Free 7th Grade Math Worksheets—Printable w/ Answers — Mashup Math - Free Printable

Let's solve each problem using the Associative Property of Multiplication, which states:

> (a × b) × c = a × (b × c)

This means that when multiplying three or more numbers, the way we group them does not change the product. We can rearrange the parentheses without changing the result.

We’ll go through each problem one by one and fill in the blanks accordingly.

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1) (__ × 7) × 5 = __ × (3 × 5)

We need to make both sides equal using the associative property.

Left side: (__ × 7) × 5
Right side: __ × (3 × 5)

We see that on the right, we have (3 × 5), so the missing number must be 3 to match the grouping.

So left becomes: (3 × 7) × 5
Right: 3 × (3 × 5)? Wait — no, that doesn’t work.

Wait — let’s look again.

We want:
(__ × 7) × 5 = __ × (3 × 5)

But 3 × 5 = 15. So right side is __ × 15.

Left side: (__ × 7) × 5 = __ × 35 (since 7×5=35)

So for both sides to be equal, we need the same value.

But unless the missing number is the same, it won't work.

Wait — maybe there's a typo? Let's re-express.

Actually, the associative property allows us to regroup, but not necessarily change values.

Let’s suppose the missing number is x.

Then:
(x × 7) × 5 = x × (7 × 5) = x × 35

But the right side is: __ × (3 × 5) = __ × 15

So unless x × 35 = y × 15, this won’t work unless we’re matching variables.

Wait — perhaps the blank is meant to be filled with the same number?

Let’s try to find a number such that both expressions are equivalent.

But actually, the pattern suggests that the same number should appear in both blanks.

Let’s assume the missing number is 3.

Then:
(3 × 7) × 5 = 21 × 5 = 105
3 × (3 × 5) = 3 × 15 = 45 → Not equal.

Not good.

Wait — maybe it's 6?

(6 × 7) × 5 = 42 × 5 = 210
6 × (3 × 5) = 6 × 15 = 90 → No.

Wait — perhaps the first blank is 3, and second is 3?

No, that gives different results.

Wait — maybe it's a typo? Or maybe we're supposed to use the same number in both places?

Let’s think differently.

The associative property says:
(a × b) × c = a × (b × c)

So if we have (x × 7) × 5, that equals x × (7 × 5) = x × 35

Now, the right side is: __ × (3 × 5) = __ × 15

So unless x × 35 = y × 15, they can't be equal unless x and y are related.

But unless the numbers are chosen properly, this won't work.

Wait — perhaps the missing number is 3, and the right side has 3?

But then:

(3 × 7) × 5 = 21 × 5 = 105
3 × (3 × 5) = 3 × 15 = 45 → Nope.

Alternatively, maybe the first blank is 3, and the second blank is 21?

(3 × 7) × 5 = 21 × 5 = 105
21 × (3 × 5) = 21 × 15 = 315 → No.

Wait — perhaps I misread.

Let me re-express:

1) (__ × 7) × 5 = __ × (3 × 5)

Let’s suppose the missing number is x on the left and y on the right.

So:
(x × 7) × 5 = y × (3 × 5)
→ x × 35 = y × 15

So x / y = 15 / 35 = 3/7

So if x = 3, y = 7 → 3 × 35 = 105, 7 × 15 = 105 → Yes!

So:

(3 × 7) × 5 = 7 × (3 × 5)

Check:
Left: (3 × 7) × 5 = 21 × 5 = 105
Right: 7 × (3 × 5) = 7 × 15 = 105 → YES!

So the answer is:
3 and 7

So:
(3 × 7) × 5 = 7 × (3 × 5)

Wait — but the blank on the right is before (3 × 5), so it's __ × (3 × 5)

So we need:
(3 × 7) × 5 = __ × (3 × 5)

We just saw: (3 × 7) × 5 = 105
And 7 × (3 × 5) = 7 × 15 = 105 → So yes!

So the blank on the right is 7

So:
(3 × 7) × 5 = 7 × (3 × 5)

But the left blank is also 3

So answer:
1) (3 × 7) × 5 = 7 × (3 × 5)

Yes — that works.

But wait — is that using the associative property?

Let’s check:
Left: (3 × 7) × 5
Right: 7 × (3 × 5)

But (3 × 7) × 5 = 3 × 7 × 5
7 × (3 × 5) = 7 × 3 × 5 = same thing — commutative and associative.

So yes, it’s valid.

So the two blanks are: 3 and 7

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2) (6 × 2) × __ = __ × (4 × 6)

Left: (6 × 2) × x = 12 × x
Right: y × (4 × 6) = y × 24

We want 12x = 24y → x = 2y

Try y = 2 → x = 4

Then:
Left: (6 × 2) × 4 = 12 × 4 = 48
Right: 2 × (4 × 6) = 2 × 24 = 48 → YES!

But what if we use associative property?

We want to regroup.

(6 × 2) × x = 6 × (2 × x)

But right side: y × (4 × 6)

So unless we can make 6 × (2 × x) = y × (4 × 6)

But 6 × (2 × x) = 12x
y × 24 = 24y

So again, 12x = 24y → x = 2y

So pick y = 4 → x = 8? Try:

(6 × 2) × 8 = 12 × 8 = 96
4 × (4 × 6) = 4 × 24 = 96 → YES

But simpler: try x = 4, y = 2

(6 × 2) × 4 = 12 × 4 = 48
2 × (4 × 6) = 2 × 24 = 48 → YES

But is there a better choice?

Wait — notice that 6 and 4 are common.

Try to make both sides have same numbers.

Suppose we set x = 4, y = 2

Then:
(6 × 2) × 4 = 6 × (2 × 4) = 6 × 8 = 48
But right: 2 × (4 × 6) = 2 × 24 = 48

But 2 × (4 × 6) = (2 × 4) × 6 = 8 × 6 = 48

So it's associative and commutative.

But the question wants us to fill in the blanks.

Let’s suppose we use 4 on the left and 2 on the right.

So:
(6 × 2) × 4 = 2 × (4 × 6)

Is that correct?

Left: (6 × 2) × 4 = 12 × 4 = 48
Right: 2 × (4 × 6) = 2 × 24 = 48 → YES

So:
2) (6 × 2) × 4 = 2 × (4 × 6)

Yes.

Alternatively, could we use other numbers?

But likely, the intended answer is 4 and 2

But let’s see if there’s a pattern.

Wait — notice that 6 and 4 are swapped.

Maybe the blank on the left is 4, and on the right is 6?

Try:
(6 × 2) × 4 = 6 × (4 × 6)?
Left: 12 × 4 = 48
Right: 6 × 24 = 144 → No.

No.

So only works if right blank is 2

So answer: 4 and 2

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3) (__ × 9) × 5 = (__ × 5) × 3

Left: (x × 9) × 5 = x × 45
Right: (y × 5) × 3 = y × 15

So x × 45 = y × 15 → x = y / 3

Try y = 3 → x = 1

Then:
(1 × 9) × 5 = 9 × 5 = 45
(3 × 5) × 3 = 15 × 3 = 45 → YES

So:
(1 × 9) × 5 = (3 × 5) × 3

But the blanks are both __, so first blank is 1, second is 3.

But is there a better way?

Try x = 3 → 3 × 45 = 135 → y = 135 / 15 = 9

Then:
(3 × 9) × 5 = 27 × 5 = 135
(9 × 5) × 3 = 45 × 3 = 135 → YES

So (3 × 9) × 5 = (9 × 5) × 3

That looks nicer.

So:
(3 × 9) × 5 = (9 × 5) × 3

Yes — this uses associative and commutative properties.

So blanks: 3 and 9

Answer: 3 and 9

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4) 7 × (__ × 10) = __ × (7 × 13)

Left: 7 × (x × 10) = 7 × 10 × x = 70x
Right: y × (7 × 13) = y × 91

So 70x = 91y → x/y = 91/70 = 13/10

So x = 13, y = 10 → 70×13 = 910, 91×10 = 910 → YES

So:
7 × (13 × 10) = 10 × (7 × 13)

Check:
Left: 7 × (13 × 10) = 7 × 130 = 910
Right: 10 × (7 × 13) = 10 × 91 = 910 → YES

So blanks: 13 and 10

Answer: 13 and 10

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5) (__ × 17) × 21 = 21 × (__ × 12)

Left: (x × 17) × 21 = x × 17 × 21
Right: 21 × (y × 12) = 21 × 12 × y

So x × 357 = 252 × y

So x/y = 252 / 357 = divide by 21: 12 / 17

So x = 12, y = 17

Then:
(12 × 17) × 21 = 204 × 21 = 4284
21 × (17 × 12) = 21 × 204 = 4284 → YES

So:
(12 × 17) × 21 = 21 × (17 × 12)

But the blank on the right is __ × 12 → so y = 17

So:
(12 × 17) × 21 = 21 × (17 × 12)

So blanks: 12 and 17

Answer: 12 and 17

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6) __ × (26 × 14) = 14 × (__ × 8)

Left: x × (26 × 14) = x × 364
Right: 14 × (y × 8) = 14 × 8 × y = 112y

So 364x = 112y → x/y = 112 / 364 = simplify: divide by 4 → 28 / 91 → divide by 7 → 4 / 13

So x = 4, y = 13

Check:
Left: 4 × (26 × 14) = 4 × 364 = 1456
Right: 14 × (13 × 8) = 14 × 104 = 1456 → YES

So:
4 × (26 × 14) = 14 × (13 × 8)

Answer: 4 and 13

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7) (29 × __) × 5 = 29 × (__ × 33)

Left: (29 × x) × 5 = 29 × 5 × x = 145x
Right: 29 × (y × 33) = 29 × 33 × y = 957y

So 145x = 957y

Divide both sides by... let's find ratio.

145x = 957y → x/y = 957 / 145

Calculate: 957 ÷ 145 ≈ 6.6 → try y = 5 → 957×5 = 4785, 145×x = 4785 → x = 4785 / 145 = 33

145 × 33 = 4785? 145 × 30 = 4350, 145 × 3 = 435 → total 4785 → YES

So x = 33, y = 5

So:
(29 × 33) × 5 = 29 × (5 × 33)

Check:
Left: (29 × 33) × 5 = 957 × 5 = 4785
Right: 29 × (5 × 33) = 29 × 165 = 4785 → YES

So blanks: 33 and 5

Answer: 33 and 5

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8) (__ × 6) × 5 = (__ × 4) × 5

Left: (x × 6) × 5 = x × 30
Right: (y × 4) × 5 = y × 20

So 30x = 20y → 3x = 2y → x = 2, y = 3

Try:
(2 × 6) × 5 = 12 × 5 = 60
(3 × 4) × 5 = 12 × 5 = 60 → YES

So:
(2 × 6) × 5 = (3 × 4) × 5

But is there a better choice?

Could we use same number?

If x = y, then 30x = 20x → x = 0 → only zero.

So not possible.

But 2 and 3 work.

But wait — maybe we can use 4 and 6?

(4 × 6) × 5 = 24 × 5 = 120
(6 × 4) × 5 = 24 × 5 = 120 → YES

Oh! But that would be:
(4 × 6) × 5 = (6 × 4) × 5 → but the right side is (__ × 4) × 5

So if we put 6 on the right: (6 × 4) × 5 → yes

So:
(4 × 6) × 5 = (6 × 4) × 5

But the blanks are:
First: __ × 6 → so 4
Second: __ × 4 → so 6

So: 4 and 6

But earlier I had 2 and 3 — but 4 and 6 is simpler.

But both work.

But is 4 and 6 better?

Yes — because it shows that 4×6 = 6×4, and multiplied by 5.

But the associative property is about grouping, not order.

But since multiplication is commutative, it's fine.

But the problem says "use the associative property", so we should keep the same numbers.

But here, the numbers are different.

Wait — in this case, we can choose any pair where x × 6 = y × 4

So x/y = 4/6 = 2/3

So x = 2k, y = 3k

So simplest: k=1 → x=2, y=3

Or k=2 → x=4, y=6

Either works.

But probably the intended answer is 4 and 6, because it’s symmetric.

But let’s see the pattern.

In previous problems, we used similar numbers.

But here, both are acceptable.

But let’s check the equation:

(__ × 6) × 5 = (__ × 4) × 5

So if we set both blanks to the same number, say x:

Then: (x × 6) × 5 = (x × 4) × 5 → 30x = 20x → x=0

Only zero.

So must be different.

So best: 4 and 6

Because 4×6 = 6×4

So:
(4 × 6) × 5 = (6 × 4) × 5

Yes.

So answer: 4 and 6

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9) (__ × 46) × 78 = (__ × 27) × 78

Left: (x × 46) × 78 = x × 46 × 78
Right: (y × 27) × 78 = y × 27 × 78

Since both have ×78, we can cancel it (assuming ≠0)

So: x × 46 = y × 27

So x/y = 27/46

So smallest integers: x = 27, y = 46

Then:
(27 × 46) × 78 = (46 × 27) × 78 → same thing

So:
(27 × 46) × 78 = (46 × 27) × 78

So blanks: 27 and 46

Answer: 27 and 46

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10) __ × (4 × 22) = (4 × 25) × __

Left: x × (4 × 22) = x × 88
Right: (4 × 25) × y = 100 × y

So 88x = 100y → 22x = 25y → x/y = 25/22

So x = 25, y = 22

Then:
25 × (4 × 22) = 25 × 88 = 2200
(4 × 25) × 22 = 100 × 22 = 2200 → YES

So:
25 × (4 × 22) = (4 × 25) × 22

Answer: 25 and 22

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Final Answers:

1) 3 × 7) × 5 = 7 × (3 × 5)
2) (6 × 2) × 4 = 2 × (4 × 6)
3) (3 × 9) × 5 = (9 × 5) × 3
4) 7 × (13 × 10) = 10 × (7 × 13)
5) (12 × 17) × 21 = 21 × (17 × 12)
6) 4 × (26 × 14) = 14 × (13 × 8)
7) (29 × 33) × 5 = 29 × (5 × 33)
8) (4 × 6) × 5 = (6 × 4) × 5
9) (27 × 46) × 78 = (46 × 27) × 78
10) 25 × (4 × 22) = (4 × 25) × 22

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Summary of Blanks:

| Problem | First Blank | Second Blank |
|--------|-------------|--------------|
| 1 | 3 | 7 |
| 2 | 4 | 2 |
| 3 | 3 | 9 |
| 4 | 13 | 10 |
| 5 | 12 | 17 |
| 6 | 4 | 13 |
| 7 | 33 | 5 |
| 8 | 4 | 6 |
| 9 | 27 | 46 |
| 10 | 25 | 22 |

✔ All solved using the associative and commutative properties of multiplication.

Let me know if you'd like an explanation for any specific one!