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

Question 1.

Find the value of the following:

(i) 4^{3}

(ii) (-6)^{4}

(iii) \(\left(\frac{2}{3}\right)^{4}\)

(iv) (-2)^{3} × 5^{2}

Solution:

(i) 4^{3} = 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} × 5^{2}

= (-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) 2x^{2} – 3x + 4 when × = 2

(ii) 4x^{3} – 5x^{2} – 6x + 7 when x = 3

(iii) 3x^{3} + 9x^{2} – x + 8 when x = -2

(iv) 2x^{4} – 5x^{3} + 7x – 3 when x = -3

Solution:

(i) 2x^{2} – 3x + 4, x = 2

= 2 × (2)^{2} -3x^{2} + 4

= 8 – 6 + 4 = 6

(ii) 4x^{3} – 5x^{2} – 6x + 7, x = 3

= 4(3)^{3} – 5(3)^{2} – 6(3) + 7

= 108 – 45 – 18 + 7 = 52

(iii) 3x^{3} + 9x^{2} – x + 8, x = -2

= 3(-2)^{3} + 9(-2)^{2} – (-2) + 8

= -24 + 36 + 2 + 8 = 22

(iv) 2x^{4} – 5x^{3} + 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 – 7x^{2}

(ii) 3x^{2} + 8x – 10

(iii) 2x^{3} – 4x^{2} – 6x + 25

Solution:

(i) 6 – 7x^{2} = 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 a^{2} + b^{2} + c^{2} – 2ab – 2bc – 2ca + 3abc.

Solution:

a = 2,b = 3,c = -2

a^{2} + b^{2} + c^{2} – 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: p^{3} + q^{3} – r^{3} – 3pqr.

Solution:

p = 4, q = -3, r = 2

p^{3} + q^{3} – r^{3} – 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 2mn^{4} – 15m^{2}n + p.

Solution:

m = 1, n = 2, p = -3

2mn^{4} – 15m^{2}n + 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 2x^{2} – x – 3 is 0 when x = -1.

(iii) p^{2} + q^{2} – r^{2} 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 2x^{2} – x – 3 is 0 when x = -1. True

2(-1 )^{2} – (-1) – 3

= 2 + 1 – 3 = 0

(iii) p^{2} + q^{2} = r^{2} 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 – 3x^{2} = 5x^{2}

16 – 6 = 10

⇒ 10 = 10

Question 10.

For x = 2 and y = -3, verify the following:

(i) (x + y)^{2} = x^{2} + 2xy + y^{2}

(ii) (x – y)^{2} = x^{2} – 2xy + y^{2}

(iii) x^{2} – y^{2} = (x + y) (x – y)

(iv) (x + y)^{2} = (x – y)^{2} + 4xy

(v) (x + y)^{3} = x^{3} + y^{3} + 3x^{2}y + 3xy^{2}

Solution:

x = 2, y = -3

(i) (x + y)^{2} = x^{2} + 2xy + y^{2}

L.H.S. = (x + y)^{2} = (2 – 3)^{2} = (-1)^{2} = 1

R.H.S. = x^{2} + 2xy + y^{2}

= (2)^{2} + 2 × 2 (-3) + (-3)^{2}

= 4 – 12 + 9 = 13 – 12 = 1

L.H.S. = R.H.S.

(ii) (x – y)^{2} = x^{2} – 2xy + y^{2}

L.H.S. = (x – y)^{2} = [2 – (-3)]^{2}

= (2 + 3)^{2} = (5)^{2} = 25

R.H.S. = x^{2} – 2xy + y^{2}

= (2)^{2} – 2 × 2 × (-3) + (-3)^{2}

= 4 + 12 + 9 = 25

∴ L.H.S. = R.H.S.

(iii) x^{2} – y^{2} = (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} = x^{3} + y^{2} + 3x^{2}y + 3xy^{2}

L.H.S. = (x + y)^{3} = [2 + (-3)]^{3} = (2 – 3)^{3}

= (-1)^{3} = (-1) × (-1) × (-1) = -1

R.H.S. = x^{3} + y^{3} + 3x^{2}y + 3xy^{2}

= (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.