EN हिं

NCERT Solutions for Class 8th Maths Chapter 6 .1 Some Properties of Multiplication — Figure it Out

Book page 1426 Updated on2026-09-19

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.
×12345678910
112345678910
22468101214161820
336912151821242730
4481216202428323640
55101520253035404550
66121824303642485460
77142128354249566370
88162432404856647280
99182736455463728190
10102030405060708090100
The multiplication grid, page 142. The nine framed numbers are the 3 × 3 frame.
3 × 53 × 63 × 7
4 × 54 × 64 × 7
5 × 55 × 65 × 7
   
 pq 
   
The same nine numbers written as products (left), and the blank frame whose middle number is pq (right).
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 – 1column qcolumn q + 1
row p – 1(p – 1)(q – 1)(p – 1) q(p – 1)(q + 1)
row pp (q – 1)pqp (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 = 153 × 6 = 183 × 7 = 21
4 × 5 = 204 × 6 = 244 × 7 = 28
5 × 5 = 255 × 6 = 305 × 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.

ab = 2a + 4ab(a + 2)(b – 4)
1663 × 2 = 6 ✓
28164 × 4 = 16 ✓
310305 × 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.
Was this helpful?