Division Word Problems Worksheets for Grade 5 - Free Printable

Let’s solve each problem one by one. We’ll use division because we’re splitting a total into equal groups.

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Problem 1:
Janet has 48 boxes and 480 crayons total.
We want to know how many crayons are in each box.
So, divide total crayons by number of boxes:
→ 480 ÷ 48

Let’s calculate:
48 × 10 = 480 → So, 480 ÷ 48 = 10

✔ Each box has 10 crayons.

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Problem 2:
Green has 11 boxes and 485 erasers total.
Divide total erasers by number of boxes:
→ 485 ÷ 11

Let’s do the division:
11 × 44 = 484 → That’s 1 less than 485.
So, 485 ÷ 11 = 44 with a remainder of 1

But the question asks “how many erasers are there in each box?” — it doesn’t say anything about leftovers. Since we can’t split an eraser, we assume each box gets the same whole number, and maybe one is left over. But typically in these problems, if it divides evenly or they expect a whole number answer, we check again.

Wait — let me double-check:
11 × 44 = 484 → 485 - 484 = 1 → so yes, remainder 1.

But perhaps the problem expects us to report the quotient only? Or maybe I made a mistake?

Actually, let’s try long division:

11 into 48 → 4 times (44), subtract → 4
Bring down 5 → 45
11 into 45 → 4 times (44), subtract → 1

So, 44 R1 → but since the problem says “how many erasers are there in each box”, and you can’t have a fraction of an eraser per box unless specified, the answer should be 44, and 1 eraser is left over. But the problem doesn’t ask for remainder — just “in each box”. In school math, sometimes they expect the exact division even if it’s not whole — but here it’s not.

Wait — maybe I misread? Let me check the numbers again.

“Green has 11 boxes of erasers. She has 485 erasers in total.”

Hmm… 485 ÷ 11 = 44.0909... — not a whole number.

But looking back at Problem 1, it divided evenly. Maybe this one does too? Did I miscalculate?

Wait — 11 × 44 = 484 → correct.
11 × 45 = 495 → too big.

So no, it doesn’t divide evenly. But perhaps the problem meant 484? Or maybe we’re supposed to write the decimal? But for Grade 5, usually they give problems that divide evenly.

Wait — let me check the original image text again (even though I’m not supposed to describe it, I need to verify the numbers).

Actually, in the user’s input, it says:
“Green has 11 boxes of erasers. She has 485 erasers in total.”

That’s what’s written. So we must go with that.

In real life, you’d say each box has 44 erasers, and 1 is left over. But since the question is “how many erasers are there in each box?”, and assuming all boxes have the same amount, then it’s 44, and one extra.

But in many textbooks, if it doesn’t divide evenly, they might still expect the quotient as the answer, ignoring remainder unless asked.

Alternatively, maybe I should express it as a mixed number? But Grade 5 may not have learned that yet.

Wait — let’s look at other problems. Problem 3: 180 ÷ 15 = 12 — nice.
Problem 4: 1500 ÷ 25 = 60 — nice.
Problem 5: 2880 ÷ 32 — let’s see later.

Perhaps Problem 2 is intended to be 484? But it says 485.

Maybe it’s a trick? Or maybe I should just compute it as is.

Another thought: perhaps “boxes of erasers” means something else? No, probably not.

I think for accuracy, I’ll state: 485 ÷ 11 = 44 with remainder 1, so each box has 44 erasers, and 1 is left over. But since the question doesn’t mention remainder, and in context, likely they want the number per box, which is 44.

But let’s hold on — maybe I made a calculation error.

Do 11 × 44 = 484 → yes.
485 - 484 = 1 → yes.

So, technically, the answer is 44 r1, but for the purpose of this worksheet, since others divide evenly, perhaps it's a typo, but I have to go with given numbers.

Wait — let me calculate 485 ÷ 11 properly.

11 * 44 = 484
So 485 / 11 = 44 + 1/11 ≈ 44.09, but that’s not helpful.

I recall that in some curricula, they accept the quotient as the answer when remainder is small, but strictly speaking, the problem might expect us to say 44.

Looking ahead, Problem 5: 2880 ÷ 32.

32 * 90 = 2880 → exactly 90. Nice.

So Problems 1,3,4,5 divide evenly. Only Problem 2 does not.

Perhaps it's 484? But the text says 485.

Maybe "Green" is a name, and it's correct.

I think I should proceed with the calculation as is, and note that it's not even, but for the answer, since the question is "how many in each box", and boxes are identical, it's 44, and one left out.

But to match the format, perhaps the expected answer is 44.

Let me check online or think differently.

Another idea: perhaps "boxes of erasers" means she has 11 boxes, each containing some erasers, total 485, so average per box is 485/11, but again, not integer.

I think for consistency, I'll calculate it as 44, and move on, but I'm uneasy.

Wait — let's do exact division: 485 ÷ 11.

11 goes into 48 four times (44), remainder 4. Bring down 5, make 45. 11 goes into 45 four times (44), remainder 1. So 44 R1.

So the quotient is 44.

In many elementary contexts, they teach that the answer to "how many in each group" is the quotient, and remainder is separate.

So I'll go with 44 for now.

But let's confirm with Problem 3.

Problem 3:
Amelia has 180 books to share with 15 friends.
So, 180 ÷ 15.

15 × 12 = 180 → so 12 books per friend.

Easy.

Problem 4:
Sophie has to write 1500 emails, 25 per day.
Days needed = 1500 ÷ 25.

25 × 60 = 1500 → so 60 days.

Good.

Problem 5:
Jessica inviting 32 friends, has 2880 cookies.
Cookies per friend = 2880 ÷ 32.

Let me calculate: 32 × 90 = 2880 → yes, because 30×90=2700, 2×90=180, total 2880. So 90.

All others are clean except Problem 2.

Perhaps in Problem 2, it's 484? But the user wrote 485.

Looking back at the user's message: "Green has 11 boxes of erasers. She has 485 erasers in total."

It's 485.

Maybe it's a test to see if students recognize remainders, but the question doesn't ask for remainder.

I think for the sake of completing, I'll put 44 for Problem 2, noting that there's a remainder, but since the instruction is to provide final answer, and others are whole numbers, perhaps it's acceptable.

To be precise, let's see if 485 is divisible by 11.

Sum of digits alternating: 4 - 8 + 5 = 1, not divisible by 11, so no.

So, I'll proceed.

For Problem 2, the answer is 44 erasers per box, with 1 left over, but since the question likely wants the number per box, it's 44.

Now, compiling all answers.

Final Answers:

1. 10
2. 44
3. 12
4. 60
5. 90

But for Problem 2, I feel unsure. Let me double-check the division once more.

485 ÷ 11:

11 * 40 = 440
485 - 440 = 45
11 * 4 = 44
45 - 44 = 1
So 40 + 4 = 44, remainder 1. Yes.

Perhaps the problem is designed to have a remainder, and students should write 44 R1, but the question says "how many erasers are there in each box?" implying a single number.

In some interpretations, if you distribute equally, each box gets 44, and one is not distributed, so answer is 44.

I think it's fine.

So, final answers:

1. 10
2. 44
3. 12
4. 60
5. 90

Now, to present them.

Since the user didn't specify which problem, but the image has 5 problems, I should solve all.

The user said "solve the problem", but there are five. Probably all.

In the response, I'll list them.

But according to the format, I need to provide Final Answer after reasoning.

Since ExplainAnswer is false, I don't need explanation section.

TeachTopic and GeneratePractice are also false.

So only Final Answer section.

But the Final Answer should be for the task. Since there are multiple problems, I should list all answers.

Perhaps number them.

Looking at the format: "Final Answer: [Provide only the final correct answer]"

But for multiple problems, I think it's okay to list them.

I'll write:

Final Answer:
1. 10
2. 44
3. 12
4. 60
5. 90

But for Problem 2, to be accurate, let's see if there's another way.

Perhaps "Green has 11 boxes" and "485 erasers", but maybe the boxes are not all full? But the question is "how many in each box", assuming equal distribution.

I think 44 is correct as the quotient.

Some might argue for 44.09, but that's not practical.

I recall that in Grade 5, they might not have decimals yet, so whole number quotient is expected.

So I'll stick with that.

Another thought: in Problem 5, it's 2,880 with comma, but 2880 is fine.

All good.

So, final decision.