect value.

Hints

- Check every row total and column total. - The corrected value must satisfy both its row total and its column total.

Solution

1. Team B's Round 2 value must combine with \(7\) to make the row total \(16\). 2. The correct value is \(16 - 7 = 9\). 3. This also makes the Round 2 column total correct because \(8 + 9 = 17\).

Answer

The Team B, Round 2 entry should be \(9\), not \(10\).

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