NCERT Solutions Ganita Prakash (Part 1) Chapter 6 .1 Some Properties of Multiplication — Figure it Out

Book page 1426 Updated on2026-09-05

Q1.

Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.

Answer

In the frame shown, the middle entry is 4 × 6 = 24, so p is the row number and q is the column number of the centre. The row above is p – 1 and the row below is p + 1; the column to the left is q – 1 and to the right is q + 1.

	column q – 1	column q	column q + 1
row p – 1	(p – 1)(q – 1)	(p – 1) q	(p – 1)(q + 1)
row p	p (q – 1)	pq	p (q + 1)
row p + 1	(p + 1)(q – 1)	(p + 1) q	(p + 1)(q + 1)

Check against the printed frame, where p = 4 and q = 6:

3 × 5 = 15	3 × 6 = 18	3 × 7 = 21
4 × 5 = 20	4 × 6 = 24	4 × 7 = 28
5 × 5 = 25	5 × 6 = 30	5 × 7 = 35

Why it happens: in a multiplication grid the entry in row i and column j is i × j. Moving one step up or down changes only the row factor by 1; moving one step left or right changes only the column factor by 1. That is why every neighbour of pq is one of the products in the table.

Try This: expand the two diagonal pairs. Down-diagonal: (p – 1)(q – 1) × (p + 1)(q + 1); up-diagonal: (p – 1)(q + 1) × (p + 1)(q – 1). Both equal (p² – 1)(q² – 1), so the two diagonal products of the frame are always equal.

Q2.

Expand the following products. (i) (3 + u) (v – 3) (ii) 2⁄3 (15 + 6a) (iii) (10a + b) (10c + d) (iv) (3 – x) (x – 6) (v) (–5a + b) (c + d) (vi) (5 + z) (y + 9)

Answer

Each term of the first bracket multiplies each term of the second; the integer sign rules fix the signs.

(i) (3 + u)(v – 3) = 3v – 9 + uv – 3u
= uv + 3v – 3u – 9

(ii) 2⁄3 (15 + 6a) = 2⁄3 × 15 + 2⁄3 × 6a
= 10 + 4a

(iii) (10a + b)(10c + d) = 100ac + 10ad + 10bc + bd

(iv) (3 – x)(x – 6) = 3x – 18 – x² + 6x
= –x² + 9x – 18

(v) (–5a + b)(c + d) = –5ac – 5ad + bc + bd

(vi) (5 + z)(y + 9) = 5y + 45 + yz + 9z

Spot checks: (i) at u = 1, v = 4: 4 × 1 = 4 and 4 + 12 – 3 – 9 = 4 ✓. (iv) at x = 2: 1 × (–4) = –4 and –4 + 18 – 18 = –4 ✓.

Did you know? Part (iii) is the whole of two-digit multiplication. A two-digit number with digits a, b is 10a + b, so (10a + b)(10c + d) = 100ac + 10(ad + bc) + bd — hundreds, tens and units of the usual written method.

Q3.

Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.

Answer

Set the change to zero and solve.

(a + 2)(b – 4) = ab
ab – 4a + 2b – 8 = ab
2b = 4a + 8
b = 2a + 4

So any pair in which the second number is 4 more than twice the first will work.

a	b = 2a + 4	ab	(a + 2)(b – 4)
1	6	6	3 × 2 = 6 ✓
2	8	16	4 × 4 = 16 ✓
3	10	30	5 × 6 = 30 ✓

Why it happens: raising the first factor by 2 gains 2b, lowering the second by 4 loses 4a, and the corner loses another 8. The product survives only when the gain 2b exactly pays for the loss 4a + 8.

Q4.

Expand (i) (a + ab – 3b²) (4 + b), and (ii) (4y + 7) (y + 11z – 3).

Answer

(i) Break up the second bracket:

(a + ab – 3b²)(4 + b) = 4(a + ab – 3b²) + b(a + ab – 3b²)
= 4a + 4ab – 12b² + ab + ab² – 3b³
= 4a + 5ab + ab² – 12b² – 3b³

(4ab and ab are like terms: 4ab + ab = 5ab.)

(ii)

(4y + 7)(y + 11z – 3) = 4y(y + 11z – 3) + 7(y + 11z – 3)
= 4y² + 44yz – 12y + 7y + 77z – 21
= 4y² + 44yz – 5y + 77z – 21

(–12y and 7y are like terms: –12y + 7y = –5y.)

Test (ii) at y = 1, z = 1: LHS = 11 × 9 = 99; RHS = 4 + 44 – 5 + 77 – 21 = 99 ✓

Q5.

Expand (i) (a – b) (a + b), (ii) (a – b) (a² + ab + b²) and (iii) (a – b)(a³ + a²b + ab² + b³), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?

Answer

(i) (a – b)(a + b) = a² + ab – ab – b² = a² – b²

(ii) (a – b)(a² + ab + b²)
= a³ + a²b + ab² – a²b – ab² – b³ = a³ – b³

(iii) (a – b)(a³ + a²b + ab² + b³)
= a⁴ + a³b + a²b² + ab³ – a³b – a²b² – ab³ – b⁴ = a⁴ – b⁴

The pattern: multiplying (a – b) by the sum of all products a^i b^j whose degrees add to n – 1 gives aⁿ – bⁿ. The next identity is therefore

(a – b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ – b⁵

Check by expanding:

a(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ + a⁴b + a³b² + a²b³ + ab⁴
–b(a⁴ + a³b + a²b² + ab³ + b⁴) = –a⁴b – a³b² – a²b³ – ab⁴ – b⁵
Adding: a⁵ – b⁵ ✓

Numerical test at a = 3, b = 2: LHS = 1 × (81 + 54 + 36 + 24 + 16) = 211; RHS = 243 – 32 = 211 ✓

Why it happens: every middle term appears twice — once with a + sign from the “a” half and once with a – sign from the “–b” half — so the whole middle cancels in pairs (this is called telescoping). Only the very first term a·a⁴ and the very last (–b)·b⁴ survive.

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