RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions

RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 5 Powers and Exponents Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 7
Subject Maths
Chapter Chapter 5
Chapter Name Powers and Exponents
Exercise Additional Questions
Number of Questions 23
Category RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 5 Powers and Exponents Additional Questions

Multiple Choice Questions

Question 1
Value of 42 will be :
(A) 16
(B) 15
(C) 18
(D) 4

Question 2
Number 625 is written in (RBSESolutions.com) exponential form as :
(A) 52
(B) 54
(C) 56
(D) none

RBSE Solutions

Question 3
Value of (-1)3 will be :
(A) 1
(B) 0
(C) -1
(D) none

Question 4
Value (RBSESolutions.com) of 35 : 33 will be :
(A) 9
(B) 10
(C) 16
(D) 25

Question 5
Value of 33 x 35 x 32 will be :
(A) 38
(B) 310
(C) 318
(D) 30

Question 6
Value of (3)5 ÷ (2)5 will be :
RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions Q6

Question 7
Value (RBSESolutions.com) of (5 x 6)4will be :
(A) 48 x 64
(B) 54 x 64
(C) 34 x 24
(D) none
Answer:
1. (A), 2. (B), 3. (C), 4. (A), 5. (B), 6. (A), 7. (B)

Fill in the blanks:
(i) Exponential form (RBSESolutions.com) of \(\frac { 2 }{ 3 } \times\frac { 2 }{ 3 } \times\frac { 2 }{ 3 } \times\frac { 2 }{ 3 } \times\frac { 2 }{ 3 }\) will be ………
(ii) Value of (10)° will be ………
(iii) Expended form of 22 x 32 will be ………
Answer:
(i) (\(\frac { 2 }{ 3 }\))5
(ii) 1,
(iii) 2 x 2 x 2 x 3 x 3

RBSE Solutions

True/False
(i) Value of 2 x 32 will be 18.
(ii) Value (RBSESolutions.com) of \(\frac { 2 }{ 3 } \div \frac { 2 }{ 3 }\) will be 2.
(iii) Standard form of 2500 is 2.5 x 103
(iv) The value of \(\frac { { 2 }^{ 3 }\times{ 3 }^{ 3 } }{ { 2 }^{ 3 }\times{ 3 }^{ 2 } }\) is \(\frac { 3 }{ 2 }\)
Answer:
(i) T
(ii) F
(iii) T
(iv) T

Very Short Answer Type Questions

Question 1
Simplify –
(i) 2 x 103
(ii) 72 x 22
(iii) 23 x 5
Solution:
(i) 2 x 103 = 2 x 10 x 10 x 10 = 2000
(ii) 72 x 22 = 7 x 7 x 2 x 2 = 196
(iii) 23 x 5 = 2 x 2 x 2 x 5 = 40

Question 2
Express the following (RBSESolutions.com) in exponent form
(i) 32 x 34 x 38
(ii) (52)3 ÷ 53
Solution:
(i) 32 x 34 x 38 = 32 + 4 + 8 = 314
(ii) (52)3÷ 53= (5)6 ÷ 53 = \(\frac { { 5 }^{ 6 } }{ { 5 }^{ 3 } }\)= 56 – 3 = 53

Question 3
Expend in exponential (RBSESolutions.com) form by using the power of 10
(i) 172,
(ii) 5643
Solution:
(i) 172 = 1 x 100 +7 x 10 + 2 x 1
= 1 x 102 +7 x 101 + 2 x 100
(ii) 5643 = 5 x 1000 +6 x 100 +4 x 10 + 3 x 1
= 5 x 103 + 6 x 102 + 4 x 101 + 3 x 100

Short Answer Type Questions

Question 1
Find the greater number (RBSESolutions.com) in the following
(i) 43 or 34
(ii) 53 or 35
Solution:
(i) 43 = 4 x 4 x 4 = 64 and 34 = 3 x 3 x 3 x 3= 81
Clearly 81 > 64
∴ 34 is greater.

(ii) 53 = 5 x 5 x 5 = 125
and 35 = 3 x 3 x 3 x 3 x 3 = 243
Clearly 243 > 125
∴ 35 is a greater.

RBSE Solutions

Question 2
Express each of the following in the (RBSESolutions.com) form of powers of prime factors as product.
(i) 648
(ii) 405
Solution:
On factorizing by
RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions
∴ 648 = 2 x 2 x 2 x 3 x 3 x 3 x 3
= 2x 34

(ii) On factorizing by division (RBSESolutions.com) method vision method
RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions
∴ 405 = 3 x 3 x 3 x 3 x 5
= 34 x 5

Question 3
Write the following in (RBSESolutions.com) expended from
a3b2, a2b3, b2a3 and b3a2
Solution:
a3b2 = a3x b2 = (a x a x a) x (b x b)
a2b3 = a2 x a3 = (a x a ) x (b x b x b)
b2a3 = b2 x a3 = (b x b) x (a x a x a)
b3a2 = b3 x a2 = (b x b x b) x (a x a)

Long Answer Type Questions

Question 1
Simplify :
RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions
Solution:
RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Additional Questions

Question 2
Express the following (RBSESolutions.com) numbers in standard form
(i) 5,00,00,000
(ii) 70,00,000
Solution:
(i) 5,00,00,000 = 5 x 10000000 = 5 x 107
(ii) 70,00,000 = 7 x 1000000 = 7 x 106

Question 3
A drop of water of 1.8 gm weight has 60,230,000,000,000,000,000,000 molecules. Express (RBSESolutions.com) it in standard form.
Solution:
Number of molecules in a dot of water of weight 1.8 gm
= 60. 230, 000, 000, 000, 000, 000, 000
= 6.023 x 1000000000000000000000
= 6.023 x 1022

RBSE Solutions

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