## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 9 Algebra Ex 9.4

Question 1.
Find the value of the following:
(i) 43
(ii) (-6)4
(iii) $$\left(\frac{2}{3}\right)^{4}$$
(iv) (-2)3 × 52
Solution:
(i) 43 = 4 × 4 × 4 = 64
(ii) (-6)4 = (-6) × (-6) × (-6) × (-6) = 1296
(iii) $$\left(\frac{2}{3}\right)^{4}=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}$$
$$=\frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}=\frac{16}{81}$$
(iv) (-2)3 × 52
= (-2) × (-2) × (-2) × 5 × 5
= (-8) × 25 = -200

Question 2.
Find the value of:
(i) 3x + 2y when x = 3 and y – 2
(ii) 5x – 3y when x = 2 and y = -5
(iii) a + 2b – 5c when a = 2, b = -3 and c = 1
(iv) 2p + 3q + 4r + pqr when p = -1, q = 2 and r = 3
(v) 3ab + 4bc – 5ca when a = 4, 6 = 5 and c = -2.
Solution:
(i) 3x + 2y, x = 3,y = 2
(3 × 3) + (2 × 2) = 9 + 4 = 13
(ii) 5x – 3y, x = 2, y = -5
(5 × 2) – (3 × -5) = 10 + 15 = 25
(iii) a + 2b – 5c, a =2,b = -3, c = 1
2 + (2 × -3) -5 × (1)
= 2 – 6 – 5 = -9
(iv) 2p + 3q + 4r + pqr, p = -1, q = 2, r = 3
= (2 × -1) + (3 × 2) + (4 × 3) + (-1) × 2 × 3
= -2 + 6 + 12 – 6= 10
(v) 3ab + 4bc – 5ca, a = 4, b = 5, c = -2 (3 × 4 × 5) + (4 × 5 × -2) -5 × -2 × 4
= 60 – 40 + 40 = 60

Question 3.
Find the value of:
(i) 2x2 – 3x + 4 when × = 2
(ii) 4x3 – 5x2 – 6x + 7 when x = 3
(iii) 3x3 + 9x2 – x + 8 when x = -2
(iv) 2x4 – 5x3 + 7x – 3 when x = -3
Solution:
(i) 2x2 – 3x + 4, x = 2
= 2 × (2)2 -3x2 + 4
= 8 – 6 + 4 = 6
(ii) 4x3 – 5x2 – 6x + 7, x = 3
= 4(3)3 – 5(3)2 – 6(3) + 7
= 108 – 45 – 18 + 7 = 52
(iii) 3x3 + 9x2 – x + 8, x = -2
= 3(-2)3 + 9(-2)2 – (-2) + 8
= -24 + 36 + 2 + 8 = 22
(iv) 2x4 – 5x3 + 7x – 3, x = -3
= 2(-3)4 – 5(-3)3 + 7(-3) – 3
= 162 + 135 – 21 – 3 = 273

Question 4.
If x = 5, find the value of:
(i) 6 – 7x2
(ii) 3x2 + 8x – 10
(iii) 2x3 – 4x2 – 6x + 25
Solution:
(i) 6 – 7x2 = 6 – 7(5)2 = 6 – 7(25)
= 6 – 175 = -169
(ii) 3(5)2 + 8(5) – 10
= 3(25) + 40 – 10
= 75 + 40 – 10 = 75 + 30 = 105
(iii) 2(5)3 – 4(5)2 – 6(5) + 25
= 2(125) – 4(25) – 30 + 25
= 250 – 100 – 30 + 25= 145

Question 5.
If x = 2, y = 3 and z = -1, find the values of:
(i) x + y
(ii) $$\frac{x y}{z}$$
(iii) $$\frac{2 x+3 y-4 z}{3 x-z}$$
Solution: Question 6.
If a = 2, b = 3 and c = -2, find the value of a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc.
Solution:
a = 2,b = 3,c = -2
a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc
= (2)2 + (3)2 + (-2)2 – 2 × 2 × 3 – 2 × 3 × – 2 – 2 × -2 × 2 + 3 × 2 × 3 × -2
= 4 + 9 + 4 – 12 + 12 + 8 – 36
= 25 – 36 = -11

Question 7.
If p = 4, q = -3 and r = 2, find the value of: p3 + q3 – r3 – 3pqr.
Solution:
p = 4, q = -3, r = 2
p3 + q3 – r3 – 3pqr
= (4)3 + (-3)3 – (2)3 – 3 × 4 × -3 × 2
= 64 – 27 – 8 + 72
= 136 – 35 = 101

Question 8.
If m = 1, n = 2 and p = -3, find the value of 2mn4 – 15m2n + p.
Solution:
m = 1, n = 2, p = -3
2mn4 – 15m2n + p
= 2(1 )(2)4 – 15(1 )2(2) + (-3)
= 32 – 30 – 3 = —1

Question 9.
State true or false:
(i) The value of 3x – 2 is 1 when x = 0.
(ii) The value of 2x2 – x – 3 is 0 when x = -1.
(iii) p2 + q2 – r2 when p = 5, q = 12 and r = 13.
(iv) 16 – 3x = 5x when x = 2.
Solution:
(i) The value of 3x – 2 is 1 when x = 0. False
Correct :
∵ 3 × 0 – 2 = -2
(ii) The value of 2x2 – x – 3 is 0 when x = -1. True
2(-1 )2 – (-1) – 3
= 2 + 1 – 3 = 0
(iii) p2 + q2 = r2 when p = 5, q = 12 and r = 13. True
(5)2 + (12)2 = (13)2
= 25 + 144 = 169
⇒ 169= 169
(iv) 16 – 3x = 5x when x = 2. True
16 – 3x2 = 5x2
16 – 6 = 10
⇒ 10 = 10

Question 10.
For x = 2 and y = -3, verify the following:
(i) (x + y)2 = x2 + 2xy + y2
(ii) (x – y)2 = x2 – 2xy + y2
(iii) x2 – y2 = (x + y) (x – y)
(iv) (x + y)2 = (x – y)2 + 4xy
(v) (x + y)3 = x3 + y3 + 3x2y + 3xy2
Solution:
x = 2, y = -3
(i) (x + y)2 = x2 + 2xy + y2
L.H.S. = (x + y)2 = (2 – 3)2 = (-1)2 = 1
R.H.S. = x2 + 2xy + y2
= (2)2 + 2 × 2 (-3) + (-3)2
= 4 – 12 + 9 = 13 – 12 = 1
L.H.S. = R.H.S.

(ii) (x – y)2 = x2 – 2xy + y2
L.H.S. = (x – y)2 = [2 – (-3)]2
= (2 + 3)2 = (5)2 = 25
R.H.S. = x2 – 2xy + y2
= (2)2 – 2 × 2 × (-3) + (-3)2
= 4 + 12 + 9 = 25
∴ L.H.S. = R.H.S.

(iii) x2 – y2 = (x + y)(x – y)
L.H.S. = (x)2 – (y)2 = (2)2 – (-3)2
= 4 – 9 = -5
R.H.S. = (x + y)(x – y)
= (2 – 3) [2 – (-3)]
= (2 – 3) (2 + 3) = -1 × 5 = -5
∴ L.H.S. = R.H.S.

(iv) (x + y)2 = (x – y)2 + 4xy
L.H.S. = (x + y)2 = [2 + (-3)]2
= (2 – 3)2 = (-1)2 = 1
R.H.S. = (x – y)2 + 4xy
= [2 – (-3)]2 + 4 × 2 × (-3)
= (2 + 3)2 + 4 × 2 × (-3)
= (5)2 – 24 = 25 – 24 = 1
∴ L.H.S. = R.H.S.

(v) (x + y)3 = x3 + y2 + 3x2y + 3xy2
L.H.S. = (x + y)3 = [2 + (-3)]3 = (2 – 3)3
= (-1)3 = (-1) × (-1) × (-1) = -1
R.H.S. = x3 + y3 + 3x2y + 3xy2
= (2)3 + (-3)3 + 3 (2)2 (-3) + 3 × 2 (-3)2
=8 – 27 + 3 × 4 × (-3) + 6 (9)
= 8 – 27 – 36 + 54 = 62 – 63 = -1
∴ L.H.S. = R.H.S.