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NCERT Solutions for Class 7th Maths Chapter 2 In-text Questions — Multiplication of Integers

Book page 39–41 Updated on2026-09-19

Q1.
What is the value of the expression 5 × – 3 × 4? Does it matter whether we multiply 5 × – 3 and then multiply the product with 4, or if we multiply – 3 × 4 first and then multiply the product with 5?
Answer

The value is –60, and no, the grouping does not matter.

(5 × (–3)) × 4
= (–15) × 4
= –60

5 × ((–3) × 4)
= 5 × (–12)
= –60
Why it happens: both routes multiply the same three magnitudes 5, 3 and 4, giving 60, and both contain exactly one negative number, so the sign is negative. Since neither the magnitude nor the sign depends on how you bracket, the answer cannot depend on it either.
Q2.
Take a few more examples of multiplication of 3 integers and check this property. What do you observe?
Answer

The grouping never changes the answer. Integer multiplication is associative.

ExpressionGrouped one wayGrouped the other way
(–2) × 6 × (–5)(–12) × (–5) = 60(–2) × (–30) = 60
(–4) × (–3) × (–2)12 × (–2) = –24(–4) × 6 = –24
7 × (–1) × (–9)(–7) × (–9) = 637 × 9 = 63

So for any three integers a, b and c,

a × (b × c) = (a × b) × c
Why it happens: associativity already holds for positive numbers, and it fixes the magnitude. The sign is decided by counting how many of the three numbers are negative — a count that brackets cannot alter. Both halves of the answer are therefore untouched by regrouping.
Q3.
In the expression 5 × – 3 × 4, try to multiply 5 and 4 first and then multiply the product with – 3: (5 × 4) × – 3. Are there orders in which 5 × – 3 × 4 can be evaluated? Will the product be the same in all these cases?
Answer

Yes — every order gives –60.

(5 × 4) × (–3) = 20 × (–3) = –60

The three numbers can be arranged in 6 orders, and each arrangement can be bracketed in 2 ways.

OrderFirst productFinal product
5, –3, 45 × (–3) = –15–60
5, 4, –35 × 4 = 20–60
–3, 5, 4(–3) × 5 = –15–60
–3, 4, 5(–3) × 4 = –12–60
4, 5, –34 × 5 = 20–60
4, –3, 54 × (–3) = –12–60
Why it happens: commutativity lets you reorder the three numbers and associativity lets you rebracket them. Used together, they say that a product of three or more integers has one value, no matter how it is written. That is why we may drop the brackets and simply write 5 × (–3) × 4.
Q4.
Multiply the expression 25 × – 6 × 12 in all the different orders and check if the product is the same in all cases.
Answer

Every order gives –1800.

(25 × (–6)) × 12 = (–150) × 12 = –1800
25 × ((–6) × 12) = 25 × (–72) = –1800
(25 × 12) × (–6) = 300 × (–6) = –1800
GroupingStep 1Step 2Value
(25 × –6) × 12–150–150 × 12–1800
25 × (–6 × 12)–7225 × (–72)–1800
(25 × 12) × –6300300 × (–6)–1800
(–6 × 25) × 12–150–150 × 12–1800
Why it happens: the magnitude is 25 × 6 × 12 = 1800 however it is grouped, and there is exactly one negative factor, so the sign is negative. The third grouping is also the easiest to do in your head, which is the practical use of these properties — reorder a product to make the arithmetic simple.
Tip: multiplying 25 × 12 first is smart because 25 × 12 = 300 is a round number. Choosing a friendly pair to multiply first is allowed precisely because the order does not matter.
Q5.
Look at the following series of multiplications: – 1 × – 1 = 1, – 1 × – 1 × – 1 = – 1, – 1 × – 1 × – 1 × – 1 = 1, – 1 × – 1 × – 1 × – 1 × – 1 = 1. When –1 is multiplied 2 or 4 times the product is positive. When it is multiplied 3 or 5 times the product is negative. Can you generalise these statements further?
Answer

Yes. An even number of –1s gives +1; an odd number of –1s gives –1.

(–1) × (–1) = 1   (2 factors, even)
(–1) × (–1) × (–1) = –1   (3 factors, odd)
(–1) × (–1) × (–1) × (–1) = 1   (4 factors, even)
(–1) × (–1) × (–1) × (–1) × (–1) = –1   (5 factors, odd)
Note on the book: the printed line for five –1s shows the answer as 1. That is a slip — the sentence just below it correctly says that multiplying –1 three or five times gives a negative product. Five –1s multiply to –1.
Why it happens: the –1s can be paired off, and each pair (–1) × (–1) is worth 1. If the count is even, every –1 finds a partner and the whole product collapses to 1. If the count is odd, one lonely –1 is left over after all the pairing, and it drags the answer to –1.
Q6.
Using this understanding of multiplication of many integers, can you give a simple rule to find the sign of the product of many integers?
Answer

Count the negative factors.

  • An even number of negative factors → the product is positive.
  • An odd number of negative factors → the product is negative.
  • If any factor is 0, the product is 0.
(–2) × (–1) × (–5) × (–3): four negatives, even → positive
= 2 × 1 × 5 × 3 = 30

(–7) × 4 × (–1) × (–2): three negatives, odd → negative
= –(7 × 4 × 1 × 2) = –56
Why it happens: pull a factor of –1 out of every negative number. If there are k negatives, the product becomes (–1) multiplied k times, times a product of positive magnitudes. The magnitudes give a positive number, and the k copies of –1 give +1 when k is even and –1 when k is odd. That is the whole rule.
Tip: you never have to work out the sign step by step. Glance at the expression, count the minus signs on the factors, and you know the answer’s sign before doing any arithmetic.
Q7.
Now, consider the expression 5 × (4 + (– 2)). As in the case of positive integers, is this expression equal to 5 × 4 + 5 × (– 2)?
Answer

Yes. Both sides come to 10.

5 × (4 + (–2))
= 5 × 2
= 10

5 × 4 + 5 × (–2)
= 20 + (–10)
= 10

This is the distributive property, and it works for integers just as it did for whole numbers.

Why it happens: in the token picture, 5 × (4 + (–2)) is five copies of a set holding 4 greens and 2 reds. You may count the bag in two ways — cancel inside each set first and then repeat it five times, or gather all the greens together and all the reds together and count them separately. Counting the same bag differently cannot give a different total.
Q8.
Check if the distributive property holds for (– 2) × (4 + (– 3)) (that is, if this expression equals (– 2) × 4 + (– 2) × (– 3)), and for a few other such expressions of your choice. What do you observe? Will this always happen?
Answer

It holds. Both sides give –2.

(–2) × (4 + (–3))
= (–2) × 1
= –2

(–2) × 4 + (–2) × (–3)
= (–8) + 6
= –2

Two more checks of your own choosing:

ExpressionBracket firstDistribute first
(–5) × (7 + (–10))(–5) × (–3) = 15(–35) + 50 = 15
6 × ((–4) + (–1))6 × (–5) = –30(–24) + (–6) = –30

Yes, this will always happen. For any integers a, b and c,

a × (b + c) = (a × b) + (a × c)
Why it happens: the rectangular token arrangement shown in the book proves it. A block of a rows, each holding b greens and c reds, can be read as a copies of (b + c), or split down the middle into a × b and a × c. The same tokens are being counted, so the two readings must agree — for any signs of b and c.
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