Aimathic
Login | English | Deutsch

Grade 3 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Two-step word problems

Learning goals

  • Identifies the relevant quantities and the unknown in a context.
  • Chooses operations or equations that represent the situation.
  • Carries out the two calculations needed in a word problem.
  • Interprets the result in context, including units when appropriate.
  • Checks the answer for reasonableness.

Click problems to add them to your worksheet.

Two-step word problems 1
Theo has \(5\) bags with \(3\) marbles in each. He gives away \(4\) marbles. What must he calculate before subtracting the marbles he gave away? Explain why.

Hints

- What total do you need before you can find what remains? - Which numbers tell you how many marbles Theo starts with?

Solution

1. Theo needs the starting total of marbles before he subtracts what he gives away. 2. The first calculation is \(5\times3\).

Answer

\(5\times3\), because he must find the total number of marbles before subtracting \(4\).
Two-step word problems 2
Two toy train cars each hold \(3\) bears. One more bear waits at the station. Lena writes \(2\times3=6\), then \(6+1=7\). What do \(6\) and \(7\) count? Explain why she needs both steps.

Hints

- Picture the two cars and the station as separate places. - Which number would you use if you were counting only the riders? - Check whether the final number counts every bear in the story.

Solution

1. The \(6\) counts the bears riding in the two cars: \(2\times3=6\). 2. Adding the bear at the station gives \(6+1=7\) bears altogether. 3. The first step counts the equal groups; the second includes the extra bear.

Answer

\(6\) is the number of bears in the cars. \(7\) includes the bear at the station. The steps count the groups first and then the extra bear.
Two-step word problems 3
Ava has \(3\) boxes with \(2\) crayons in each box. She gives away \(1\) crayon. Ava finds \(3\times2=6\) first. Finish her work to find how many crayons she has left. Tell what the \(6\) counts.

Hints

- Picture the crayons just before Ava gives any away. - Try checking your final answer by putting the given-away crayon back.

Solution

1. The \(6\) counts all the crayons before Ava gives any away. 2. The remaining crayons are \(6-1=5\).

Answer

\(6\) is the starting number of crayons; \(6-1=5\), so \(5\) crayons remain.
Two-step word problems 4
Ben says, “When I multiply my number by \(3\), I get \(30\).” Mia says, “My number is \(4\) greater than Ben's number.” What number is Mia thinking of?

Hints

- Find Ben's number first. - Use division to undo multiplication by \(3\). - Then add \(4\).

Solution

1. Find Ben’s number: \(30 \div 3 = 10\). 2. Mia’s number is \(10 + 4 = 14\).

Answer

Mia is thinking of \(14\).
Two-step word problems 5
Start with \(340\). a) What number is \(60\) greater than \(340\)? b) What number is \(400\) greater than your answer to part a)?

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

a) \(400\) b) \(800\)
Two-step word problems 6
A colored-pencil case has \(16\) pencils in the top section and \(13\) in the bottom section. Four pencils are added to the top section. How many pencils are in the case afterward?

Hints

- Update the section that changes first. - The bottom section keeps its original number of pencils. - Add the two section amounts after the change.

Solution

1. The top section then has \(16+4=20\) pencils. 2. Add the unchanged bottom section: \(20+13=33\) pencils.

Answer

There are \(33\) colored pencils in the case afterward.
Two-step word problems 7
A bin has \(7\) red cubes. It also has \(5\) small blue cubes and \(2\) large blue cubes. How many red and blue cubes are in the bin altogether?

Hints

- Combine the two sizes of blue cubes first. - The red-cube total is already given. - Add the two color totals.

Solution

1. Find the number of blue cubes: \(5+2=7\). 2. Add red and blue cubes: \(7+7=14\).

Answer

There are \(14\) red and blue cubes altogether.
Two-step word problems 8
A school collects \(135\) cans on Monday and \(120\) cans on Tuesday. Its goal is \(300\) cans. How many more cans must the school collect to reach the goal?

Hints

- First find how many cans have already been collected altogether. - Then compare that subtotal with the goal. - Check that the collected amount plus your answer equals \(300\).

Solution

1. Combine the two collection days: \(135+120=255\) cans. 2. Find the amount still needed: \(300-255=45\) cans.

Answer

The school must collect \(45\) more cans.
Two-step word problems 9
A classroom has \(4\) boxes with \(6\) markers in each box and \(9\) loose markers. How many markers are there altogether?

Hints

- First find the number of markers in the boxes. - Then add the loose markers. - Keep the two steps separate until you have found the product.

Solution

1. Find the markers in the boxes: \(4 \times 6=24\). 2. Add the loose markers: \(24+9=33\).

Answer

There are \(33\) markers altogether.
Two-step word problems 10
Ms. Rivera has \(50\) stickers. She sets aside \(14\) stickers, then shares the rest equally among \(6\) students. How many stickers does each student receive?

Hints

- The stickers set aside are not shared. - Find the amount that remains before forming equal shares. - The final answer is the size of one equal share.

Solution

1. Find the number left to share: \(50-14=36\). 2. Share equally among six students: \(36 \div 6=6\).

Answer

Each student receives \(6\) stickers.
Two-step word problems 11
Mr. Kim shares \(30\) pencils equally among \(5\) tables. Then he places \(2\) extra pencils on each table. How many pencils does each table have now?

Hints

- First find the equal share before any extra pencils are added. - The same number of extra pencils is then added to each table. - The question asks for the new amount on one table, not the total number of extra pencils.

Solution

1. The equal share is \(30 \div 5=6\) pencils per table. 2. Add the extra pencils: \(6+2=8\) pencils per table.

Answer

Each table has \(8\) pencils.
Two-step word problems 12
Camila has \(5\) packs with \(8\) stickers in each pack. Camila uses \(6\) stickers. Write one equation using \(n\) for the number of stickers left, then find \(n\).

Hints

- Decide what \(n\) represents before writing the equation. - Represent the equal packs first, then show what happens to that total. - Check that your equation follows the order of events in the story.

Solution

1. Camila starts with \(5\times8\) stickers, and then \(6\) are used. 2. An equation is \(n=5\times8-6\). 3. \(5\times8=40\), and \(40-6=34\), so \(n=34\).

Answer

\(n=5 \times 8-6\), so \(n=34\).
Two-step word problems 13
In picture a), each row stands for one plate of cookies. Picture b) shows loose cookies. How many cookies are there altogether? Write two calculations. Explain why the loose cookies are added only once.
Figure for problem 558458

Hints

- Could equal groups help you count picture a) without counting every dot? - Which cookies do you still need to count after finding the total on the plates?

Solution

1. The three plates have four cookies each, giving \(3\times4\). 2. The plates have \(3\times4=12\) cookies. 3. The two loose cookies make \(12+2=14\) cookies in all. They are not on every plate.

Answer

\(3\times4=12\); \(12+2=14\) cookies. The two loose cookies are separate from the plates, so add them once.
Two-step word problems 14
A fruit stand has \(4\) boxes of \(6\) pears. It sells \(5\) pears. How many pears are left? Write both calculations.

Hints

- What do the boxes tell you about how the pears are arranged? - Which pears belong in the amount left at the end?

Solution

1. First find the starting number of pears: \(4\times6=24\). 2. Subtract those sold: \(24-5=19\).

Answer

\(4\times6=24\); \(24-5=19\). There are \(19\) pears left.
Two-step word problems 15
\(18\) ribbons are shared equally among \(3\) tables. Then each table gets \(2\) more ribbons. How many ribbons does each table have now?

Hints

- If the ribbons were shared evenly, how would you work out one table's share? - What does “each table” tell you about the extra ribbons?

Solution

1. Each table gets \(18\div3=6\) ribbons at first. 2. Two more makes \(6+2=8\) ribbons per table.

Answer

\(8\) ribbons per table.
Two-step word problems 16
A gardener fills \(4\) baskets with \(6\) apples in each. Then the apples are shared equally among \(3\) neighbors. How many apples does each neighbor get?

Hints

- What does the count for one basket tell you about the harvest? - What amount is being shared among the neighbors?

Solution

1. The baskets hold \(4\times6=24\) apples altogether. 2. Each neighbor gets \(24\div3=8\) apples.

Answer

Each neighbor gets \(8\) apples.
Two-step word problems 17
The triangles mark fence posts in a straight line. Each neighboring pair is \(3\,\text{m}\) apart. How far is it from the first post to the last post?
Figure for problem 558463

Hints

- How many posts are marked? Is that also the number of spaces between them? - Which parts of the picture each have the given distance?

Solution

1. Five posts make \(5-1=4\) spaces. 2. The distance is \(4\times3=12\,\text{m}\).

Answer

\(12\,\text{m}\).
Two-step word problems 18
A shelf has \(5\) rows of \(4\) books. Two whole rows are removed. How many books remain? Show two calculations.

Hints

- Which books are still on the shelf? - Can the equal rows help you count what is left?

Solution

1. First find the rows left: \(5-2=3\). 2. The remaining books are \(3\times4=12\).

Answer

\(5-2=3\) rows; \(3\times4=12\) books remain.
Two-step word problems 19
Noah finds \(7\) shells on the beach and \(5\) more in a bucket. He shares all the shells equally among \(3\) cousins. How many shells does each cousin get? Show both steps and tell what you find first.

Hints

- Imagine moving all the shells onto a table before sharing them. - What would a sketch of three equal shares need to show? - Check your result by putting the shares back together.

Solution

1. Find the total number of shells: \(7+5=12\). 2. Share the total among the three cousins: \(12\div3=4\). 3. Each cousin gets \(4\) shells.

Answer

First \(7+5=12\) shells in all. Then \(12\div3=4\) shells per cousin.
Two-step word problems 20
A pet shop sells \(4\) fish. After the sale, the fish that are left are placed in \(3\) tanks with \(5\) fish in each tank. How many fish did the shop have before the sale? Show both steps.

Hints

- Does the tank count describe the fish before or after the sale? - Think about how the number changed when the fish were sold. - Check your starting number by taking away the fish sold.

Solution

1. After the sale, \(3\times5=15\) fish remain. 2. Add back the \(4\) fish sold to find the starting number: \(15+4=19\).

Answer

\(3\times5=15\) fish remained; \(15+4=19\) fish were there before the sale.
Two-step word problems 21
Fatima has \(4\) bags of seeds. Each bag has \(10\) seeds, but \(3\) seeds in each bag are too old to plant. How many good seeds are there in all? Write both calculations.

Hints

- Are the old seeds all in one bag, or in every bag? - How many seeds from one bag can Fatima plant? - How can you find the number for all the bags without counting one by one?

Solution

1. Each bag has \(10-3=7\) good seeds. 2. With \(4\) bags, the total is \(4\times7=28\) good seeds.

Answer

\(10-3=7\) good seeds in each bag; \(4\times7=28\) good seeds in all.
Two-step word problems 22
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.
Two-step word problems 23
A school event begins with \(120\) bottles of apple juice and \(215\) bottles of water. By the end of the event, \(217\) bottles have been used. How many full bottles remain?

Hints

- First combine the two starting groups of bottles. - Then decide how the total changes when bottles are used. - Keep the two steps separate and check that the final amount is less than the starting amount.

Solution

1. Find the starting total: \(120 + 215 = 335\) bottles. 2. Subtract the bottles used: \(335 - 217 = 118\) bottles.

Answer

There are \(118\) full bottles remaining.
Two-step word problems 24
A bird park has \(42\) penguins. It has \(15\) more flamingos than penguins. The park also has \(124\) parakeets. How many flamingos and parakeets are there altogether?

Hints

- Use the penguin count to find the number of flamingos first. - Then focus only on the two bird groups named in the question. - Check that the flamingo count is greater than \(42\) before finding the final total.

Solution

1. Find the number of flamingos: \(42 + 15 = 57\). 2. Combine the flamingos and parakeets: \(57 + 124 = 181\).

Answer

There are \(181\) flamingos and parakeets altogether.
Two-step word problems 25
Students are logging miles in a school bike challenge. Class 3A rides \(267\) miles and Class 3B rides \(284\) miles. a) How many miles do the two third-grade classes ride altogether? b) The fourth-grade classes ride \(686\) miles altogether. How many more miles do they ride than the two third-grade classes?

Hints

- Combine the two third-grade distances first. - For “how many more,” compare the larger total with the smaller total. - Check that your difference added to the third-grade total reaches the fourth-grade total.

Solution

1. Add the two third-grade distances: \(267 + 284 = 551\) miles. 2. Compare the totals: \(686 - 551 = 135\) miles.

Answer

a) The two third-grade classes ride \(551\) miles altogether. b) The fourth-grade classes ride \(135\) more miles.
Two-step word problems 26
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 \(\$265\). How much money will he have left after the purchase?

Hints

- First combine the two purchase prices. - Then compare the total cost with the amount Paul has. - The money left should be less than \(\$265\).

Solution

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

Answer

Paul will have \(\$12\) left.
Two-step word problems 27
Ms. Brooks has \(\$75\) for a class party. Drinks cost \(\$36\) and snacks cost \(\$45\). Is the budget enough? If not, how much more is needed?

Hints

- Combine the two costs first. - Compare the total cost with the budget before deciding whether there is enough money. - If the cost is larger, the amount needed is the difference between the two amounts.

Solution

1. Find the total cost: \(\$36 + \$45 = \$81\). 2. Since \(\$81 > \$75\), find the shortfall: \(\$81 - \$75 = \$6\).

Answer

No. Ms. Brooks needs \(\$6\) more.
Two-step word problems 28
A sports club has \(\$250\). Soccer balls cost \(\$125\), and practice vests cost \(\$130\). Maya adds the two costs and compares the total with \(\$250\). Jordan subtracts \(\$125\) from \(\$250\), then compares what is left with \(\$130\). Use both methods. Do they give the same result? How much more money does the club need?

Hints

- In Maya’s method, compare the total purchase cost with the budget. - In Jordan’s method, compare the second purchase with what remains after the first purchase. - Think about why both methods measure the same gap between cost and available money.

Solution

1. Maya’s method: \(\$125 + \$130 = \$255\), then \(\$255 - \$250 = \$5\). The club is short by \(\$5\). 2. Jordan’s method: \(\$250 - \$125 = \$125\), then \(\$130 - \$125 = \$5\). The club is short by \(\$5\). 3. Both methods compare the same two purchase costs with the same budget, so they measure the same shortfall.

Answer

Both methods show that the club needs \(\$5\) more.
Two-step word problems 29
Tim and Sarah walk to a playground on \(5\) days in one week. Tim's walk there and back is \(120\,\text{m}\) in total each day. Sarah's walk there and back is \(160\,\text{m}\) in total each day. How many more meters does Sarah walk than Tim during the week?

Hints

- Compare the two daily distances first. - The difference is the same on each walking day. - The question asks for the difference over the whole week.

Solution

1. Find the difference for one day: \(160-120=40\,\text{m}\). 2. The difference repeats for \(5\) days: \(5\times40=200\,\text{m}\).

Answer

Sarah walks \(200\,\text{m}\) more than Tim during the week.
Two-step word problems 30
During recess, Mia runs \(320\,\text{m}\). She then walks for \(9\) minutes at \(60\,\text{m}\) per minute. How many meters does Mia travel altogether?

Hints

- Find the walking distance from the minutes and the distance per minute. - Then combine that distance with the running distance already given. - Check that the final distance is greater than each part.

Solution

1. Find the walking distance: \(9 \times 60\,\text{m} = 540\,\text{m}\). 2. Add the running and walking distances: \(320\,\text{m} + 540\,\text{m} = 860\,\text{m}\).

Answer

Mia travels \(860\,\text{m}\) altogether.
Two-step word problems 31
A long-distance hiking club plans a three-day hike that is \(78\) miles long. Members 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 is the distance greatest?

Hints

- Which day's distance do you still need to know? - How must the three daily distances relate to the whole hike? - When you have all three distances, check whether your choice is greater than each of the others.

Solution

1. Add the first two distances: \(24+29=53\) miles. 2. The third-day distance is \(78-53=25\) miles. 3. Compare \(24\), \(29\), and \(25\). Day 2 is the longest.

Answer

Day 2, with \(29\) miles.
Two-step word problems 32
A cycling club rides over three days. <table><tr><th colspan="3">Total distance: \(167\) miles</th></tr><tr><td>Day 1: \(52\) miles</td><td>Day 2: \(64\) miles</td><td>Day 3: ? miles</td></tr></table> How many miles does the club ride on Day 3? Which day is the shortest?

Hints

- Find the distance covered on the first two days together. - The three days must add up to the total distance. - Compare all three daily distances after finding the missing one.

Solution

1. Add the known distances: \(52+64=116\) miles. 2. Subtract: \(167-116=51\) miles on Day 3. 3. Compare \(52\), \(64\), and \(51\). Day 3 is shortest.

Answer

The club rides \(51\) miles on Day 3. Day 3 is the shortest.
Two-step word problems 33
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 \(24\,\text{kg}\).
Two-step word problems 34
A farm packs \(45\,\text{kg}\) of potatoes equally into \(9\) bags. A family buys \(4\) bags. How many kilograms of potatoes does the family buy?

Hints

- First find the equal amount in one bag. - Then use the number of bags the family buys. - Check that the amount for \(4\) bags is less than the full \(45\,\text{kg}\).

Solution

1. Find the weight of one bag: \(45\,\text{kg} \div 9 = 5\,\text{kg}\). 2. Find the weight of \(4\) bags: \(4 \times 5\,\text{kg} = 20\,\text{kg}\).

Answer

The family buys \(20\,\text{kg}\) of potatoes.
Two-step word problems 35
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?

Hints

- First find the cost of one serving. - Use that price for one serving for the new number of servings. - Check that buying more than \(4\) servings costs more than \(\$12\).

Solution

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

Answer

a) One serving costs \(\$3\). b) Seven servings cost \(\$21\).
Two-step word problems 36
A gardener plants \(27\) tulips equally in \(3\) rows. a) How many tulips are in one row? b) The gardener has \(45\) more tulips. How many additional rows of the same size can be planted?

Hints

- Find the size of one equal row first. - Use that row size to group the additional tulips. - The second answer is a number of rows, not a number of tulips.

Solution

1. Find the number of tulips in one row: \(27 \div 3 = 9\). 2. Find how many rows of \(9\) fit into \(45\): \(45 \div 9 = 5\).

Answer

a) One row contains \(9\) tulips. b) The gardener can plant \(5\) additional rows.
Two-step word problems 37
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 \(\$2\).
Two-step word problems 38
Lucas buys a poster for \(\$14\). Two identical picture frames together cost \(\$6\) more 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 \(\$10\).
Two-step word problems 39
Ms. Okafor buys \(8\) identical packages of markers. She pays with \(\$50\) and receives \(\$18\) in change. How much does one package cost?

Hints

- First find how much Ms. Okafor 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 \(\$4\).
Two-step word problems 40
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.
Two-step word problems 41
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.
Two-step word problems 42
Ms. Weber pays \(\$25\) for some notebooks and receives \(\$4\) in change. Each notebook costs \(\$3\). How many notebooks does Ms. Weber buy?

Hints

- First find how much money was actually spent after the change was returned. - Then use the price of one notebook to form equal groups. - Check that the number of notebooks times \(\$3\) gives the amount spent.

Solution

1. Find the amount spent: \(\$25 - \$4 = \$21\). 2. Divide by the price of one notebook: \(\$21 \div \$3 = 7\).

Answer

Ms. Weber buys \(7\) notebooks.
Two-step word problems 43
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.
Two-step word problems 44
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.
Two-step word problems 45
A store has \(9\) school-supply sets. Each set originally contains \(6\) items. The store adds \(1\) highlighter to each set. How many items are in the \(9\) sets now?

Hints

- First find the new number of items in one set. - Then treat the \(9\) sets as equal groups. - Check that adding one item to every set makes the final total larger than \(9 \times 6\).

Solution

1. Each changed set has \(6 + 1 = 7\) items. 2. Nine equal sets contain \(9 \times 7 = 63\) items.

Answer

The \(9\) sets contain \(63\) items.
Two-step word problems 46
A school receives \(100\) wide-ruled notebooks. It also orders \(3\) packages with \(50\) graph-paper notebooks in each package. How many notebooks does the school receive altogether?

Hints

- Find the number of graph-paper notebooks from the equal packages. - Then combine that amount with the wide-ruled notebooks already given. - The final total should be greater than both separate amounts.

Solution

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

Answer

The school receives \(250\) notebooks altogether.
Two-step word problems 47
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?

Hints

- First find the equal amount in one flower bed. - Use that bed size for the new number of beds. - Check that \(8\) beds require more tulips than \(3\) beds.

Solution

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

Answer

a) \(6\) tulips b) \(48\) tulips
Two-step word problems 48
Six boxes of apples have the same weight. Together they weigh \(30\,\text{kg}\). A garden cart can carry at most \(48\,\text{kg}\). Can the cart carry \(10\) boxes at once? Show how you know.

Hints

- Find the weight of one equal box first. - Use that weight to find the weight of \(10\) boxes. - Compare the result with the cart’s maximum load.

Solution

1. Find the weight of one box: \(30\,\text{kg} \div 6 = 5\,\text{kg}\). 2. Ten boxes weigh \(10 \times 5\,\text{kg} = 50\,\text{kg}\). 3. Since \(50\,\text{kg} > 48\,\text{kg}\), the cart cannot carry \(10\) boxes at once.

Answer

No. Ten boxes weigh \(50\,\text{kg}\), which is more than the \(48\,\text{kg}\) limit.
Two-step word problems 49
Luke uses \(8\) ounces of flour to bake \(4\) muffins. How many ounces of flour does he need for \(6\) muffins?

Hints

- First find the amount of flour for one muffin. - Use that amount for the new number of muffins. - Check that the flour amount increases because the batch has more muffins.

Solution

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

Answer

Luke needs \(12\,\text{oz}\) of flour for \(6\) muffins.
Two-step word problems 50
Three equal packages contain \(24\) colored pencils altogether. How many colored pencils are in \(7\) packages of the same size?

Hints

- First find the number of colored pencils in one package. - Use that package size for the new number of packages. - Check that \(7\) packages contain more pencils than \(3\) packages.

Solution

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

Answer

Seven packages contain \(56\) colored pencils.
Two-step word problems 51
During physical education class, Coach Ramirez 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.
Two-step word problems 52
For a school event, Class 3A buys \(4\) equal bags with \(32\) oranges altogether. Class 3B buys \(3\) more bags of the same size. How many oranges are in the \(3\) extra bags?

Hints

- First use Class 3A’s total and bag count to find how many oranges are in one bag. - The question asks only about the additional bags, so no subtraction of bag counts is needed. - After finding the amount in one bag, scale it to the stated number of additional bags.

Solution

1. Find the number of oranges in one bag: \(32 \div 4=8\). 2. Multiply the \(3\) additional bags by \(8\) oranges per bag: \(3 \times 8=24\). 3. The additional bags contain \(24\) oranges.

Answer

The \(3\) additional bags contain \(24\) oranges.
Two-step word problems 53
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.
Two-step word problems 54
Four equal packages of colored pencils cost \(\$32\) altogether. How much do \(9\) packages cost?

Hints

- Find the cost of one package first. - Use that price for one package for \(9\) equal packages. - Check that \(9\) packages cost more than \(4\) packages.

Solution

1. Find the cost of one package: \(\$32 \div 4 = \$8\). 2. Nine packages cost \(9 \times \$8 = \$72\).

Answer

Nine packages cost \(\$72\).
Two-step word problems 55
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.
Two-step word problems 56
A school cafeteria has \(85\) apples. It has \(25\) fewer pears than apples. It has as many bananas as apples and pears together. How many bananas does the cafeteria have?

Hints

- Make a note or sketch for each type of fruit. Which amounts do you know? - Imagine as many pears as apples. What changes because there are fewer pears? - Should the banana count be greater than the apple count? Use the story to check.

Solution

1. Find the pears: \(85-25=60\). 2. There are as many bananas as apples and pears together: \(85+60=145\).

Answer

\(145\) bananas.
Two-step word problems 57
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.
Two-step word problems 58
The diagram shows apple trees planted in one straight row. Each marker represents one tree. Neighboring trees are \(4\,\text{m}\) apart. What is the distance from the first tree to the last tree?
Figure for problem 517850

Hints

- Count the spaces between neighboring tree markers rather than counting the markers as spaces. - How many spaces are made by \(8\) objects in one row? - Use the same distance for every space.

Solution

1. The diagram shows \(8\) trees, so there are \(8 - 1 = 7\) 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 \(28\,\text{m}\).
Two-step word problems 59
A \(12\,\text{m}\)-long string of pennant flags has a flag at each end. The diagram shows all the equally spaced flags. What is the distance between each pair of neighboring flags?
Figure for problem 517851

Hints

- Count the spaces between the flag markers, not just the flags. - The full \(12\,\text{m}\) is shared equally among those spaces. - Choose the operation that splits one total into equal parts.

Solution

1. The diagram shows \(5\) flags, so there are \(5 - 1 = 4\) 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 \(3\,\text{m}\).
Two-step word problems 60
Ms. Miller already has \(56\) glitter stickers. She buys \(9\) packs with \(5\) star stickers in each pack. How many stickers does she have altogether?

Hints

- Find the number of star stickers in the equal packs. - Then combine that amount with the glitter stickers already given. - Check that the final total is greater than each separate amount.

Solution

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

Answer

Ms. Miller has \(101\) stickers altogether.
Two-step word problems 61
A school receives \(100\) new notebooks. It gives out \(60\) of them. The notebooks left are shared equally among \(5\) study groups. How many notebooks does each group receive?

Hints

- First find how many notebooks remain after the ones already given out. - Then share the remaining notebooks equally among the groups. - Check that \(5\) equal groups of your answer use all \(40\) remaining notebooks.

Solution

1. Find the number of notebooks remaining: \(100 - 60 = 40\). 2. Share the \(40\) notebooks equally: \(40 \div 5 = 8\).

Answer

Each study group receives \(8\) notebooks.
Two-step word problems 62
Emma buys \(4\) notebooks for \(\$3\) each. After paying, she has \(\$24\) left. How much money did Emma have before shopping?

Hints

- First find how much Emma spends on the equal-price notebooks. - Her starting amount is the money spent together with the money left. - Check by subtracting the purchase cost from your starting amount.

Solution

1. Find the cost of the notebooks: \(4 \times \$3 = \$12\). 2. Add the amount spent and the money left: \(\$12 + \$24 = \$36\).

Answer

Emma had \(\$36\) before shopping.
Two-step word problems 63
A class buys \(6\) cases of drinks for \(\$9\) each. After paying, \(\$18\) remains in the class fund. How much money was in the fund before the purchase?

Hints

- First find the total cost of the equal-price cases. - The starting fund includes both the amount spent and the money left. - Check by subtracting the purchase cost from your starting amount.

Solution

1. Find the cost of the drinks: \(6 \times \$9 = \$54\). 2. Add the amount spent and the money left: \(\$54 + \$18 = \$72\).

Answer

The class fund originally contained \(\$72\).
Two-step word problems 64
Three child tickets for a school play cost \(\$5\) each. A group ticket for three children costs \(\$12\). How much is saved by buying the group ticket instead of three individual tickets?

Hints

- Find the cost of the three individual tickets first. - Savings is the difference between the larger price and the smaller price. - Check that the group ticket costs less than the three individual tickets together.

Solution

1. Find the cost of three individual tickets: \(3 \times \$5 = \$15\). 2. Compare that cost with the group ticket: \(\$15 - \$12 = \$3\).

Answer

Buying the group ticket saves \(\$3\).
Two-step word problems 65
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.
Two-step word problems 66
A sprinkler uses exactly \(45\,\text{L}\) of water in \(5\) minutes. It uses the same amount of water each minute. How much water does it use in \(8\) minutes?

Hints

- First find how much water is used in one minute. - Use that constant amount for \(8\) minutes. - Check that the amount for \(8\) minutes is greater than the amount for \(5\) minutes.

Solution

1. Find the amount used per minute: \(45\,\text{L} \div 5 = 9\,\text{L}\). 2. Find the amount used in eight minutes: \(8 \times 9\,\text{L} = 72\,\text{L}\).

Answer

The sprinkler uses \(72\,\text{L}\) in \(8\) minutes.
Two-step word problems 67
A school copier makes \(10\) copies in \(20\) seconds. It keeps working at the same speed. How many copies can it make in one minute, or \(60\) seconds?

Hints

- Compare \(60\) seconds with the given \(20\)-second interval. - Find the number of equal \(20\)-second intervals in one minute. - The copier makes the same number of copies during each equal interval.

Solution

1. Find how many \(20\)-second intervals fit in \(60\) seconds: \(60 \div 20 = 3\). 2. The copier makes \(10\) copies in each interval, so it makes \(3 \times 10 = 30\) copies.

Answer

The copier can make \(30\) copies in one minute.
Two-step word problems 68
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.
Two-step word problems 69
Jonas plants \(7\) flowers each minute. Mia plants \(3\) more flowers each minute than Jonas. How many flowers does Mia plant in \(5\) minutes?

Hints

- First find how many flowers Mia plants in one minute. - Then use that amount for \(5\) equal minutes. - Check that the final total is five times Mia’s one-minute amount.

Solution

1. Mia plants \(7+3=10\) flowers each minute. 2. In \(5\) minutes, she plants \(10 \times 5 = 50\) flowers.

Answer

Mia plants \(50\) flowers in \(5\) minutes.
Two-step word problems 70
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.
Two-step word problems 71
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.
Two-step word problems 72
A class has a \(9\)-yard ribbon. The students cut off \(3\) yards for a large poster. Then they use exactly half of the remaining ribbon to wrap gifts. How many yards of ribbon remain at the end?

Hints

- First find how much ribbon remains after the poster piece is cut off. - The students use exactly half of that remaining amount. - If half is used, what fraction of that same amount is left?

Solution

1. After the first cut, \(9-3=6\) yards remain. 2. The students use half of those \(6\) yards, so the other half remains: \(6 \div 2 = 3\) yards.

Answer

\(3\) yards of ribbon remain.
Two-step word problems 73
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.
Two-step word problems 74
Leon has \(48\) stickers. Sophie has \(7\) packs with \(7\) stickers in each pack. Sophie says, “I have more stickers than Leon.” Is she correct? How many more stickers does she have, if any?

Hints

- Find Sophie’s total from the equal packs. - Compare Sophie’s total with Leon’s given total. - The difference tells how many more stickers the larger total has.

Solution

1. Find Sophie’s total: \(7 \times 7 = 49\) stickers. 2. Compare and find the difference: \(49 - 48 = 1\).

Answer

Yes. Sophie has \(1\) more sticker than Leon.
Two-step word problems 75
Ms. Miller wants to bake at least \(60\) muffins for a school fair. She already has \(36\) muffins. Then she bakes \(3\) trays with \(7\) muffins on each tray. Does she reach her goal? Explain.

Hints

- Find how many muffins are in the new trays. - Combine that amount with the muffins already baked. - Compare the final total with the goal; no extra calculation is needed for the comparison.

Solution

1. Find the number in the new trays: \(3 \times 7 = 21\) muffins. 2. Add them to the muffins already baked: \(36 + 21 = 57\). 3. Since \(57 < 60\), she does not reach the goal.

Answer

No. Ms. Miller has \(57\) muffins, which is fewer than \(60\).
Two-step word problems 76
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.
Two-step word problems 77
A game figure costs \(\$7\). Noah wants to buy \(3\) figures but has only \(\$15\). How much more money does he need?

Hints

- First find the total cost of the three equal-price figures. - Compare that cost with the money Noah already has. - The difference is the additional amount he needs.

Solution

1. Find the cost of three figures: \(3 \times \$7 = \$21\). 2. Find the shortfall: \(\$21 - \$15 = \$6\).

Answer

Noah needs \(\$6\) more.
Two-step word problems 78
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.
Two-step word problems 79
A rain barrel can hold \(400\,\text{L}\). It starts with \(280\,\text{L}\) of water. After some water is used, it takes \(230\,\text{L}\) to fill the barrel again. How many liters of water were used?

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

\(110\,\text{L}\)
Two-step word problems 80
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.
Two-step word problems 81
At the Star Theater, \(5\) tickets cost \(\$45\). At the Moon Theater, \(4\) tickets cost \(\$40\). At which theater does one ticket cost less? Show the price of one ticket at each theater.

Hints

- A larger total price does not always mean a higher price per ticket. - Imagine that the tickets in each theater's bundle all cost the same amount. - Make sure you are comparing the cost of the same number of tickets.

Solution

1. One ticket at the Star Theater costs \(\$45\div5=\$9\). 2. One ticket at the Moon Theater costs \(\$40\div4=\$10\). 3. Since \(9<10\), one ticket costs less at the Star Theater.

Answer

Star Theater: \(\$9\) per ticket. Moon Theater: \(\$10\) per ticket. One ticket costs less at the Star Theater.
Two-step word problems 82
Grandpa Henry is \(65\) years old today. His grandson Ben is \(56\) years younger. How old will Ben be \(7\) years from now?

Hints

- First use the age difference to find Ben’s age today. - Then move Ben’s age forward by the stated number of years. - Check that both people would be \(7\) years older after \(7\) years.

Solution

1. Find Ben’s age today: \(65-56=9\). 2. Seven years from now, Ben will be \(9+7=16\) years old.

Answer

Ben will be \(16\) years old.
Two-step word problems 83
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.
Two-step word problems 84
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.
Two-step word problems 85
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.
Two-step word problems 86
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.
Two-step word problems 87
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.
Two-step word problems 88
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.
Two-step word problems 89
A school supply store begins the day with \(356\) blue notebooks and \(289\) red notebooks. At the end of the day, \(218\) notebooks remain. How many notebooks were sold?

Hints

- Combine the two starting notebook amounts first. - Then compare the starting total with the number left at the end of the day. - Check that sold notebooks plus remaining notebooks equal the starting total.

Solution

1. Find the starting total: \(356+289=645\). 2. Subtract the number remaining: \(645-218=427\).

Answer

The store sold \(427\) notebooks.
Two-step word problems 90
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.
Two-step word problems 91
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.
Two-step word problems 92
A produce seller has \(312\) apples. There are \(125\) fewer pears than apples. How many apples and pears are at the stand altogether?

Hints

- First find the number of pears from the “fewer than” relationship. - Then combine the two fruit amounts. - Check that the pear count is smaller than the apple count by \(125\).

Solution

1. Find the number of pears: \(312-125=187\). 2. Add the apples and pears: \(312+187=499\).

Answer

There are \(499\) apples and pears altogether.
Two-step word problems 93
A school garden compares this year’s harvest with last year’s harvest. <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></td><td></td><td>\(77\,\text{kg}\)</td></tr> </table> How much did the apple harvest increase? How much did the apple and pear harvests increase altogether?

Hints

- For apples, compare this year’s amount with last year’s amount. - The pear increase is already given in the table. - Combine the two increases for the overall change.

Solution

1. Apples increased by \(412-345=67\,\text{kg}\). 2. Add the two increases: \(67+77=144\,\text{kg}\).

Answer

Apples increased by \(67\,\text{kg}\). Altogether, the apple and pear harvests increased by \(144\,\text{kg}\).
Two-step word problems 94
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.
Two-step word problems 95
A craft group has \(6\) students. Each student needs \(8\) white beads and \(2\) gold beads for a necklace. How many beads are needed altogether?

Hints

- First find how many beads one student needs. - Then use the number of students as equal groups. - Check that the final total is six times the amount for one student.

Solution

1. Each student needs \(8+2=10\) beads. 2. Six students need \(6 \times 10=60\) beads altogether.

Answer

The group needs \(60\) beads altogether.
Two-step word problems 96
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.
Two-step word problems 97
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.
Two-step word problems 98
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.
Two-step word problems 99
A wildlife park has \(24\) meerkats in one habitat and \(19\) in another. The park sends \(7\) meerkats to another zoo. How many meerkats remain in the park?

Hints

- Combine the two starting habitats first. - Then subtract the animals that leave the park. - Check that the final number is smaller than the starting total.

Solution

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

Answer

\(36\) meerkats remain in the park.
Two-step word problems 100
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.
Two-step word problems 101
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.
Two-step word problems 102
Students sell \(350\) raffle tickets on Friday. On Saturday, they sell \(125\) more tickets than on Friday. Jordan says, “We sold more than \(800\) tickets over the two days.” Is Jordan 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. The students sold \(825\) tickets over the two days, and \(825>800\).
Two-step word problems 103
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.
Two-step word problems 104
A bakery makes \(240\) plain rolls in the morning. It makes \(70\) more whole-wheat rolls than plain rolls. How many plain and whole-wheat rolls does the bakery make altogether?

Hints

- First use the comparison to find the number of whole-wheat rolls. - Then combine the two kinds of rolls. - Check that the whole-wheat count is \(70\) greater than the plain-roll count.

Solution

1. Find the number of whole-wheat rolls: \(240+70=310\). 2. Add the two kinds: \(240+310=550\).

Answer

The bakery makes \(550\) plain and whole-wheat rolls altogether.
Two-step word problems 105
A school library has \(450\) books. Of these, \(180\) are fiction books and \(130\) are nonfiction books. All the remaining books are picture books. How many picture books are in the library?

Hints

- Combine the two categories whose counts are given. - The picture books are what remains from the full library total. - Check that all three categories add to \(450\).

Solution

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

Answer

There are \(140\) picture books in the library.
Two-step word problems 106
Three third-grade classes collect \(820\,\text{kg}\) of paper altogether. Ms. Lee’s class collects \(245\,\text{kg}\), and Mr. Grant’s class collects \(275\,\text{kg}\). How many kilograms does Ms. Patel’s class collect?

Hints

- Combine the two class amounts that are known. - The third class collected the part of the total that is left. - Check that all three class amounts add to \(820\,\text{kg}\).

Solution

1. Add the first two class amounts: \(245+275=520\,\text{kg}\). 2. Subtract from the total: \(820-520=300\,\text{kg}\).

Answer

Ms. Patel’s class collects \(300\,\text{kg}\) of paper.
Two-step word problems 107
A mail carrier drives \(18\) miles during the first hour of a route. During the second hour, the carrier drives \(4\) miles farther than during the first hour. How far does the carrier drive during the two hours altogether?

Hints

- First use the comparison to find the second-hour distance. - Then combine the distances from the two hours. - Check that the second-hour distance is \(4\) miles greater than the first-hour distance.

Solution

1. Find the second-hour distance: \(18+4=22\) miles. 2. Add the two hours: \(18+22=40\) miles.

Answer

The carrier drives \(40\) miles altogether.
Two-step word problems 108
Lucas is saving for a bicycle that costs \(\$450\). He saves \(\$135\) in May and \(\$185\) in June. How much more must he save to have enough money for the bicycle?

Hints

- Combine the two savings amounts first. - Then compare the saved amount with the bicycle price. - Check that the amount saved plus the amount still needed equals \(\$450\).

Solution

1. Find the amount already saved: \(\$135+\$185=\$320\). 2. Subtract from the bicycle price: \(\$450-\$320=\$130\).

Answer

Lucas must save \(\$130\) more.
Two-step word problems 109
At a field day, three third-grade classes complete \(840\) laps altogether. Ms. Lee’s class completes \(260\) laps, and Mr. Grant’s class completes \(220\) laps. Ms. Patel’s class completes the remaining laps. How many laps does Ms. Patel’s class complete, and which class completes the most laps?

Hints

- Combine the two known class totals. - Subtract their combined amount from the overall total. - After finding the remaining class amount, compare the three numbers without another calculation.

Solution

1. Add the first two class totals: \(260+220=480\) laps. 2. Find the remaining laps: \(840-480=360\) laps. 3. Compare \(360\), \(260\), and \(220\). Ms. Patel’s class completes the most laps.

Answer

Ms. Patel’s class completes \(360\) laps and completes the most laps.
Two-step word problems 110
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.
Two-step word problems 111
A school library needs to shelve \(850\) books. Volunteers shelve \(260\) books on Monday and \(215\) books on Tuesday. How many books still need to be shelved after Tuesday?

Hints

- Combine the numbers shelved on Monday and Tuesday. - Then compare that subtotal with the full number of books. - Check that shelved books plus remaining books equal \(850\).

Solution

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

Answer

There are \(375\) books left to shelve.
Two-step word problems 112
A walking trail is \(1000\,\text{m}\) long. The first section is \(340\,\text{m}\) long, and the second section is \(220\,\text{m}\) long. How long is the third section?

Hints

- Combine the lengths of the two known sections first. - The three sections together make the full trail length. - Check that the three section lengths add to \(1000\,\text{m}\).

Solution

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

Answer

The third section is \(440\,\text{m}\) long.
Two-step word problems 113
A gardener has \(54\) tulip bulbs. He wants to plant \(10\) full rows with exactly \(6\) bulbs in each row. How many more bulbs does he need?

Hints

- First find how many bulbs are needed for all \(10\) rows. - Then compare that total with the \(54\) bulbs already available. - Check that adding your answer to \(54\) reaches the needed total.

Solution

1. Ten rows require \(10 \times 6=60\) bulbs. 2. Compare the required number with the bulbs he has: \(60-54=6\).

Answer

The gardener needs \(6\) more bulbs.
Two-step word problems 114
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 \(461\,\text{L} > 389\,\text{L}\).
Two-step word problems 115
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.
Two-step word problems 116
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.
Two-step word problems 117
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 \(260\).
Two-step word problems 118
The picture shows trees planted in equal rows. After a storm, all the trees in the shaded bottom row fell. How many trees are still standing? Show two ways to calculate.
Figure for problem 537369

Hints

- Use the picture to find the number of rows and the number of trees in one row. - One complete bottom row is affected. - You can subtract the fallen row or multiply using only the rows that remain.

Solution

1. The picture shows \(6\) rows with \(8\) trees in each row, so the original total is \(6 \times 8=48\) trees. Subtract the fallen row: \(48-8=40\). 2. Alternatively, \(5\) rows remain standing, so \(5 \times 8=40\).

Answer

\(40\) trees are still standing. Two methods are \(6 \times 8-8=40\) and \(5 \times 8=40\).
Two-step word problems 119
The picture shows all the seats in a small movie theater. The shaded seats are already reserved. Without counting the available seats one by one: a) write a multiplication equation for the total number of seats; b) find the number of reserved seats from the picture; c) write the subtraction that gives the number of available seats.
Figure for problem 537371

Hints

- Use the rectangular arrangement to find the total rather than counting individual seats. - The shaded block gives the number that must be removed from the total. - Show both the multiplication equation and the subtraction equation.

Solution

1. The picture has \(7\) rows with \(10\) seats in each row, so \(7 \times 10=70\) seats are shown. 2. There are \(13\) shaded reserved seats. 3. Subtract the reserved seats: \(70-13=57\). 4. There are \(57\) available seats.

Answer

a) \(7 \times 10=70\) b) \(13\) reserved seats c) \(70-13=57\) available seats
Two-step word problems 120
A market stand begins the day with \(72\) apples arranged in rows of \(8\). During the morning, it sells \(3\) complete rows. How many apples remain?

Hints

- First find how many apples are in the three complete rows that were sold. - Then subtract the sold apples from the starting amount. - Check that the number sold and the number remaining add to \(72\).

Solution

1. Find the number sold in three rows: \(3 \times 8=24\). 2. Subtract from the starting amount: \(72-24=48\).

Answer

\(48\) apples remain.
Two-step word problems 121
A theater sells \(24\) child tickets in the morning and \(19\) child tickets in the afternoon. It sells \(38\) adult tickets altogether. Were more child tickets or adult tickets sold, and how many more?

Hints

- Combine the morning and afternoon child-ticket counts. - Compare that total with the adult-ticket total already given. - Subtract the smaller total from the larger total to find how many more.

Solution

1. Find the child-ticket total: \(24+19=43\). 2. Compare with the adult-ticket total: \(43-38=5\).

Answer

\(5\) more child tickets were sold.
Two-step word problems 122
Paul reads \(25\) pages on Monday and \(17\) pages on Tuesday. Rayan reads \(39\) pages over the same two days. Who reads more altogether, and by how many pages?

Hints

- Find Paul’s two-day total first. - Rayan’s two-day total is already given. - Compare the two totals by finding their difference.

Solution

1. Paul reads \(25+17=42\) pages altogether. 2. Compare the totals: \(42-39=3\).

Answer

Paul reads \(3\) more pages than Rayan.
Two-step word problems 123
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. - Check the completed row and column totals against the table.

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: \(8\)
Two-step word problems 124
Team North has \(25\) points. During a review, it receives \(6\) more points and then loses \(2\) points because of a penalty. Team South has \(28\) points and its score does not change. What is Team North’s final score, and which team has more points afterward?

Hints

- Apply the score changes in the order they occur. - The second change acts on the score after the first change. - Compare the final North score with South’s unchanged score only after both operations.

Solution

1. Add the awarded points: \(25+6=31\). 2. Apply the penalty: \(31-2=29\). 3. Team North finishes with \(29\) points, which is greater than Team South’s \(28\) points.

Answer

Team North finishes with \(29\) points, so Team North has more points afterward.
Two-step word problems 125
A tile design has an area of \(18\) square units. Six square units are white, and \(8\) square units are blue. The rest is yellow. What is the yellow area?

Hints

- Combine the areas of the two known colored parts. - The three colors together make the whole design. - Subtract the known area from the total area.

Solution

1. The white and blue parts cover \(6 + 8 = 14\) square units. 2. Subtract the known parts from the whole: \(18 - 14 = 4\) square units.

Answer

The yellow area is \(4\) square units.
Two-step word problems 126
The diagram shows an upper seating section and a lower seating section. Every row has the same number of seats. Without counting seats one by one: a) How many more rows does the upper section have? b) Use that row difference to find how many more seats the upper section has. Show both steps.
Figure for problem 550364

Hints

- Compare the sections by rows before thinking about individual seats. - The diagram shows the same number of seats in every row. - Show the row subtraction first, then multiply the extra rows by the number of seats in each row.

Solution

1. The upper section has \(4\) rows and the lower section has \(2\) rows, so the upper section has \(4-2=2\) extra rows. 2. Each row has \(7\) seats, so \(2 \times 7=14\). 3. The upper section has \(14\) more seats.

Answer

a) The upper section has \(2\) more rows. b) \(2 \times 7=14\), so the upper section has \(14\) more seats.
Two-step word problems 127
A club has \(18\) red badges and \(12\) blue badges. It divides all the badges equally among \(5\) teams. Write one equation using \(n\) for the number of badges each team receives, then find \(n\).

Hints

- The teams share all badges, so the two badge amounts must be combined before sharing. - Let \(n\) represent the size of one equal share. - Use parentheses so the equation shows which operation happens first.

Solution

1. First combine the two badge colors: \(18+12=30\). 2. The equation is \(n=(18+12) \div 5\). 3. \(30 \div 5=6\), so \(n=6\).

Answer

\(n=(18+12) \div 5\), so \(n=6\).
Two-step word problems 128
A school has \(248\) books in one reading room and \(367\) books in another. It sends \(190\) books to classrooms. a) How many books remain in the two reading rooms altogether? b) Round each starting number and the number sent to the nearest hundred. Use the rounded numbers to check whether your exact answer is reasonable.

Hints

- Find the exact two-step answer first. - For the estimate, round each number to the nearest hundred before calculating. - Compare the estimate with the exact answer, not for equality but for a similar size.

Solution

1. Combine the starting amounts: \(248+367=615\) books. 2. Subtract the books sent out: \(615-190=425\) books. 3. To the nearest hundred, \(248\approx200\), \(367\approx400\), and \(190\approx200\). 4. The estimate is \(200+400-200=400\), which is close to \(425\). The exact answer is reasonable.

Answer

a) \(425\) books b) Rounded numbers: \(200\), \(400\), and \(200\). The estimate is about \(400\) books, so \(425\) is reasonable.
Two-step word problems 129
A museum shop starts with \(325\) postcards. It sells \(178\) postcards, then receives \(92\) new postcards. a) How many postcards are in the shop now? b) Round each number to the nearest hundred and use the rounded calculation to decide whether your exact answer is reasonable.

Hints

- Follow the events in order for the exact calculation. - Round the three quantities separately for the estimate. - Compare the exact answer with the estimated total.

Solution

1. After the sale, \(325-178=147\) postcards remain. 2. After the delivery, \(147+92=239\) postcards are in the shop. 3. To the nearest hundred, \(325\approx300\), \(178\approx200\), and \(92\approx100\). 4. The estimate is \(300-200+100=200\). This is close enough to \(239\) to support the exact answer.

Answer

a) \(239\) postcards b) Rounded numbers: \(300\), \(200\), and \(100\). The estimate is about \(200\) postcards, so \(239\) is reasonable.
Two-step word problems 130
A baker puts \(5\) cookies on each of \(8\) trays. Then \(12\) cookies are eaten. Noah writes \((8+12) \times 5\) to find how many cookies remain. Explain Noah’s error and find the correct number of cookies remaining.

Hints

- What do the \(8\), \(5\), and \(12\) mean in the story? - Find how many cookies there are before any are eaten. - Then show what happens when \(12\) cookies are eaten.

Solution

1. Noah adds the cookies eaten to the number of trays before finding how many cookies there were at the start. Those numbers should not be added. 2. First find the starting number of cookies: \(8 \times 5=40\). 3. Then subtract the cookies eaten: \(40-12=28\).

Answer

Noah should multiply to find the starting number of cookies before subtracting the cookies that were eaten. The correct calculation is \(8\times5-12=28\), so \(28\) cookies remain.
Two-step word problems 131
After \(4\) packs with \(6\) stickers in each pack are added to a collection, the collection has \(38\) stickers. How many stickers were in the collection before the packs were added?

Hints

- First find how many stickers were added in the \(4\) packs. - The \(38\) stickers are the amount after the packs were added, so work backward. - Check by adding the pack total to your starting amount.

Solution

1. Find how many stickers were added: \(4 \times 6=24\). 2. Work backward from the final total: \(38-24=14\).

Answer

There were \(14\) stickers in the collection before the packs were added.
Two-step word problems 132
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.
Two-step word problems 133
A library box contains some nonfiction books at the start of the day. A class checks out \(15\) books and later returns \(11\) of them. 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

- First find the number of books left that stayed checked out. - Work backward from the ending number of books. - Check by starting with your answer, removing \(15\), and returning \(11\).

Solution

1. The class keeps \(15-11=4\) books checked out. 2. Add those \(4\) books back to the \(28\) remaining books: \(28+4=32\).

Answer

There were \(32\) books in the box at the start of the day.
Two-step word problems 134
At the end of the day, a school library has \(650\) books. Earlier that day, students checked out \(45\) books, and later the library received \(25\) new books. How many books were in the library at the start of the day?

Hints

- The ending amount is known, so undo the events in reverse order. - What operation undoes receiving \(25\) books? - What operation undoes checking out \(45\) books?

Solution

1. Work backward through the last event. Before the \(25\) new books arrived, there were \(650 - 25 = 625\) books. 2. Before \(45\) books were checked out, there were \(625 + 45 = 670\) books. 3. Check forward: \(670 - 45 + 25 = 650\).

Answer

There were \(670\) books in the library at the start of the day.
Two-step word problems 135
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 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 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.
Two-step word problems 136
A bakery sells \(145\) loaves in its store and delivers \(120\) loaves to a restaurant. After that, the bakery has \(265\) loaves left. How many loaves did the bakery have at the start of the morning?

Hints

- First find how many loaves left the bakery. - The number left is the same as that amount. - Add the loaves that left and the loaves that remained.

Solution

1. The bakery sold and delivered \(145+120=265\) loaves. 2. It also had \(265\) loaves left. 3. So it started with \(265+265=530\) loaves.

Answer

\(530\) loaves
Two-step word problems 137
The chart shows the numbers of marbles in Box A and Box B. 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 the two bars. - Think about how moving one marble changes both boxes at once. - Check that the two boxes have the same number after the move.

Solution

1. Box A contains \(24\) marbles and Box B contains \(16\), so the difference is \(24-16=8\). 2. Each marble moved lowers the difference by \(2\): one leaves Box A and one is added to Box B. 3. Therefore, \(8 \div 2=4\) marbles must be moved.

Answer

\(4\) marbles must be moved from Box A to Box B.
Two-step word problems 138
The chart shows student votes for cocoa and tea. Some students switch their vote from cocoa to tea. How many students must switch so that cocoa and tea have equal numbers of votes?
Figure for problem 538204

Hints

- Compare the lengths and values of the two bars. - One switch changes both vote totals at the same time. - Check that the two new vote totals are equal.

Solution

1. Cocoa has \(14\) votes and tea has \(8\), so the difference is \(14-8=6\). 2. Each switch lowers the difference by \(2\), because cocoa loses one vote while tea gains one. 3. Therefore, \(6 \div 2=3\) students must switch.

Answer

\(3\) students must switch from cocoa to tea.
Two-step word problems 139
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 \(10\).
Two-step word problems 140
A classroom has \(9\) rows with \(6\) chairs in each row. Two chairs are removed from the room. Leo writes \(9 \times (6 - 2)\) to find how many chairs remain. Explain Leo's error and find the correct number of chairs remaining.

Hints

- Decide whether the \(2\) removed chairs come from every row or from the room altogether. - Compare what \(6 - 2\) would mean inside the multiplication expression with what the story says. - Find the original total before applying the removal.

Solution

1. Leo's expression removes \(2\) chairs from every row, which would remove \(18\) chairs altogether. 2. The room starts with \(9 \times 6 = 54\) chairs. 3. Only \(2\) chairs are removed from the whole room, so \(54 - 2 = 52\) chairs remain.

Answer

Leo's expression removes \(2\) chairs from each row instead of \(2\) chairs total. The correct answer is \(52\) chairs.
Two-step word problems 141
At the end of an art lesson, the supply cabinet has \(87\) markers. Earlier, Ms. Davis gave out \(48\) markers, and later \(15\) unused markers were returned. How many markers were in the cabinet before the lesson?

Hints

- Work backward from the ending amount in reverse event order. - First undo the markers that were returned. - Then restore the markers that had been given out.

Solution

1. Work backward by undoing the return: \(87-15=72\). 2. Undo the markers that were given out: \(72+48=120\).

Answer

There were \(120\) markers in the cabinet before the lesson.
Two-step word problems 142
A supply room has \(72\) pencils packed equally with \(8\) pencils in each bag. Five full bags are given to classrooms. How many pencils remain? Solve in two ways: first by finding the pencils given away, and second by reasoning with the number of full bags. Explain why the methods agree.

Hints

- One route can work with individual pencils; the other can work with whole bags. - How many pencils are in the five bags that leave? - How many full bags are represented by \(72\) pencils?

Solution

1. Method 1: five bags contain \(5 \times 8 = 40\) pencils, so \(72 - 40 = 32\) pencils remain. 2. Method 2: \(72 \div 8 = 9\) full bags. After giving away \(5\) bags, \(9 - 5 = 4\) bags remain. Then \(4 \times 8 = 32\) pencils remain. 3. Both methods describe removing the same five equal groups from the same starting amount.

Answer

Method 1: \(5\times8=40\), then \(72-40=32\). Method 2: \(72\div8=9\) bags, \(9-5=4\) bags, then \(4\times8=32\). Both methods remove the same \(5\) bags, so \(32\) pencils remain.
Two-step word problems 143
A class has \(63\) stickers. Mr. Lewis keeps \(14\) stickers for a prize box and shares the rest equally among \(7\) students. Sam subtracts first and then divides. Priya says she can first notice that \(14\) is \(2\) groups of \(7\), then reason about how many groups of \(7\) remain. Use both approaches and explain why they agree.

Hints

- In Sam's method, find how many stickers remain before sharing. - In Priya's method, write \(63\) and \(14\) as groups of \(7\). - Explain how both methods remove the same \(14\) stickers.

Solution

1. Sam's method: \(63 - 14 = 49\), then \(49 \div 7 = 7\). Each student gets \(7\) stickers. 2. Priya's method: \(63 = 9 \times 7\) and \(14 = 2 \times 7\). Removing \(2\) groups from \(9\) groups leaves \(7\) groups of \(7\), so each student gets \(7\) stickers. 3. Both methods remove the same \(14\) stickers before finding the equal share.

Answer

Each student gets \(7\) stickers. Sam removes \(14\) stickers first and then divides. Priya sees \(63\) as \(9\) groups of \(7\) and removes \(2\) groups of \(7\). Both methods remove the same \(14\) stickers before sharing, so they agree.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Building free math worksheets