## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.2

Question 1.
Find the product of:
(i) 4x3 and -3xy
(ii) 2xyz and 0
(iii) –$$\frac{2}{3}$$p2q, $$\frac{3}{4}$$pq2 and 5pqr
(iv) -7ab,-3a3 and –$$\frac{2}{7}$$ab2
(v) –$$\frac{1}{2}$$x2 – $$\frac{3}{5}$$xy, $$\frac{2}{3}$$yz and $$\frac{5}{7}$$xyz
Solution:
Product of
(i) 4x3 and -3xy = 4x3 × (-3xy) = -12x3+1 y = -12x4y
(ii) 2xyz and 0 = 2xyz × 0 = 0

Question 2.
Multiply:
(i) (3x – 5y + 7z) by – 3xyz
(ii) (2p2 – 3pq + 5q2 + 5) by – 2pq
(iii) ($$\frac{2}{3}$$a2b – $$\frac{4}{5}$$ab2 + $$\frac{2}{7}$$ab + 3) by 35ab
(iv) (4x2 – 10xy + 7y2 – 8x + 4y + 3) by 3xy
Solution:
(i) – 3xyz (3x – 5y + 7z)
= (- 3xyz) × 3x + (- 3xyz) × (- 5y) + (- 3xyz) × (7z)
= – 9x2yz + 15xyz2 – 21xyz2

(ii) -2pq (2p2 – 3pq + 5q2 + 5)
= (-2pq) × 2p2 + (-2pq) × (-3pq) + (- 2pq) × (5q2) + (-2pq) × 5
= -4p3q + 6p2q2 – 10pq3 – 10pq

(iii) $$\left(\frac{2}{3} a^{2} b-\frac{4}{5} a b^{2}+\frac{2}{7} a b+3\right)$$ by 35ab
= $$\frac{2}{3}$$a2b × 35ab – $$\frac{4}{5}$$ab2 × 35ab = $$\frac{2}{7}$$ab × 35ab + 3 × 35ab
= $$\frac{70}{3}$$a3b2 – 28a2b3 + 10a2b2 + 105ab

(iv) (4x2 – 10xy + 7y2 – 8x + 4y + 3) by 3xy
4x2 × 3xy – 10xy × 3xy + 7y2 × 3xy – 8x × 3xy + 4y × 3xy + 3 × 3xy
= 12x3y – 30x2y2 + 21xy3 – 24x2y + 12xy2 + 9xy

Question 3.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
(i) (p2q, pq2)
(ii) (5xy, 7xy2)
Solution:
(i) Sides of a rectangle are p2q and pq2
Area = p2q × pq2 = p2+1q2+1 = p3q3

(ii) Sides are 5xy and 7xy2
Area = 5xy × 7xy2 = 35x1+1 × y1+2 = 35x2y3

Question 4.
Find the volume of rectangular boxes with the following length, breadth and height respectively:
(i) 5ab, 3a2b, 7a4b2
(ii) 2pq, 4q2, 8rp
Solution:
Length, breadth and height of a rectangular box are
(i) 5ab, 3a2b, 7a4b2
∴ Volume = Length × breadth × height
= 5ab × 3a2b × 7a4b2
= 5 × 3 × 7 × a1+2+4 × b1+1+2
= 105a7b4

(ii) 2pq, 4q2, 8rp
∴ Volume = 2pq × 4q2 × 8rp
= 2 × 4 × 8 × p1+1 × q1+2 × r
= 64p2q3r

Question 5.
Simplify the following expressions and evaluate them as directed:
(i) x2(3 – 2x + x2) for x = 1; x = -1; x = $$\frac{2}{3}$$ and x = –$$\frac{1}{2}$$
(ii) 5xy(3x + 4y – 7) – 3y(xy – x2 + 9) – 8 for x = 2, y = -1
Solution:
(i) x2(3 – 2x + x2)
for x = 1; x = -1; x = – 1 x = $$\frac{2}{3}$$ and x = –$$\frac{1}{2}$$
x2(3 – 2x + x2) = 3x2 – 2x3 + x4
(a) x = 1, then
3x2 – 2x3 + x4 = 3(1)2 – 2(1 )3 + (1)4
= 3 × 1 – 2 × 1 + l
=3 – 2 + 1 = 2

(b) x = -1
3x2 – 2x3 + x4 = 3(-1)2 – 2(-1)3 + (-1)4
= 3 × 1 – 2 × (-1) + 1 = 3 + 2 + 1 = 6

(c) x = $$\frac{2}{3}$$

(ii) 5xy(3x + 4y – 7) – 3y(xy – x2 + 9) – 8
= 15x2y + 20xy2 – 35xy – 3xy2 + 3 x2y – 21y – 8
= 18x2y + 17xy2 – 35xy – 27y – 8
When x = 2, y = -1
= 18(2)2 × (-1) + 17(2) (-1)2 – 35(2) (-1) – 27(-1) – 8
= 18 × 4 × (-1) + 17 × 2 × 1 – 35 × 2 × (-1) – 27 × (-1) – 8
= -74 + 34 + 70 + 27 – 8
= 131 – 80 = 51

Question 6.
Add the following:
(i) 4p(2 – p2) and 8p3 – 3p
(ii) 7xy(8x + 2y – 3) and 4xy2(3y – 7x + 8)
Solution:
Add
(i) 4p(2 – p2) and 8p3 – 3p
= 8p – 4p3 + 8p3 – 3p
= 5p + 4p3
= 4p3 + 5p

(ii) 7xy(8x + 2y – 3) and 4xy2(3y – 7x + 8)
= 56x2y + 14xy2 – 21xy + 12xy3 – 28x2y2 + 32xy2
= 12xy3 – 28x2y2 + 56x2y +46xy2 = 21xy

Question 7.
Subtract:
(i) 6x(x – y + z)- 3y(x + y – z) from 2z(-x + y + z)
(ii) 7xy(x2 -2xy + 3y2) – 8x(x2y – 4xy + 7xy2) from 3y(4x2y – 5xy + 8xy2)
Solution:
(i) 6x(x – y + z) – 3y(x + y – z) from 2z(-x + y + z)
6x2 – 6xy + 6xz – 3xy – 3y2 + 3yz from – 2xz + 2yz + 2z2
= (-2xz + 2yz + 2z2) – (6x2 – 6xy + 6xz – 3xy – 3y2 + 3yz)
= – 2xz + 2yz + 2z2 – 6x2 + 6xy – 6xz + 3xy + 3y2 – 3yz
= 9xy – yz – 8zx – 6x2 + 3y2 + 2z2

(ii) 7xy(x2 – 2xy + 3y2) – 8x(x2y – 4xy + 7xy2) from 3y(4x2y – 5xy + 8xy2)
7x3y – 14x2y2 + 21xy3 – 8x3y + 32x2y – 56x2y2 from 12x2y2 – 15xy2 + 24xy3
= (12x2y2 – 15xy2 + 24xy3) – (7x3y – 14x2y2 + 21xy3 – 8x3y + 32x2y – 56x2y2
= 12x2y2 – 15xy2 + 24xy3 – 7x3y + 14x2y2 – 12xy3 + 8x3y – 32x2y + 56x2y2
= 82x2y2 + 3xy3 + x3y – 15xy2 – 32x2y