# Algebraic Expressions ICSE Class-8th Concise Selina

Algebraic Expressions ICSE Class-8th Concise Selina Mathematics  Solutions Chapter-11 . We provide step by step Solutions of Exercise / lesson-11 Algebraic Expressions for ICSE Class-8 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-11 A,  Exe-11 B, Exe-11 C,  Exe-11 D and Exe-11 E to develop skill and confidence . Visit official Websitefor detail information about ICSE Board Class-8.

## Algebraic Expressions ICSE Class-8th Concise Selina Mathematics  Solutions Chapter-11

–: Select Topics :–

Exe-11 A,

Exe-11 B,

Exe-11 C,

Exe-11 D,

Exe-11 E,

### Exercise – 11 AAlgebraic Expressions for ICSE Class-8th Mathematics

#### Question 1 :-

Separate the constants and variables from the following :

#### Question 2 :-

Write the number of terms in each of the following polynomials.
(i) 5x2 + 3 x ax
(ii) ax ÷ 4 – 7
(iii) ax – by + y x z
(iv) 23 + a x b ÷ 2.

#### Question 3 :-

Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :

#### Question 4 :-

Write the degree of each polynomial given below :

#### Question 5 :-

Write the coefficient of :
(i) ab in 7abx ,
(ii) 7a in 7abx ;
(iii) 5x2 in 5x2 – 5x ;
(iv) 8 in a2 – 8ax + a ;
(v) 4xy in x2 – 4xy + y2.

#### Question 6 :-

In $\frac { 5 }{ 7 }$ xy2z3, write the coefficient of

#### Question 7 :-

In each polynomial, given below, separate the like terms :
(i) 3xy, − 4yx2, 2xy2, 2.5x2y, −8yx, −3.2y2x and x2y

(ii) y2z3, xy2z3, −5x2yz, −4y2z3, −8xz3y2, 3x2yz and 2z3y2

(i) Like terms are 3xy, −8yx; −4yx2, 2.5x2y, and x2y; 2xy2 and −3.2y2x

(ii) Like terms are y2z3, −4y2z3, and 2z3y2; xy2zand −8xz3y2: −5x2yz and 3x2yz

### Algebraic Expressions ICSE Class-8th Mathematics Exe – 11 B

Evaluate :

#### Question 3 :-

Find the total savings of a boy who saves ₹ (4x – 6y) ; ₹ (6x + 2y) ; ₹ (4y – x) and ₹ (y – 2x) for four consecutive weeks.

Subtract :

#### Question 5 :-

(i) Take away – 3x3 + 4x2 – 5x + 6 from 3x3 – 4x2 + 5x – 6
(ii) Take m2 + m + 4 from -m2 + 3m + 6 and the result from m2 + m + 1.

#### Question 6 :-

Subtract the sum of 5y2 + y – 3 and y2 – 3y + 7 from 6y2 + y – 2.

#### Question 7  :-

What must be added to x4 – x3 + x2 + x + 3 to obtain x4 + x2 – 1 ?

x4        + x2        – 1
+x4 – x3 + x2 + x + 3
–     +     –      –     –
x3         – x  – 4

#### Question 8 :-

(i) How much more than 2x2 + 4xy + 2y2 is 5x2 + 10xy – y2 ?
(ii) How much less 2a2 + 1 is than 3a2 – 6 ?

(i)

5x2  + 10xy  –    y2
+2x2  +   4xy  + 2y2
–        –           –
3x2 +    6xy  – 3y

(ii)

3a2 –  6
+ 2a2 +  1
–         –
a2  –  7

#### Question 9 :-

If x = 6a + 86 + 9c ; y = 2b – 3a – 6c and z = c – b + 3a ; find
(i) x + y + z
(ii) x – y + z
(iii) 2x – y – 3z
(iv) 3y – 2z – 5x

(i)                     x =     6a + 8b + 9c
y = − 3a + 2b − 6c
z =  +3a  −  b +   c
Adding x + y + z =    6a  + 9b + 4c

(ii) x − y + z = (6a + 8b + 9c) − (2b − 3a − 6c) + (c − b + 3a)

= 6a + 8b + 9c − 2b + 3a + 6c + c − b + 3a

= 6a + 3a + 3a + 8b − 2b − b + 9c + 6c + c

= 12a + 5b + 16c

(iii) 2x − y − 3z = 2(6a + 8b + 9c) − (2b − 3a − 6c) − 3(c −b + 3a)

= 12a + 16b + 18c − 2b + 3a + 6c − 3c +3b − 9a

= 12a + 3a − 9a + 16b + 3b − 2b + 18c + 6c −3c

= 6a + 17b + 21c

(iv) 3y − 2z − 5x = 3(2b − 3a − 6c) − 2(c − b + 3a) − 5(6a + 8b + 9c)

= 6b − 9a − 18c − 2c + 2b − 6a − 30a − 40b − 45c

= − 9a − 6a − 30a + 6b + 2b − 40b − 18c − 2c − 45c

= − 45a − 32b − 65c

#### Question 10 :-

The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.

Required perimeter = Sum of three sides

= x2 − 3xy + 8 + 4x2 + 5xy − 3 + 6 − 3x2 + 4xy

= x2 + 4x2 − 3x2 − 3xy + 5xy + 4xy + 8 − 3 + 6

= 2x2 + 6xy + 11

#### Question 11 :-

The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.

Perimeter of the triangle = Sum of three sides

= 8y2 – 9y + 4

Sum of two sides = 3y2 – 5y  + 4y2 + 12

= 7y2 − 5y + 12

∴ (8y – 9y + 4)  – (7y2  – 5y + 12)

= 8y2  – 9y + 4  – 7y2 + 5y  – 12

= y2  – 4y  – 8

Hence third side = y2  – 4y  – 8

#### Question 12 :-

The two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2. Find its perimeter.

Adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2

∴ Perimeter = 2(2x2 – 5xy + 3z2 + 4xy – x2 – z2)

= 4x2 – 10xy + 6z2 + 8xy – 2x2 – 2z2

= 2x2 – 2xy + 4z2

#### Question 13 :-

What must be subtracted from 19x4 + 2x3 + 30x – 37 to get 8x4 + 22x3 – 7x – 60 ?

The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)

= 22x – 20y – 13z + 26 – 15x + 18y – 19z

= 7x – 2y – 32z + 26

#### Question 14 :-

How much smaller is 15x – 18y + 19z than 22x – 20y – 13z + 26 ?

The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)

= 22x – 20y – 13z + 26 – 15x + 18y – 19z

= 7x – 2y – 32z + 26

#### Question 15 :-

How much bigger is 15x2y2 – 18xy2 – 10x2y than -5x2 + 6x2y – 7xy ?

The required result,
(5x2y2 – 18xy2 – 10x2y) – (–5x2 + 6x2y – 7xy)

= 5x2y2 – 18xy2 – 10x2y + 5x2 + 6x2y – 7xy

= 5x2y2 – 18xy2 – 16x2y + 5x2 + 7xy

### ICSE Class-8th Concise Selina Maths Exe – 11 C

Multiply :

#### Question 2 :-

Multiply :

(i)

5x− 8xy + 6y2 − 3 × −3xy

= −15x3y3 + 24x2y− 18xy+ 9xy

(ii)

3-(2/3)xy + (5/7)xy2 – (16/21)x2y × (-21x2y2)

= −63x2y2 + 14x3y3 − 15x3y4 + 16x4y3

(iii)

6x3 − 5x    + 10
×    4    − 3x
24x3 − 20x   + 40
− 18x5 + 15x3 − 30x2
18x5 + 39x− 30x− 20x + 40

(iv)

2y  − 4y3 +  6y5
×   y2 +   y   −  3
2y3 − 4y5 +  6y7
+ 2y2 − 4y4 +   6y6
− 6y  + 12y− 18y5

6y7 + 6y−(4+18)y5 − 4y4 + (2+12)y3 + 2y2 − 6y

= 6y7 + 6y6 − 22y5 − 4y4 + 14y3 + 2y2 − 6y

(v)

5p2    + 25pq   + 4q2
×  2p2    − 2pq     +  3q2
10p4    + 50p3q  + 8p2q2
−10p3q  − 50p2q2 − 8pq3
+15p2q2 + 75pq3  + 12q4
10p4    + 40p3q  − 27p2q2 + 67pq3 + 12q4

#### Question 3 :-

Simplify :
(i) (7x – 8) (3x + 2)
(ii) (px – q) (px + q)
(iii) (5a + 5b – c) (2b – 3c)
(iv) (4x – 5y) (5x – 4y)
(v) (3y + 4z) (3y – 4z) + (2y + 7z) (y + z)

(i)

(7x – 8) (3x + 2) = 7x(3x + 2) −8(3x+ 2)

= 21x2 + 14x – 24x – 16

= 21x2 – 10x – 16

(ii)

(px – q) (px + q) = px(px + q) – q(px + q)

= p2x2 + pxq – pqx – q2

= p2x2 – q2

(iii)

(5a + 5b – c) (2b – 3c)

= 5a(2b –  3c) + 5b(2b – 3c) – c(2b – 3c)

= 10ab – 15ac + 10b2 – 15ac – 2bc + 3c2

= 10ab + 10b2 – 17bc – 15ac + 3c2

(iv)

(4x – 5y) (5x – 4y)

= 4x(5x – 4y) – 5y(5x – 4y)

= 20x2 – 16xy –  25xy + 20y2

= 20x2 –  41xy + 20y2

(v)

(3y + 4z) (3y – 4z) + (2y + 7z) (y + z)

= 3y (3y – 4z) + 4z (3y – 4z) + 2y (y + z) (y + z)

= 9y2 – 12yz + 12yz – 16z2 + 2y2 + 2yz + 7yz + 7z2

= (9+2)y+ (–12 + 12 + 2 + 7)yz + (–16 + 7)z2

= 11y2 + 9yz – 9z2

#### Question 4 :-

The adjacent sides of a rectangle are x2 – 4xy + 7y2 and x3 – 5xy2. Find its area.

Reqd. area = (x2 – 4xy + 7y2) (x3 – 5xy2)

= x2 (x–  5xy2) – 4xy (x3 –  5xy2) + 7y2 (x3 –  5xy2)

= x5 – 5x3y2 – 4x4y + 20x2y3 + 7x3y– 35xy4

= x+ 2x3y2 –  4x4y + 20x2y– 35xy4

= (x5 –  4x4y + 2x3y2 + 20x2y– 35xy4) sq. unit.

#### Question 5 :-

The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.

Reqd. Area = 1/2 (base) x (altitude)

=1/2 (3x − 4y) (6x + 5y)

=1/2 (18x2 + 15xy − 24xy − 20y2)

=1/2 (18x2 − 9xy − 20y2) sq. unit.

#### Question 6 :-

Multiply -4xy3 and 6x2y and verify your result for x = 2 and y= 1.

(−4xy3) × (6x2y) = (−4 × 6) (x x x2) (y3 x y)

= −24x3y4
For x = 2 and y = 1

(−4xy3) × (6x2y) = (−4 × 2 × 13) × (6 × 22 × 1)

= (−8) × 24 = −192

And, −24x3y4 = −24 × 23 × 14

= −24 × 8 × 1 = −192

∴ For x = 2 and y = 1, it is verified that
(−4xy3) × (6x2y) = −24x3y4

#### Question 7 :-

Find the value of (3x3) x (-5xy2) x (2x2yz3) for x = 1, y = 2 and z = 3.

For x = 1, y = 2 and z = 3

(3x3) × (−5xy2) × (2x2yz3)

(3 × 13) × (−5 × 1 × 22) × (2 × 12 × 2 × 33)

3 × (−5 × 4) × (2 × 1 × 2 × 27)

3 × (−20) × 108

= −6480

#### Question 8 :-

Evaluate (3x4y2) (2x2y3) for x = 1 and y = 2.

(3x4y2) (2x2y3)

(3 × 14 × 22) × (2 × 12 × 23)

(3 × 1 × 4) × (2 × 1 × 8)

= 12 × 16

= 192

#### Question 9 :-

Evaluate (x5) × (3x2) × (-2x) for x = 1.

For x = 1

(x5) × (3x2) × (−2x)

(15) × (3 × 12) × (−2 × 1)

1 × 3 × (−2)

= −6

#### Question 10 :-

If x = 2 and y = 1; find the value of (-4x2y3) × (-5x2y5).

For x = 2 and y = 1

(−4x2y3) × (−5x2y5)

(−4 × 22 × 13) × (−5 × 22 × 15)

(−4 × 4 × 1) × (−5 −4 × 1)

−16 × −20

= 320

#### Question 11 :-

Evaluate:
(i) (3x – 2)(x + 5) for x = 2.
(ii) (2x – 5y)(2x + 3y) for x = 2 and y = 3.
(iii) xz (x2 + y2) for x = 2, y = 1 and z= 1.

(i)

For x = 2

(3x – 2) (x + 5)

(3 × 2 − 2) (2 + 5)

(6 − 2) × 7

4 × 7

= 28

(ii)

(2x – 5y) (2x + 3y)

= 2x × 2x − 5y × 2x + 2x × 3y − 5y × 3y

= 4x− 10xy + 6xy − 15y2

Hence, x = 2 and y = 3

4x2 − 10xy + 6xy − 15y2

= 4(2)2 − 10(2)(3) + 6(2)(3) − 15(3)2

= 16 − 60 + 36 − 135

= 52 − 195

= −143

(iii)

For x = 2, y = 1 and z = 1

xz (x2 + y2)

2(1) × (22 + 12)

2 × (4 + 1)

2 × 5

= 10

#### Question 12 :-

Evaluate:
(i) x(x – 5) + 2 for x = 1.
(ii) xy2(x – 5y) + 1 for x = 2 and y = 1.
(iii) 2x(3x – 5) – 5(x – 2) – 18 for x = 2.

(i)

For x = 1

x(x – 5) + 2

1(1 − 5) + 2

−4 + 2

= −2

(ii)

For x = 2 and y = 1

xy2(x – 5y) + 1

2 × 1(2 − 5 × 1) + 1

2 × (2 − 5) + 1

2 × (−3) + 1

= −5

(iii)

For x = 2

2x(3x – 5) – 5(x – 2) – 18

2 × 2(3 × 2 − 5) − 5(2 − 2) − 18

4(6 − 5) − 5 × 0 − 18

4 − 18

= − 14

#### Question 13 :-

Multiply and then verify :
-3x2y2 and (x – 2y) for x = 1 and y = 2.

#### Question 14 :-

Multiply:
(i) 2x2 – 4x + 5 by x2 + 3x – 7
(ii) (ab – 1)(3 – 2ab)

#### Question 15 :-

Simplify : (5 – x)(6 – 5x)(2 -x).

### Algebraic Expressions Solved Questions of Exe-11 D for ICSE Class-8th Concise Selina Mathematics

#### Question 1 :-

Divide :

(iv)

(v)

= −7a4−2 . b5−4 . c6−3

= −7a2bc3

(vi)

(ix)

a+4)a2+7a+12¯(a+3
a2 +4a
−      −
3a + 12
3a + 12
−      −
×

∴ a = 3

(x)

x-6)x2+3x-54¯(x+9
x2 -6x
−      +
9x − 54
+  9x − 54
−      +
×

∴ x + 9

(xi)

3x+4y)12x2+7xy-12y2¯(4x-3y
12x2 +16xy
−         −
−9xy − 12y2
−9xy − 12
+       +
×

∴ 4x − 3y

(xii)

x2-2 ) x6-8       ¯(x4+2x2+4
x6 -2×4
−    +
2x4 − 8
2x4        − 4x2
−         +
4x2 − 8
4x2 − 8
−     +
×

∴  x4 + 2x2 + 4

(xiii)

2x3-x-6)6x3-13x2-13x+30¯(3x-5
6x3-3x2-18x
−     +        +
−10x2 + 5x + 30
−10x+ 5x + 30
+        −      −
×

∴  3x − 5

(xiv)

2a+3b+5c)4a2+12ab+9b2-25c2¯(2a+3b-5c
4a2 + 6ab                         + 10ca
−    −                                −
6ab + 9b2 − 25c2   − 10ca
6ab + 9b2                  +15bc
−       −                        −
−10ca −25c2 −15bc
−10ca −25c2 −15bc
+        +        +
×

∴ 2a + 3b − 5c

(xv)

-x3+x+4)+ x6-2x-8 x3+ x2+8x+16¯(- x3+x+4
+ x6   −x4   − 4x3
−       +      +
−x4 −4x3 + x2 + 8x +16
−x4          + x2 + 4x
+              −     −
−4x3 + 4x + 16
−4x3 + 4x + 16
+      −      −
×

∴ −x3 + x + 4

#### Question 2 :-

.
Find the quotient and the remainder (if any) when :
(i) a− 5a2 + 8a + 15 is divided by a + 1. verify your answer.

(ii) 3x4 + 6x3 − 6x2 + 2x − 7 is divided by x − 3. verify your answer.

(iii) 6×2 + x  − 15 is divided by 3x +5. verify your answer.

(i)

a+1) a3-5 a2+8a+15¯( a2-6a+14
a3 + a
−   −
−6a2 + 8a + 15
−6a2 − 6a
+       +
14a + 15
14a + 14
1

∴ Quotient = a2 − 6a + 14 and reminder = 1

Verification :

Dividiend = Quotient × Divisor + Reminder

= (a2 − 6a + 14) × (a + 1) + 1

= a3 − 6a2 + 14a + a2 − 6a + 14 + 1

= a3 − 5a2 + 8a + 15 which is given

(ii)

-x-3)3x4+6x3-6x2+2x-7¯(3x3+15x2+39x+119
3x4 − 9x3
−    +
15x3 − 6x2 + 2x − 7
15x3 − 45x2
−     +
39x2 + 2x − 7
39x2 − 117x
−      +
119x − 7
119x − 357
−   +
350

∴ Quotient = 3x3 + 15x2 + 39x + 119 and reminder = 350

Verification :

Dividend = Quotient × Divisor + Reminder

= (3x3 + 15x2 + 39x + 119) (x − 3) + 350

= 3x4 + 15x3 + 39x2 + 119x − 9x3 − 45x2 − 117x − 357 + 350

= 3x4 + 6x3 − 6x2 + 2x − 7 which is given

(iii)

3x+5)6x2 +x−15¯(2x−3
6x2  + 10x
−     −
−9x − 15
−9x − 15
+     +
×

∴ Quotient = 2x − 3 and remainder = 0

Verification :

Dividend = Quotient × Divisor + Reminder

= (2x − 3) (3x + 5) + 0

= 6x2 + 10x − 9x − 15 + 0

= 6x2 − x − 15 which is given

#### Question 3 :-

The area of a rectangle is x3 – 8x2 + 7 and one of its sides is x – 1. Find the length of the adjacent side.

Area = x3 − 8x + 7

One side = x − 1

∴ Adjacent side = (x3 − 8x + 7) ÷ (x − 1)

x-1)x3-8x+7¯(x-7x-7

x3 − x2
−       +
− 7x2 + 7x
− 7x2 + 7x
+       −
− 7x + 7
− 7x + 7
+      −
×

∴ Other side = x2 − 7x − 7

#### Question 4 :-

The product of two numbers is 16x4 – 1. If one number is 2x – 1, find the other.

Product of two numbers = 16x4 − 1

One number = 2x − 1

Then the second number =16×4-12x-1

= 8x3 + 4x2 + 2x + 1

2x-1)16x4                       -1¯(8x3 +4x2 +2x+1
16x4 − 8x3
−        +
8x3
8x3 − 4x2
−    +
4x2
4x2 − 2x
−    +
2x − 1
2x − 1
−    +
×

∴ Other side = 8x3 + 4x2 + 2x + 1

#### Question 5 :-

Divide x6 – y6 by the product of x2 + xy + y2 and x – y.

Product of (x+ xy + y2) and (x − y)

= (x − y) (x2 + xy + y2)

= x(x2 + xy + y2) − y(x2 + xy + y2)

= x3 + x2y + xy2 − x2y − xy2 − y3

= x3 − y3

Now, (x6 − y6) ÷ (x3 − y3)

= x3 + y3

x3-y3)x6   -x6¯(x3+y3
x6 − x3y3
−   +
x3y3 − y6
x3y3 − y
−      +
×

### Exercise – 11 E Algebraic Expressions

#### Question 1 :-

Simplify :

a2 − 2a + {5a2 − (3a + 4a2)}

= a2 − 2a + {5a2 − 3a + 4a2}

= a− 2a + {9a − 3a}

= a− 2a + 9a2 − 3a

= 10a2 − 5a

#### Question 2 :-

= x − y − {x − y − (x + y) − x + y}

= x − y − {x − y − x − y − x + y}

= x − y − x + y + x + y + x − y

= 2x

#### Question 3 :-

3 (1 − x2) − 2{x2 − (3 − 2x2)}

−3 (1 − x2) − 2 {x− (3 − 2x2)}

= −3 + 3x2 − 2 {x2 − 3 + 2x2}

= −3 + 3x2 − 2{3x2 − 3}

= −3 + 3x2 − 6x2 + 6

= 3 − 3x2

#### Question 4 :-

= 2{m − 3 (n + m − 2n)}

= 2{m − 3 (m − 3 (m − n)}

= 2{m − 3m + 3n}

= 2{3n − 2m}

= 6n − 4m

#### Question 5 :-

= 3x − [3x − {3x − (3x − 3x + y)}]

= 3x − [3x − {3x − y}]

= 3x − [3x − 3x + y]

= 3x − y

#### Question 6 :-

= p2x − 2 {px − 3x (x2 − 3a + x2)}

= p2x − 2{px − 3x (2x2 − 3a)}

= p2x − 2{px − 6x3 + 9ax}

= p2x − 2px + 12x3 − 18ax

#### Question 7 :-

= 2[6 + 4 {m − 6 (7 − n − p) + q }]

= 2[6 + 4 {m − 42 + 6n + 6p + q}]

= 2[6 + 4m − 168 + 24n + 24p + 4q]

= 2[4m + 24n + 24p + 4q − 162]

= 8m + 48n + 48p + 8q − 324

#### Question 8 :-

= a − [a − b − a − {a − (a − b + a)}]

= a − [− b − {a − a + b − a}]

= a − [− b − b + a]

= a + b + b − a

= 2b

#### Question 9 :-

= 3x  − [4x − 3x + 5 − 3 {2x − (3x − 2x + 3y)}]

= 3x − [4x − 3x + 5y − 3 {2x − (x + 3y)}]

= 3x − [4x − 3x + 5y − 3 {2x − x − 3y}]

= 3x − [x + 5y − 6x + 3x + 9y]

= 3x − [−2x + 14y]

= 3x + 2x − 14y

= 5x − 14y

a5 ÷ a+ 3a × 2a

= a5−3 + 3a × 2a

= a2 + 6a2

= 7a2

#### Question 11 :-

x5 ÷ (x2 × y2) × y3

x5 ÷ (x2 × y2) × y3

=( x5 x2 /y2) ×y3

= x5−2 − y3−2

= x3y

#### Question 12 :-

(x5 ÷ x2) × y× y3

(x5 ÷ x2) × y× y3

= x5−2 × y2+3

= x3y5

#### Question 13 :-

(y3 − 5y2) ÷ y × (y − 1)

(y3 − 5y2) ÷ y × (y − 1)

=(y3-5y2)/y×y-1

= (y2 − 5y) × (y − 1)

= y(y − 1) − 5y (y − 1)

= y− y2 − 5y2 + 5y

= y3 − 6y2 + 5y

#### Question 14 :-

= 3a × [2b − 6 {a − 7a + 3b}]

= 3a × [2b − 6 {−6a + 3b}]

= 3a × [2b + 36a − 18b]

= 3a × [36a − 16b]

= 108a2 − 48ab

#### Question 15 :-

7x + 4 {x2 ÷ (5x ÷ 10)} − 3 {2 − x3 ÷ (3x2 ÷ x)}

7x + 4 {x2 ÷ (5x ÷ 10)} − 3 {2 − x3 ÷ (3x2 ÷ x)}

= 7x + 8x − 6 +x

= x2 + 15x − 6

— End of Algebraic Expressions Solutions :–

Thanks