Division word problems with remainders

Learning goals

Interprets a division remainder to answer a word problem.
Rounds a group count up when every item must be accommodated.
Finds how much is needed to complete another group.
Uses a nonzero remainder to reject a whole-number solution.
Checks that divisor × quotient + remainder equals the dividend.

Difficulty: 1 – 5

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Division word problems with remainders 1

A farmer collects $34$ eggs. Six are cracked and cannot be packed. The rest go into cartons holding $6$ eggs each. How many full cartons can the farmer fill, and how many good eggs are left over?

Hints

- Picture the collection after the cracked eggs are set aside. - Could a partly filled carton count as a full one? - Check that the packed eggs and the loose eggs account for every good egg.

Solution

1. The number of good eggs is $34 - 6 = 28$. 2. Four full cartons hold $4 \times 6 = 24$ eggs, leaving $28 - 24 = 4$ good eggs.

Answer

$4$ full cartons, with $4$ good eggs left over.

Division word problems with remainders 2

Lucas has $82$ stickers. He puts $9$ stickers on each full page of his album. a) How many pages can he fill completely? b) How many stickers will be on the next page? c) How many more stickers does Lucas need to fill that page?

Hints

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

Solution

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

Answer

a) Lucas can fill $9$ pages completely. b) $1$ sticker will be on the next page. c) Lucas needs $8$ more stickers to fill that page.

Division word problems with remainders 3

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

Hints

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

Solution

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

Answer

He has spent $\$57$ and can buy $2$ comic books with the $\$13$ left.

Division word problems with remainders 4

A fruit seller begins the day with $60\,\text{kg}$ of apples. Before lunch, the seller sells $8$ small crates containing $4\,\text{kg}$ each. The remaining apples are packed into bags holding $3\,\text{kg}$ each. How many bags can be filled completely, and how many kilograms remain?

Hints

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

Solution

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

Answer

The seller can fill $9$ bags, and $1\,\text{kg}$ remains.

Division word problems with remainders 5

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

Hints

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

Solution

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

Answer

a) The baker can fill $19$ bags completely. b) $2$ rolls will be left over. c) The baker needs $5$ more rolls to fill another bag.

Division word problems with remainders 6

Max wants to buy sticker packs. Large packs cost $\$7$, and small packs cost $\$4$. Max is exactly $\$5$ short of having enough money for $4$ large packs. How many small packs can he buy instead, and how much money will he have left?

Hints

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

Solution

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

Answer

Max can buy $5$ small packs and will have $\$3$ left.

Division word problems with remainders 7

A garden center uses $35$ quarts of potting soil to fill $5$ equal planters. a) How many quarts are needed to fill $8$ planters? b) Ms. Miller has $60$ quarts of potting soil. How many planters can she fill completely, and how many quarts will remain?

Hints

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

Solution

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

Answer

a) $56$ quarts b) She can fill $8$ planters completely, and $4$ quarts will remain.

Division word problems with remainders 8

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

Hints

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

Solution

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

Answer

The flour lasts for $9$ complete days altogether, and $1\,\text{kg}$ remains.

Division word problems with remainders 9

Leonie has collected $135$ stickers. She places $15$ stickers on each of the first $4$ pages of her album. She puts $9$ stickers on each additional page. How many additional pages can she fill, and how many stickers remain?

Hints

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

Solution

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

Answer

Leonie can fill $8$ additional pages, and $3$ stickers remain.

Division word problems with remainders 10

A baker makes $245$ pretzels on Saturday morning. In the afternoon, the baker makes $85$ fewer pretzels than in the morning. All the pretzels will be packed in bags that can hold up to $8$ pretzels each. The baker has $50$ empty bags. Are there enough bags? Justify your answer with calculations.

Hints

- Determine the afternoon amount and then the complete number of pretzels. - Compare that total with the maximum capacity of the available bags. - If the available bags are short, consider whether one additional partly filled bag is enough.

Solution

1. The baker makes $245-85=160$ pretzels in the afternoon. 2. The total number of pretzels is $245+160=405$. 3. Fifty bags can hold up to $50 \times 8=400$ pretzels. 4. Since $405>400$, fifty bags are not enough. One additional bag is needed, for $51$ bags altogether.

Answer

No. The baker makes $405$ pretzels, but $50$ bags can hold only $400$. One more bag is needed.

Division word problems with remainders 11

Lucas has saved $\$42$. He buys $4$ card games for $\$6$ each. With the remaining money, he buys as many comic books as possible for $\$5$ each. How many comic books can Lucas buy, and how much money remains?

Hints

- First find how much Lucas spends on the card games. - How much money remains after that purchase? - How many times does the comic-book price fit into the remaining amount? - Does any money remain after buying the greatest possible number?

Solution

1. Find the cost of the card games: $4 \times \$6 = \$24$. 2. Find the money remaining: $\$42 - \$24 = \$18$. 3. Divide by the cost of one comic book: $\$18 \div \$5 = 3$ remainder $\$3$.

Answer

Lucas can buy $3$ comic books, and $\$3$ remains.

Division word problems with remainders 12

Lucas has saved $95$ cents. Dinner rolls cost $24$ cents each. a) Find the prices of $1$, $2$, $3$, and $4$ rolls. b) Can Lucas buy $4$ rolls? Use your results from part a to explain. c) How much change will he receive if he buys $3$ rolls?

Hints

- Build the prices by adding another $24$ cents each time. - Compare the price of four rolls with $95$ cents. - Change is the amount paid minus the cost.

Solution

1. Find the prices: $1 \times 24 = 24$ cents, $2 \times 24 = 48$ cents, $3 \times 24 = 72$ cents, and $4 \times 24 = 96$ cents. 2. Four rolls cost $96$ cents. Since $96 > 95$, Lucas cannot buy four rolls. 3. Three rolls cost $72$ cents. The change is $95 - 72 = 23$ cents.

Answer

a) The prices are $\$0.24$, $\$0.48$, $\$0.72$, and $\$0.96$. b) No. Four rolls cost $\$0.96$, which is more than $\$0.95$. c) Lucas receives $\$0.23$ in change.

Division word problems with remainders 13

At a craft store, one glitter sticker costs $35$ cents. A special pack of $3$ stickers costs $90$ cents. a) How much would $3$ individual stickers cost? Which option costs less, and what is the difference? b) Ms. Meyer has $\$2.00$. What is the greatest number of special packs she can buy, and how much money will remain?

Hints

- Find the cost of three individual stickers. - Compare that cost with the pack price. - Determine how many $90$-cent packs fit within $\$2.00$, then find the remainder.

Solution

1. Three individual stickers cost $3 \times 35 = 105$ cents, or $\$1.05$. 2. The pack costs $90$ cents, so it costs $105 - 90 = 15$ cents less. 3. Two packs cost $2 \times 90 = 180$ cents. Three packs would cost $270$ cents, which is more than $200$ cents. 4. After buying two packs, $200 - 180 = 20$ cents remain.

Answer

a) Three individual stickers cost $\$1.05$. The pack costs $\$0.15$ less. b) Ms. Meyer can buy $2$ packs and will have $\$0.20$ left.

Division word problems with remainders 14

A sports club wants to buy new balls. A soccer ball costs $\$12$, and a basketball costs $\$8$. The club wants to buy $7$ soccer balls but is $\$11$ short. 1. How much money is in the club's account? 2. Can the club buy $10$ basketballs with that money? Justify your answer. 3. What is the greatest number of basketballs the club can buy, and how much money will remain?

Hints

- First find the total cost of seven soccer balls. - What does being “short” mean for finding the available amount? - Compare the cost of ten basketballs with the available money. - Use division with a remainder to find the greatest possible number.

Solution

1. Find the cost of seven soccer balls: $7 \times \$12 = \$84$. 2. Find the amount in the account: $\$84 - \$11 = \$73$. 3. Check the cost of ten basketballs: $10 \times \$8 = \$80$. Since $\$80 > \$73$, the club cannot buy ten. 4. Divide the available money by the basketball price: $\$73 \div \$8 = 9$ remainder $\$1$.

Answer

1. The club has $\$73$. 2. No. Ten basketballs would cost $\$80$, which is more than $\$73$. 3. The club can buy at most $9$ basketballs, and $\$1$ will remain.

Division word problems with remainders 15

Classes A and B are taking a camping trip. A boys’ tent holds at most $5$ campers, and a girls’ tent holds at most $4$ campers. The table shows the number of students. <table> <tr> <th></th> <th>Girls</th> <th>Boys</th> </tr> <tr> <td>Class A</td> <td>$12$</td> <td>$13$</td> </tr> <tr> <td>Class B</td> <td>$14$</td> <td>$12$</td> </tr> </table> Boys and girls sleep in separate tents, but students from the two classes may share tents. How many tents are needed altogether?

Hints

- Combine the two classes' counts for girls and boys separately. - A nonzero remainder means another tent is needed. - Add the numbers of girls' and boys' tents.

Solution

1. There are $12+14=26$ girls. 2. Since $26\div4=6$ remainder $2$, the girls need $7$ tents. 3. There are $13+12=25$ boys. 4. Since $25\div5=5$, the boys need $5$ tents. 5. Altogether, $7+5=12$ tents are needed.

Answer

$12$ tents

Division word problems with remainders 16

A wall is $12\,\text{ft}$ wide and $8\,\text{ft}$ high. One wallpaper roll is $2\,\text{ft}$ wide and $32\,\text{ft}$ long. How many whole rolls are needed to cover the wall? Assume there is no overlap, pattern matching, or cutting waste.

Hints

- Find how many strips fit across the wall's width. - Find how many wall-height strips can be cut from one roll. - The purchase must be a whole number of rolls.

Solution

1. Find the number of vertical strips needed across the wall: $12\div2=6$ strips. 2. Each strip must be $8\,\text{ft}$ long. One roll provides $32\div8=4$ strips. 3. One roll gives 4 of the 6 needed strips, leaving 2 strips still to cover. 4. A second roll provides those remaining strips, so $2$ rolls are needed.

Answer

$2$ rolls are needed.

Division word problems with remainders 17

An orchard packs $735$ apples with $7$ apples in each bag and $852$ pears with $6$ pears in each bag. The apple bags and pear bags must stay in separate shipments. Each truck can carry up to $120$ bags. How many trucks are needed altogether for the two shipments? How many pear bags are on the final truck?

Hints

- First determine the bag count for each fruit. - Compare each bag count with one truck’s capacity. - Keep the shipments separate, then combine the numbers of trucks needed.

Solution

1. The apples use $735 \div 7=105$ bags, so they need $1$ truck. 2. The pears use $852 \div 6=142$ bags. One truck carries $120$ bags, leaving $142-120=22$ pear bags for a second truck. 3. The two separate shipments therefore need $1+2=3$ trucks altogether.

Answer

The shipments need $3$ trucks altogether. The final truck carries $22$ pear bags.

Division word problems with remainders 18

The bar graph shows how many books four Grade 4 classes collected for a school book sale. a) How many books did Classes 4A, 4B, 4C, and 4D collect altogether? b) The class that collected the most books will pack them into boxes that each hold exactly $10$ books. How many boxes will be completely filled, and how many books will be left over?

Hints

- Read each bar height using the vertical axis. - The scale includes intervals of $5$, including values between multiples of $10$. - Add all four values for part a). - Identify the tallest bar before part b). - Divide that value by $10$ and interpret the remainder.

Solution

1. Read the values from the graph: Class 4A, $40$; Class 4B, $55$; Class 4C, $30$; Class 4D, $45$. 2. Add all four values: $40 + 55 + 30 + 45 = 170$ books. 3. Class 4B collected the most books, with $55$. 4. Divide by the box capacity: $55 \div 10 = 5$ remainder $5$. Therefore, $5$ boxes are full and $5$ books remain.

Answer

a) $170$ books b) $5$ full boxes and $5$ books left over

Division word problems with remainders 19

Each dot is one student on the roster for $8$ relay teams. Unfilled dots mark absent students; filled dots mark students who will participate. How many students will participate? Can they all be placed in $8$ teams of equal size without adding students?

Hints

- Count the planned students using the rows and columns. - Use the unfilled dots to find how many are absent. - What remainder appears when the participating students are shared among the teams?

Solution

1. There are $8$ rows of $6$ dots, or $8\times6=48$ students on the roster. 2. Three dots are unfilled, so $48-3=45$ students will participate. 3. Since $45=8\times5+5$, the participating students cannot be divided into $8$ equal-size teams without a remainder.

Answer

$45$ students participate. No, they cannot form $8$ equal-size teams.

Division word problems with remainders 20

A greenhouse has $173$ seedlings to place into trays that hold $8$ seedlings each. Write an equation of the form $173=8\times q+r$, where $q$ is the number of full trays and $r$ is the number of seedlings left over. Find $q$ and $r$. Then state how many trays are needed to hold all $173$ seedlings.

Hints

- Find the greatest multiple of $8$ that does not exceed $173$. - The difference from $173$ is the remainder. - Interpret whether a nonzero remainder needs another tray in this context.

Solution

1. Since $8\times21=168$, there are $21$ full trays. 2. The remainder is $173-168=5$, so $173=8\times21+5$. 3. The $5$ remaining seedlings need one additional tray, so $22$ trays are needed altogether.

Answer

$173=8\times21+5$, so $q=21$ and $r=5$. A total of $22$ trays are needed.

Division word problems with remainders 21

A library has $938$ books to pack into boxes that hold $9$ books each. Before dividing exactly, use $900$ books as a compatible-number benchmark to decide whether the library will need fewer than, exactly, or more than $100$ boxes. Then find the exact number of boxes needed, including any partly filled box, and explain how the estimate helps check the exact answer.

Hints

- Compare the actual book count with a nearby amount that divides evenly by $9$. - Interpret the remainder in the context of boxes, not just as a number. - Use the estimate to check whether your exact box count is a reasonable size.

Solution

1. Since $900\div9=100$ and $938$ is greater than $900$, the library should expect to need more than $100$ boxes. 2. Divide exactly: $938\div9=104$ remainder $2$. 3. The $104$ full boxes hold $936$ books, and the remaining $2$ books need one more box. 4. Therefore, $105$ boxes are needed. This agrees with the benchmark prediction that the answer should be just over $100$.

Answer

The benchmark $900\div9=100$ shows that more than $100$ boxes will be needed. Exactly, $938\div9=104$ remainder $2$, so the library needs $105$ boxes including the partly filled box. The exact answer is reasonably close to the estimate of just over $100$.

Division word problems with remainders 22

A youth group of $42$ people is taking public transit. A single fare costs $\$5$. A 10-ride pass costs $\$42$ and covers $10$ people. a) Find the total cost if everyone pays a single fare. b) What is the least number of 10-ride passes needed to cover all $42$ people using only passes? Find the cost. c) Find the cost of buying $4$ 10-ride passes and single fares for the remaining people. d) How much does the least expensive option save compared with part a)?

Hints

- A partial group still requires another 10-ride pass in part b. - In part c, determine how many people remain after four passes. - Compare all totals before finding the savings.

Solution

1. a) Single fares cost $42\times\$5=\$210$. 2. b) Four passes cover $40$ people, so $5$ passes are needed to cover all $42$ people. They cost $5\times\$42=\$210$. 3. c) Four passes cost $4\times\$42=\$168$. Two people remain, and their single fares cost $2\times\$5=\$10$. The total is $\$168+\$10=\$178$. 4. d) The combination in part c is least expensive. It saves $\$210-\$178=\$32$.

Answer

a) $\$210$ b) $5$ passes for $\$210$ c) $\$178$ d) $\$32$

Division word problems with remainders 23

For a school fair, the parent-teacher organization buys cases of drinks. A case of apple juice costs $\$14$, and a case of bottled water costs $\$6$. The organization spends $\$952$ in all and buys $44$ cases of apple juice. a) How many cases of bottled water are purchased? b) A member claims that the organization could have spent exactly $\$960$ while buying the same number of apple juice cases. Use a calculation to explain why that total is impossible with the stated water price.

Hints

- Separate the known apple-juice cost from the part of each total that would have to pay for water. - In part a, the remaining money must represent equal groups of the water-case price. - In part b, ask what condition a dollar amount must satisfy to be exactly the cost of a whole number of water cases.

Solution

1. The apple juice costs $44 \times \$14=\$616$. 2. The amount spent on water is $\$952-\$616=\$336$, which buys $\$336\div\$6=56$ cases. 3. If the total were $\$960$, the amount available for water would be $\$960-\$616=\$344$. 4. Since $344=6 \times 57+2$, $\$344$ cannot be made from a whole number of $\$6$ cases, so the claim is impossible.

Answer

a) $56$ cases of bottled water b) The claim is impossible because the amount left for water would not be divisible by $6$.

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Grade 4 math worksheets

Division word problems with remainders

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