# ML Aggarwal Algebra Exe-9.4 Class 6 ICSE Maths Solutions

ML Aggarwal Algebra Exe-9.4 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-9.4 Questions for Algebra as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Algebra Exe-9.4 Class 6 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-9 Algebra Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-9.4 Questions Edition 2023-2024

### Algebra Exe-9.4

ML Aggarwal Class 6 ICSE Maths Solutions

Page-193

#### Question 1. 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.

(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 2.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

(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 3.  If x = 5, find the value of:

(i) 6 – 7x2
(ii) 3x2 + 8x – 10
(iii) 2x3 – 4x2 – 6x + 25

(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 4.If x = 2, y = 3 and z = -1, find the values of:

(i) x ÷ y
(ii) xy/z
(iii) (2x + 3y – 4z)/(3x – z)

(i) x ÷ y = 2/3
(ii) xy/z = 2 × 3/-1 = -6
(iii) (2x + 3y – 4z)/(3x – z) = (2 × 2 + 3 × 3 – 4 × -1)/(3 × 2 – (-1)) = 5/4

#### Question 5. If a = 2, b = 3 and c = 1, find the value of a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc.

a = 2,b = 3,c = -2
a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc
= (2)2 + (3)2 + (1)2 – 2 × 2 × 3 – 2 × 3 × 1 – 2 × 1 × 2 + 3 × 2 × 3 × 1
= 4 + 9 + 1 – 12 – 6 – 4 + 18
= 32 – 22

= 10

#### Question 6. If p = 4, q = -3 and r = 2, find the value of: p3 + q3 – r3 – 3pqr.

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 7.  If m = 1, n = 2 and p = -3, find the value of 2mn4 – 15m2n + p.

m = 1, n = 2, p = -3
2mn4 – 15m2n + p
= 2(1 )(2)4 – 15(1 )2(2) + (-3)
= 32 – 30 – 3 = —1

#### Question 8.  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

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.

—  : End of ML Aggarwal Algebra Exe-9.4 Class 6 ICSE Maths Solutions :–

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