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NCERT Solutions Ganita Prakash (Part 1) Chapter 6 .1 Some Properties of Multiplication — In-text Questions

Book page 1386 Updated on2026-09-05

Q1.
How do we expand this? [(a + 1) (b + 1)]
Answer

Treat the whole bracket (a + 1) as a single term and distribute the second bracket over it.

(a + 1)(b + 1) = (a + 1) b + (a + 1) 1
= (ab + b) + (a + 1)
= ab + a + b + 1

So the increase in the product ab is a + b + 1.

Why it happens: distributivity does not care what the “single number” a stands for — it can itself be a bracket. Using it once splits the second bracket, using it again splits the first, and every term of one bracket has then met every term of the other.
Q2.
What would we get if we had expanded (a + 1) (b + 1) by first taking (b + 1) as a single term? Try it?
Answer

The same answer — which is the point of trying it.

(a + 1)(b + 1) = a (b + 1) + 1 (b + 1)
= (ab + a) + (b + 1)
= ab + a + b + 1

Test at a = 5, b = 4: LHS = 6 × 5 = 30, RHS = 20 + 5 + 4 + 1 = 30 ✓

Why it happens: multiplication is commutative, so (a + 1)(b + 1) and (b + 1)(a + 1) are the same product. Whichever bracket you break up first, you end up adding the four products a×b, a×1, 1×b and 1×1.
Q3.
What happens when one of the numbers in a product is increased by 1 and the other is decreased by 1? Will there be any change in the product?
Answer

There is a change, and it is b – a – 1.

(a + 1)(b – 1) = (a + 1) b – (a + 1) 1
= ab + b – a – 1

With a = 23, b = 27:

24 × 26 = 23 × 27 + 27 – 23 – 1
= 621 + 3 = 624 ✓
Why it happens: the change is b – a – 1, not zero. So the product goes up when b is bigger than a by at least 2, stays the same when b = a + 1, and goes down when b ≤ a. Adding 1 to a factor and taking 1 off the other is not a fair swap.
Check it yourself: 9 × 11 = 99 but 10 × 10 = 100; here a = 9, b = 11, so the change is 11 – 9 – 1 = +1.
Q4.
Will the product always increase? Find 3 examples where the product decreases.
Answer

No. The change is b – a – 1, so the product decreases whenever b ≤ a.

a, bab(a + 1)(b – 1)Change b – a – 1
a = 5, b = 3156 × 2 = 123 – 5 – 1 = –3
a = 7, b = 7498 × 6 = 487 – 7 – 1 = –1
a = 10, b = 33011 × 2 = 223 – 10 – 1 = –8
Why it happens: you gain a whole column of b dots but lose a whole row of a dots, and the corner dot is lost as well. When the row you lose is at least as long as the column you gain, the total falls.
Q5.
What happens when a and b are negative integers? Check by substituting different values for a and b in each of the above cases. For example, a = –5, b = 8; a = –4, b = –5; etc.
Answer

Nothing changes — the expressions stay correct, because integers also obey distributivity: x (y + z) = xy + xz for any integers x, y, z.

a, bab(a + 1)(b + 1)ab + a + b + 1
–5, 8–40–4 × 9 = –36–40 – 5 + 8 + 1 = –36 ✓
–4, –520–3 × –4 = 1220 – 4 – 5 + 1 = 12 ✓

And for (a + 1)(b – 1) = ab + b – a – 1 with a = –5, b = 8: –4 × 7 = –28, while –40 + 8 + 5 – 1 = –28 ✓

Why it happens: an identity is a statement about expressions, not about counting dots. Once the letter-numbers are allowed to be any integers, the dot picture is only a memory aid — the guarantee comes from distributivity, which integers satisfy.
Tip: notice that with a = –5, b = 8 the “increase” a + b + 1 = 4 is positive, but with a = –4, b = –5 it is –8. The word increase is just a name for the change; the change can be negative.
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