ML Aggarwal Algebraic Expression and Identities Exe-10.2 Class 8 ICSE Ch-10 Maths Solutions

ML Aggarwal Algebraic Expression and Identities Exe-10.2 Class 8 ICSE Ch-10 Maths Solutions. We Provide Step by Step Answer of  Exe-10.2 Questions for Algebraic Expression and Identities as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

ML Aggarwal Algebraic Expression and Identities Exe-10.2 Class 8 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 8th
Chapter-10 Algebraic Expression and Identities
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-10.2 Questions
Edition 2023-2024

Algebraic Expression and Identities Exe-10.2

ML Aggarwal Class 8 ICSE Maths Solutions

Page-175

Question 1. Find the product of:

(i) 4x3 and -3xy

(ii) 2xyz and 0

(iii) –(2/3)p2q, (3/4)pq2 and 5pqr

(iv) –(1/2)x2– (3/5)xy, (2/3)yz and (5/7)xyz

Answer:

(i) 4x3 and -3xy = 4x3 × (-3xy)

= -12x3+1 y

= -12x4y

(ii) 2xyz and 0

= 2xyz × 0

= 0

(iii) –(2/3)p2q, (3/4)pq2 and 5pqr

= –(2/3)p2q x (3/4)pq2 x 5pqr

= -(2/3) x (3/4) x 5 x p2q x pq2 x pqr

= -(5/2)p4q4r

(iv) –(1/2)x2– (3/5)xy, (2/3)yz and (5/7)xyz

= –(1/2)x2 x -(3/5)xy x (2/3)yz x (5/7)xyz

= -(1/2) x (-3/5) x (2/3) x (5/7) x x2 x xy x yz + xyz

= (1/7) x4y3z2

Question 2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:

(i) (p2q, pq2)
(ii) (5xy, 7xy2)

Answer:

(i) Given, sides of a rectangle are p2q and pq2

Therefore,

Area = p2q × pq2 = p2+1 × q2+1 = p3q3

(ii) Given, sides are 5xy and 7xy2

Therefore,

Area = 5xy × 7xy2 = 35x1+1 × y1+2 = 35x2y3


Algebraic Expression and Identities Exe-10.2

ML Aggarwal Class 8 ICSE Maths Solutions

Page-176

Question 3. Find the volume of rectangular boxes with the following length, breadth and height respectively:

(i) 5ab, 3a2b, 7a4b2
(ii) 2pq, 4q2, 8rp

Answer:

(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 = Length × breadth × height

= 2pq × 4q2 × 8rp

= 2 × 4 × 8 × p1+1 × q1+2 × r

= 64p2q3r

Question 4. Multiply:

(i) (3x – 5y + 7z) by – 3xyz
(ii) (2p2 – 3pq + 5q2 + 5) by – 2pq
(iii) (2/3a2b – 4/5ab2 + 2/7ab + 3) by 35ab
(iv) (4x2 – 10xy + 7y2 – 8x + 4y + 3) by 3xy

Answer:

(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) (2/3a2b – 4/5ab2 + 2/7ab + 3) by 35ab

= (2/3)a2b × 35ab – (4/5)ab2 × 35ab + (2/7)ab × 35ab + 3 × 35ab

= (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 5. Simplify the following expressions and evaluate them as directed:

(i) x2(3 – 2x + x2) for x = 1; x = -1; x = 2/3 and x = –1/2
(ii) 5xy(3x + 4y – 7) – 3y(xy – x2 + 9) – 8 for x = 2, y = -1

Answer:

(i) x2(3 – 2x + x2)

For x = 1; x = -1; x = 2/3 and x = –1/2

x2(3 – 2x + x2) = 3x2 – 2x3 + x4

(a) For x = 1

3x2 – 2x3 + x4 = 3(1)2 – 2(1 )3 + (1)4

= 3 × 1 – 2 × 1 + l

= 3 – 2 + 1 = 2

(b) For x = -1

3x2 – 2x3 + x4 = 3(-1)2 – 2(-1)3 + (-1)4

= 3 × 1 – 2 × (-1) + 1

= 3 + 2 + 1 = 6

(c) For x = 2/3

3x2 – 2x3 + x4 = 3(2/3)2 – 2(2/3)3 + (2/3)4

= 3 × (4/9) – 2 × (8/27) + (16/81)

= (4/3) – (16/27) + (16/81)

= (108 – 48 + 16)/81

= (124 – 48)/81

= 76/81

(d) For x = -1/2

3x2 – 2x3 + x4 = 3(-1/2)2 – 2(-1/2)3 + (-1/2)4

= 3 × (1/4) – 2 × (-1/8) + (1/16)

= (3/4) + ¼ + (1/16)

= (12 + 4 + 1)/16

= 17/16

(ii) 5xy(3x + 4y – 7) – 3y(xy – x2 + 9) – 8

= 15x2y + 20xy2 – 35xy – 3xy2 + 3 x2y – 21y – 8

= 18x2y + 17xy2 – 35xy – 27y – 8

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

(ML Aggarwal Algebraic Expression and Identities Exe-10.2 Class 8)

Question 6. Add the following:

(i) 4p(2 – p2) and 8p3 – 3p
(ii) 7xy(8x + 2y – 3) and 4xy2(3y – 7x + 8)

Answer:

(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)

Answer:

(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

— End of Algebraic Expression and Identities Exe-10.2 Class 8 ICSE Maths Solutions :–

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