Exponents and Powers Class 8 RS Aggarwal Exe-2B MCQs Goyal Brothers ICSE Maths Solutions

Exponents and Powers Class 8 RS Aggarwal Exe-2B MCQs Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2.  We provide step by step Solutions of cisce prescribed textbook  In this article you would learn to solve MCQs on exponents and powers. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

Exponents and Powers Class 8 RS Aggarwal Exe-2B MCQs Goyal Brothers ICSE Maths Solutions

Exponents and Powers Class 8 RS Aggarwal Exe-2B MCQs Goyal Brothers ICSE Maths Solutions Ch-2

Board ICSE
Publications Goyal Brothers Prakshan
Subject Maths
Class 8th
Writer RS Aggarwal
Book Name Foundation
Chapter Exponents and Powers
Exe-2B MCQs
Edition 2024-2025

MCQs on Exponents and Powers

( Class 8 RS Aggarwal Exe-2B MCQs Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-2 )

Page- 37,38

Exercise- 2B

Multiple Choice Questions :
Que-1: Which of the following values are equal ?
I. 1^5    II. 5^0   III. 0^5   IV. 5^1

(a) I and II   (ii) II and III    (iii) I and III   (iv) I and IV

Solution- (a) I and II

Reason :I 1^5 = 1
II: 5^0 = 1
III: 0^5 = 0
IV: 5^1 = 5
Hence, 1^5 = 5^0 = 1

Que-2: If 5^x = 3125, then the value of 5^(x-3) is 

(a) 25    (b) 125  (c) 625  (d) 1625

Solution- (a) 25

Reason : 5^x = 3125
5^x = 5^5
x = 5.
Therefore 5^(x – 3) = 5^(5 – 3) = 5^2 = 25.

Que-3: The value of (256)^5/4 is

(a) 512   (b) 984    (c) 1024  (d) 1032

Solution- (c) 1024

Reason :  (256)^5/4 = (4^4)^5/4
= 4^(4×5/4)
= 4^5 = 1024.

Que-4: The value of (27)^5/4 lies between 

(a) 0 and 1    (b) 1 and 2   (c) 2 and 3  (d) 3 and 4

Solution- (a) 0 and 1.

Reason :

Que-5: The value of (32/243)^-4/5 is 

(a) 4/9    (b) 9/4     (c) 16/81    (d) 81/16

Solution- (d) 81/16

Reason : (32/243)^-4/5 = [1/(32/243)^4/5]
Consider (32/243)^4/5
[(2^5)^4/5]/[(3^5)^4/5] = 16/81
Therefore,
(32/243)^-4/5 = 1/(16/81)
(32/243)^-4/5 = 81/16.

Que-6: The value of [2-3(2-3)^-1]^-1 is 

(a) 5   (b) -5   (c) 1/5    (d) -1/5

Solution- (c) 1/5

Reason :  = [2 – 3(2 – 3)^-1]^-1
= {2 – [3/(-1)]}^-1
= (2 + 3)^-1
= (5)^-1
= 1/5.

Que-7: The value of (8^-1 – 9^-1) ÷ (4^-1 – 9^-1)^-1 is 

(a) 5    (b) 10   (c) 14    (d) 25

Solution- (b) 10

Reason :  (8^-1 – 9^-1)^-1 ÷ (4^-1 – 9^-1)^-1
= (1/8 – 1/9)^-1 ÷ (1/4 – 1/9)^-1
= (1/72)^-1 ÷ (5/36)^-1
= 72 ÷ 36/5
= 72 × 5/36 = 10.

Que-8:  (64)^-1/2 – (-32)^-4/5 = ?

(a) 1/8   (b) 3/8   (c) 1/16  (d) 3/16

Solution- (c) 1/16

Reason : (64)^-1/2  – (32)^-4/5
= (8)2 × -1/2 – (2)5 × -4/5
= 8-1 – 2-4
= (1/8) – (1/16)
= 1/16.

Que-9: Which of the following is the same as (-5/7)^-7 ?

(a) (5/7)^-7   (b) -(5/7)^-7   (c) (7/5)^7   (d) (-7/5)^7

Solution- (d) (-7/5)^7

Reason : (-5/7)^-7 = 1/(-5/7)^7
= (-7/5)^7.

Que-10: If 3^(x+y) = 81 and 81^(x-y) = 3^8, then the values of x and y are respectively

(a) -1,-3   (b) -1,3    (c) 1,3  (d) 3,1

Solution- (d) 3,1

Reason : x + y = 4  ………….(1)
Given : 81^(x-y) = 3^8
= (3^4)^x-y = 3^8
= 3^4(x-y) = 3^8
= 4(x-y) = 8
= x-y = 8/4
x-y = 2  ……….(2)
Adding equations (1) and (2), we get :
= (x+y) + (x-y) = 4+2
= x + x + y – y = 6
= 2x = 6
= x = 3
Substituting x = 3 in equation (1), we get :
= 3 + y = 4
= y = 4 – 3
= y =1.
x = 3, y =1.

Que-11: Out of the following, which one is greatest ?

(a) (0.008)^1/3   (b) (0.01)^1/2   (c) (0.2)^2  (d) 1/100

Solution- (d) 1/100

Reason : 

Que-12: (1000)^12 ÷ (10)^30 = ?

(a) (1000)^2    (b) 10    (c) 100    (d) (100)^2

Solution- (a) (1000)^2

Reason : (1000)^12 ÷ (10)^30
= (10^3)^12 ÷ (10)^30
= (10)^(3×12) ÷ (10)^30
= (10)^36 ÷ (10)^30
= (10)^(36−30)
= 10^6
= (10^3)^2 = (1000)^2.

Que-13: If (0.4)^2 ÷ (0.008) x (0.2)^6 = (0.2)^x, then the value of x is

(a) 5   (b) 6   (c) 8  (d) 7

Solution- (d) 7

Reason :Let (0.04)^2 ÷ (0.008) × (0.2)^6 = (0.2)^x
Then, (0.2)^x = [(0.2)^2]^2 ÷ (0.2)^3 × (0.2)^6
= (0.2)^x = (0.2)^(2×2) ÷ (0.2)^3 × (0.2)^6
= (0.2)^x = (0.2)^4 ÷ (0.2)^3 × (0.2)^6
= (0.2)^x = (0.2)^(4−3+6)
= (0.2)^x = (0.2)^7
= x = 7.

Que-14: If (25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^x, then the value of x is

(a) 8.5  (b) 13  (c) 16  (d) 17.5

Solution- (b) 13

Reason :Then, [(5^2)^7.5 x 5^2.5]/(5^3)^1.5 = 5^x
=  [(5^2)^7.5 x 5^2.5]/ (5)^(3 x 1.5) = 5^x
= (5^15 x 5^2.5)/5^4.5 = 5^x
= 5^x = 5^(15 + 2.5 – 4.5)
= 5^x = 5^13
x = 13.

Que-15:  If 2^(x+1) = 8^x, then x has the value

(a) -1/3  (b) 1/2  (c) 1   (d) 3

Solution- (b) 1/2

Reason :Given, 2^(x+1) = 8^x
2^(x+1) = (2^3)^x
= 2^(x+1) = 2^3x
Since bases are equal, their powers must also be equal.
= x+1 = 3x
= 2x = 1
= x = 1/2.

Que-16:  Given that 9^n + 9^n = 3^2013, what is the value of n ?

(a) 1005   (b) 1006  (c) 2011    (d) 6019

Solution- (b) 1006

Reason : 2⋅9^n = 3^2013
We know that 9 = 3^2, so we can rewrite 2⋅9^n as 2⋅(3^2)^n which simplifies to 2⋅3^2n
So, the equation becomes:
2⋅3^2n = 3^2013
Now, we can compare the exponents:
2n = 2013
To solve for n, divide both sides by 2:
n = 2013/2​
So, n = 1006.5

Que-17: If 3^(x-1) + 3^(x+1) = 90, then x is equal to

(a) 0     (b) 1    (c) 2   (d) 3

Solution- (d) 3

Reason : 3^(x-1) + 3^(x+1) = 90 , [ taking 3^x common ]
= (⅓ + 3) • 3^x = 90 ,
= 10/3 • 3^x = 90 ,
= 3^x = 90 • 3/10 = 270/10 = 27 ,
= 3^x = 3³
= x = 3.

Que-18: If (x/y)^(n-1) = (y/x)^(n-3) then the value of n is 

(a) 1/2  (b) 1  (c) 2  (d) 7/2

Solution- (c) 2

Reason : (x/y)^n-1 = (y/x)^n-3
(x/y)^n-1 {(x/y)^-1}^n-3
(x/y)^n-1 = (x/y)^-(n-3)
n-1 = -(n-3)
n-1 = -n+3
2n = 1+3
2n = 4
n = 2.

Que-19: If (25)^x = (125)^y, then x:y equals 

(a) 1:1   (b) 1:3   (c) 2:3   (d) 3:2

Solution- (d) 3:2

Reason :  (5^2)^x = (5^3)^y
This simplifies to:
5^2x = 5^3y
For these expressions to be equal, their exponents must be equal. So:
2x = 3y
We want the ratio x:y, so we can rearrange the equation to solve for x/y:
x/y = 3/2​
So, x:y = 3:2.

Que-20: The simplified value of
[(2.3)^(n+1) + (7.3)^(n-1)]/[3^(n+2) – 2(1/3)^(1-n)] is

(a) -1      (b) 0     (c) 1    (d) 3

Solution- (c) 1

Reason : [2 x 3^n x 3^1 + 7 x 3^n x 3^-1]/[3^n + 2 – 2 x (3^-1)^1-n]
Taking 3^n as common factor we get
3^n(2 x 3 + 7 x 3^-1)/[3^n+2 – 2 x 3^n-1]
3^n(2 x 3 + 7 x 3^-1)/[3^n x 3^2 – 2 x 3^n x 3^-1]
3^n(2 x 3 + 7 x 3^-1)/[3^n(3^2 – 2 x 3^-1)]
(6 + 7/3)/(9 – 2/3)
(25/3) / (25/3)
= 1.

— : End of Exponents and Powers Class 8 RS Aggarwal Exe-2B MCQs Goyal Brothers Prakashan ICSE Foundation Maths Solutions :–

Return to :- ICSE Class -8 RS Aggarwal Goyal Brothers Math Solutions

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