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

Book page 1406 Updated on2026-09-05

Q1.

Can you see how this identity can be used when one or both numbers are decreased?

Answer

Yes — a decrease is an increase by a negative number, so no new identity is needed.

(a + 1)(b – 1) = (a + 1)(b + (–1))
Take m = 1, n = –1 in Identity 1:
ab + (1) × b + a × (–1) + (1) × (–1)
= ab + b – a – 1

This is exactly the expression found earlier by direct expansion.

Why it happens: the rules for multiplying integers (positive × negative = negative, negative × negative = positive) put the right sign on every one of the four terms automatically. So one identity covers four cases: both up, both down, and either one up with the other down.

Q2.

Use Identity 1 to find how the product changes when (i) one number is decreased by 2 and the other increased by 3; (ii) both numbers are decreased, one by 3 and the other by 4. Verify the answers by finding the products without converting the subtractions to additions.

Answer

(i) Take m = –2, n = 3.

(a – 2)(b + 3) = ab + (–2)b + 3a + (–2)(3)
= ab + 3a – 2b – 6

Direct check, keeping the subtraction: (a – 2)(b + 3) = a(b + 3) – 2(b + 3) = ab + 3a – 2b – 6 ✓

(ii) Take m = –3, n = –4.

(a – 3)(b – 4) = ab + (–3)b + (–4)a + (–3)(–4)
= ab – 4a – 3b + 12

Direct check: (a – 3)(b – 4) = a(b – 4) – 3(b – 4) = ab – 4a – 3b + 12 ✓

Check it yourself: put a = 10, b = 10. (i) 8 × 13 = 104 and 100 + 30 – 20 – 6 = 104. (ii) 7 × 6 = 42 and 100 – 40 – 30 + 12 = 42.

Q3.

Expand (i) (a – u) (b + v), (ii) (a – u) (b – v).

Answer

Multiply each term of the first bracket by each term of the second, letting the sign rules do the work.

(i) (a – u)(b + v) = a·b + a·v + (–u)·b + (–u)·v
= ab – ub + av – uv

(ii) (a – u)(b – v) = a·b + a·(–v) + (–u)·b + (–u)(–v)
= ab – ub – av + uv

Test (ii) at a = 7, u = 2, b = 9, v = 3: LHS = 5 × 6 = 30; RHS = 63 – 18 – 21 + 6 = 30 ✓

Why it happens: in (ii) both “taken away” pieces, ub and av, overlap in the corner rectangle uv, so that corner is removed twice. Adding + uv back puts it in once — the same bookkeeping that will later give (a – b)² = a² – 2ab + b².

Q4.

Example 1: Expand 3a⁄2 (a – b + 1⁄5).

Answer

Distributivity is not limited to two terms inside a bracket — every term gets multiplied.

3a⁄2 (a – b + 1⁄5) = (3a⁄2 × a) – (3a⁄2 × b) + (3a⁄2 × 1⁄5)

Now simplify each term using exponent notation:

3a⁄2 × a = 3⁄2 × (a × a) = 3⁄2 a²
3a⁄2 × b = 3⁄2 × (a × b) = 3⁄2 ab
3a⁄2 × 1⁄5 = (3⁄2 × 1⁄5) a = 3⁄10 a

3a⁄2 (a – b + 1⁄5) = 3⁄2 a² – 3⁄2 ab + 3⁄10 a

Test at a = 2, b = 1: LHS = 3 × (2 – 1 + 0.2) = 3.6; RHS = 6 – 3 + 0.6 = 3.6 ✓

Tip: the three terms have different letter-numbers (a², ab, a), so they are unlike terms and the expression is already in its simplest form.

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