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NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Download PDF

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Download PDF

Edited By Ravindra Pindel | Updated on Feb 07, 2024 06:16 PM IST

NCERT Solutions for Exponents And Powers Class 7 - If someone asks you about the speed of bus or car, you may say its 40 km/hr but if you have asked about the speed of light in a vacuum, you will say its approximately 300,000,000 m/s. This is a very big number which is very difficult to write, interpret, and handle. To make any number to represent in an easy way there is a system called exponents. In this article, you will get CBSE NCERT solutions for Class 7 Maths chapter 13 Exponents and Powers.

Exponents can be used to represent very large numbers and small numbers as the power of a base number. In the above example, you can write 300,000,000 m/s as 3\times10^8 . Important topics like laws of exponents, multiplying and dividing the power with the same base, multiplying and dividing the powers with the same exponents and expressing large numbers in the standard form are covered in NCERT. There are many questions from the above topics in the solutions of NCERT for Class 7 Maths chapter 13 Exponents and Powers that are explained in a step-by-step manner. It will be very easy for you to understand the concepts. In this chapter, there are 17 questions in the 3 exercises of the NCERT textbook. You will get detailed explanations of all these questions in the NCERT Solutions for Class 7 . Check NCERT Solutions from Classes 6 to 12 by clicking on the above link. Here you will get solutions to three exercises of this chapter.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Important Formulae

am × an = am+n

am ÷ an = am-n

( am )n = amn

am × bm = (ab)m

am ÷ bn = (a÷b)m

a0 = 1 ( a, non-zero integer)

(-)even number = 1

(-)odd number = -1

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Important Points

Exponents: Exponents represent repeated multiplication of a number (base) by itself.

x4 where x is the base and 4 is the exponent. It is read as x raised to the power of 4 or fourth power of x.

Understanding Powers of 10:

  • Powers of 10 are commonly used in scientific notation and decimal places.
  • 10n represents a 1 followed by 'n' zeros.

Free download NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers PDF for CBSE Exam.

NCERT Solutions for Maths Chapter 13 Exponents and Powers Class 7

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NCERT Solutions for Class 7 Maths Chapter 13 Questions and Exercise

NCERT for Maths Chapter 13 Exponents and Powers class 7 Topic 13.2 Exponents

Question:(i) Express:

729 as a power of 3

Answer:

729 as a power of 3 is given as

729=3\times 3\times 3\times 3\times 3\times 3=3^6

Question:(ii) Express:

128 as a power of 2

Answer:

128 as a power of 2 can be given as

128=2\times 2\times2 \times 2\times 2\times 2\times 2=2^7

Question:(iii) Express:

343 as a power of 7

Answer:

343 as a power of 7 can be givena as

343=7\times 7\times 7=7^3

NCERT Solutions for Maths Chapter 13 Exponents and Powers Class 7Exercise 13.1

Question: 1(i) Find the value of:

(i)2^{6}

Answer:

the value of 2^6 is given by

2\times 2\times2 \times 2\times2 \times 2=64

Question: 1(ii) Find the value of:

(ii)9^{3}

Answer:

the value of (ii)9^{3} is given by

9\times 9\times 9=729

Question:1(iii) Find the value of:

(iii)11^{2}

Answer:

the value of (iii)11^{2} is given by

11\times 11=121

Question: 1(iv) Find the value of:

(iv)5^{4}

Answer:

the value of (iv)5^{4} is given by

5^4=5\times 5\times 5\times 5=625

Question: 2 Express the following in exponential form:

(i)6\times 6\times 6\times 6 (ii)t\times t (iii)b\times b\times b\times b (iv)5\times 5\times 7\times 7\times 7

(v)2\times 2\times a\times a (vi)a\times a\times a\times c\times c\times c\times c\times d

Answer:

(i)6\times 6\times 6\times 6 can be given as

6^4 .

(ii)t\times t can be given as t^2 .

(iii)b\times b\times b\times b can be given as b^4 .

(iv)5\times 5\times 7\times 7\times 7 can be given as

5^2\times 7^3 .

(v)2\times 2\times a\times a can be given as

2^2\times a^2 .

(vi)a\times a\times a\times c\times c\times c\times c\times d can be given as

a^3\times c^4\times d .

Question: 3 Express each of the following numbers using exponential notation:

(i) 512 (ii) 343 (iii) 729 (iv) 3125

Answer:

(i) 512

2
512
2
256
2
128
2
64
2
32
2
16
2
8
2
4
2
2

1

512=2\times 2\times2 \times2 \times 2\times 2\times2 \times 2\times 2=2^9

(ii) 343

7
343
7
49
7
7

1

343=7\times 7\times 7=7^3

(iii)729

3
729
3
243
3
81
3
27
3
9
3
3

1

729=3\times 3\times 3\times 3\times 3\times 3=3^6

(iv)3125

5
3125
5
625
5
125
5
25
5
5

1

3125=5\times 5\times 5\times 5\times 5=5^5

Question: 4 Identify the greater number, wherever possible, in each of the following?

(i)4^{3}or \: 3^{4} (ii) 5^{3}or \: 3^{5} (iii) 2^{8}or \: 8^{2} (iv) 100^{2}or \: 2^{100} (v) 2^{10}or \: 10^{2}

Answer:

(i)4^{3}or \: 3^{4}

4^3=4\times 4\times 4=64

3^4=3\times 3\times 3\times 3=81

since 81> 64

3^4 is greater than 4^3

(ii) 5^{3}or \: 3^{5}

5^3=5\times 5\times 5=125

3^5=3\times 3\times 3\times 3\times 3=243

since 243> 125

3^5 is greater than 5^3

(iii) 2^{8}or \: 8^{2}

2^8=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=256

8^2=8\times 8=64

since 256 > 64

2^8 is greater than 8^2

(iv) 100^{2}or \: 2^{100}

100^2=100\times 100=10000

2^{100}=2\times 2\times 2\times 2\times 2...........till\, 100\, times\, 2

since 2^{100}> 100^{2}

2^{100} is greater than 100^2

(v) 2^{10}or \: 10^{2}

2^1^0=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=1024

10^2=10\times 10=100

since 1024> 100

2^1^0 is greater than 10^2

Question: 5 Express each of the following as product of powers of their prime factors:

(i) 648 (ii) 405 (iii) 540 (iv) 3,600

Answer:

(i) 648

2
648
2
324
2
162
2
81
3
27
3
9
3
3

1

648=2^3\times 3^4

(ii) 405

5
405
3
81
3
27
3
9
3
3

1

405=5\times 3^4

(iii)540

2
540
2
270
3
135
3
45
3
15
5
5

1

540=2^2\times 3^3\times 5

(iv) 3600

2
3600
2
1800
2
900
2
450
3
225
3
75
5
25
5
5

1

3600=2^4\times 3^2\times 5 ^2


Question: 6(i) Simplify:

(i)2\times 10^{3}

Answer:

(i)2\times 10^{3}

can be simplified as

2\times 10\times 10\times 10=2000

Question: 6(ii) Simplify:

(ii)7^{2}\times 2^{2}

Answer:

(ii)7^{2}\times 2^{2}

can be simplified as

7\times 7\times 2\times 2=196

Question: 6(iii) Simplify:

(iii) 2^{3}\times 5

Answer:

(iii) 2^{3}\times 5

can be simplified as

2\times 2\times 2\times 5=40

Question: 6(iv) Simplify:

(iv) 3\times 4^{4}

Answer:

(iv) 3\times 4^{4}

can be simplified as

3 \times 4\times 4\times 4\times 4=768

Question: 6(v) Simplify:

(v)0\times 10^{2}

Answer:

(v)0\times 10^{2}

can be simplified as

0\times 10\times 10=0

Question: 6(vi) Simplify:

(vi)5^{2}\times 3^{3}

Answer:

(vi)5^{2}\times 3^{3}

can be simplified as

5\times 5\times 3\times3 \times 3=675

Question: 6(vii) Simplify:

(vii) 2^{4}\times 3^{2}

Answer:

(vii) 2^{4}\times 3^{2}

can be simplified as

2\times 2\times 2\times 2\times 3\times 3=144

Question: 6(v iii) Simplify:

(viii)3^{2}\times 10^{4}

Answer:

(viii)3^{2}\times 10^{4}

can be simplified as

3\times3 \times 10\times 10\times10 \times 10=90000

Question: 7(i) Simplify:

(i)(-4)^{3}

Answer:

(i)(-4)^{3}

can be simplified as

-4\times -4\times -4=-64

Question: 7(ii) Simplify:

(ii) (-3)\times (-2)^{3}

Answer:

(ii) (-3)\times (-2)^{3}

can be simplified as

-3\times -2\times -2\times -2=24

Question: 7(iii) Simplify:

(iii) (-3)^{2}\times (-5)^{2}

Answer:

(iii) (-3)^{2}\times (-5)^{2}

can be simplified as

(-3)\times (-3)\times (-5)\times (-5)=225

Question: 7(iv) Simplify:

(iv) (-2)^{3}\times (-10)^{3}

Answer:

(iv) (-2)^{3}\times (-10)^{3}

can be simplified as

(-2)\times (-2)\times (-2)\times- 10\times -10\times -10=8000

Question: 8 Compare the following numbers:

(i) 2.7\times 10^{12} ;1.5\times 10^{8} (ii) 4\times 10^{14} ;3\times 10^{17}

Answer:

(i) 2.7\times 10^{12} ;1.5\times 10^{8}

on comparing exponents of base 10.

2.7\times 10^{12} > 1.5\times 10^{8}

(ii) 4\times 10^{14} ;3\times 10^{17}

on comparing exponents of base 10.

4\times 10^{14} < 3\times 10^{17}

NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.1

Question:(i) Simplify and write in exponential form:

(i)2^{5}\times 2^{3}

Answer:

(i)2^{5}\times 2^{3}

can be simplified as

2^{5+3}=2^8

Question:(ii) Simplify and write in exponential form:

(ii) p^{3}\times p^{2}

Answer:

(ii) p^{3}\times p^{2}

can be simplified as

p(3+2)=p^5

Question:(iii) Simplify and write in exponential form:

(iii)\: 4^{3}\times 4^{2}

Answer:

(iii)\: 4^{3}\times 4^{2}

can be simplified as

4^{(3+2)}=4^5

Question:(iv) Simplify and write in exponential form:

(iv)\: a^{3}\times a^{2}\times a^{7}

Answer:

(iv)\: a^{3}\times a^{2}\times a^{7}

can be simplified as

a^{(3+2+7)}=a^1^2

Question:(v) Simplify and write in exponential form:

(v)\: 5^{3}\times 5^{7}\times 5^{12}

Answer:

(v)\: 5^{3}\times 5^{7}\times 5^{12}

can be simplified as

5^{(3+7+12)}=5^2^2

Question:(vi) Simplify and write in exponential form:

(vi)\: (-4)^{100}\times (-4)^{20}

Answer:

(vi)\: (-4)^{100}\times (-4)^{20}

can be simplified as

(-4)^{(100+20)}=(-4)^{120}

NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.2

Question: 1 Simplify and write in exponential form: (eg., 11^{6}\div 11^{2}= 11^{4} )

(i)\: 2^{9}\div 2^{3} (ii)\: 10^{8}\div 10^{4} (iii)\: 9^{11}\div 9^{7} (iv)\: 20^{15}\div 20^{13} (v)\: 7^{13}\div 7^{10}

Answer:

(i)\: 2^{9}\div 2^{3}

can be simplified as 2^{9-3}=2^6

(ii)\: 10^{8}\div 10^{4}

can be simplified as 10^{(8-4)}=10^4

(iii)\: 9^{11}\div 9^{7}

can be simplified as 9^{(11-7)}=9^4

(iv)\: 20^{15}\div 20^{13}

can be simplified as 20^{(15-13)}=20^2

(v)\: 7^{13}\div 7^{10}

can be simplified as 7^{(13-10)}=7^3

NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.3

Question: Simplify and write the answer in exponential form:

(i)\: (6^{2})^{4} (ii)\: (2^{2})^{100} (iii)\: (7^{50})^{2} (iv)\: (5^{3})^{7}

Answer:

(i)\: (6^{2})^{4}

can be simplified as 6^{(2\times 4)}=6^8

(ii)\: (2^{2})^{100}

can be simplified as 2^{(2\times 100)}=2^{200}

(iii)\: (7^{50})^{2}

can be simplified as 7^{(50\times 2)}=7^{100}

(iv)\: (5^{3})^{7}

can be simplified as 5^{(3\times 7)}=5^{21}

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Topic 13.3.4

Question: Put into another form using a^{m}\times b^{m}= (ab)^{m} :

(i)\: 4^{3}\times 2^{3} (ii)\: 2^{5}\times b^{5} (iii)\: a^{2}\times t^{2} (iv)\: 5^{6}\times (-2)^{6} (v)\: (-2)^{4}\times (-3)^{4}

Answer:

(i)\: 4^{3}\times 2^{3}

can be simplified as (4\times 2)^3=8^3

(ii)\: 2^{5}\times b^{5}

can be simplified as (2\times b)^5=(2b)^5

(iii)\: a^{2}\times t^{2}

can be simplified as (a\times t)^2=at^2

(iv)\: 5^{6}\times (-2)^{6}

can be simplified as (5\times (-2))^6=-10^6

(v)\: (-2)^{4}\times (-3)^{4}

can be simplified as ((-2)\times (-3))^4= 6^4

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Topic 13.3.5

Question: Put into another form using a^{m}\div b^{m}= (\frac{a}{b})^{m} :

(i)\: 4^{5}\div 3^{5} (ii)\: 2^{5}\div b^{5} (iii)\: (-2)^{3}\div b^{3} (iv)\: p^{4}\div q^{4} (v)\: 5^{6}\div (-2)^{6}

Answer:

(i)\: 4^{5}\div 3^{5}

can be simplified as

\left ( \frac{4}{3} \right )^{5}

(ii)\: 2^{5}\div b^{5}

can be simplified as

\left ( \frac{2}{b} \right )^{5}

(iii)\: (-2)^{3}\div b^{3}

can be simplified as

\left ( \frac{-2}{b} \right )^{3}

(iv)\: p^{4}\div q^{4}

can be simplified as

\left ( \frac{p}{q} \right )^{4}

(v)\: 5^{6}\div (-2)^{6}

can be simplified as

\left ( \frac{5}{-2} \right )^{6}

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2

Question: 1 Using laws of exponents, simplify and write the answer in exponential form:

(i)\: 3^{2}\times 3^{4}\times 3^{8} (ii)\: 6^{15}\div 6^{10} (iii)\: a^{3}\times a^{2} (iv)\: 7^{x}\times 7^{2}

(v)\:(5^{2})^{3}\div 5^{3} (vi)\:2^{5}\times 5^{5} (vii)\:a^{4}\times b^{4} (viii)\:(3^{4})^{3}

(ix)\:(2^{20}\div 2^{15})\times 2^{3} (x)\:8^{t}\div 8^{2}

Answer:

(i)\: 3^{2}\times 3^{4}\times 3^{8}

can be simplified as 3^{(2+4+8)}=3^{14}

(ii)\: 6^{15}\div 6^{10}

can be simplified as 6^{(15-10)}=6^{5}

(iii)\: a^{3}\times a^{2}

can be simplified as a^{(3+2)}=a^{5}

(iv)\: 7^{x}\times 7^{2}

can be simplified as 7^{(x+2)}=7^{(x+2)}

(v)\:(5^{2})^{3}\div 5^{3}

can be simplified as 5^{(2\times 3)}\div 5^{(3)}=5^6\div 5^3=5^{6-3}=5^3

(vi)\:2^{5}\times 5^{5}

can be simplified as

(2\times5)^{5}=10^5

(vii)\:a^{4}\times b^{4}

can be simplified as (ab)^4

(viii)\:(3^{4})^{3}

can be simplified as 3^{4\times 3}=3^{12}

(ix)\:(2^{20}\div 2^{15})\times 2^{3}

can be simplified as 2^{(20-15)}\times 2^3=2^5\times 2^3=2^{(5+3)}=2^8

(x)\:8^{t}\div 8^{2}

can be simplified as 8^{(t-2)}

Question: 2(ii) Simplify and express each of the following in exponential form:

(ii)\: ((5^{2})^{3}\times 5^{4})\div 5^{7}

Answer:

(ii)\: ((5^{2})^{3}\times 5^{4})\div 5^{7}

can be simplified as

=[5^{(2\times 3)}\times 5^4]\div 5^7

=[5^{6}\times 5^4]\div 5^7

=[5^{(6+4)}]\div 5^7

=[5^{10}]\div 5^7

=5^{10-7}

=5^3

Question: 2(iii) Simplify and express each of the following in exponential form:

(iii)\: 25^{4}\div 5^{3}

Answer:

(iii)\: 25^{4}\div 5^{3}

can be simplified as

=(5^2)^4\div 5^3

=5^{(2\times 4)}\div 5^3

=5^8\div 5^3

=5^{(8-3)}

=5^{5}

Question: 2(iv) Simplify and express each of the following in exponential form:

(iv)\: \frac{3\times 7^{2}\times 11^{8}}{21\times 11^{3}}

Answer:

(iv)\: \frac{3\times 7^{2}\times 11^{8}}{21\times 11^{3}}

can be simplified as

= \frac{3\times 7^{2}\times 11^{8}}{3\times 7\times 11^{3}}

=3^{(1-1)}\times 7^{(2-1)}\times 11^{(8-3)}

=3^{0}\times 7^{1}\times 11^{5}

=7^{1}\times 11^{5}

Question: 2(v) Simplify and express each of the following in exponential form:

(v)\: \frac{3^{7}}{3^{4}\times 3^{3}}

Answer:

(v)\: \frac{3^{7}}{3^{4}\times 3^{3}}

can be simplified as

=\frac{3^7}{3^{4+3}}

=\frac{3^7}{3^{7}}

=3^{(7-7)}

=3^0

=1

Question: 2(vi) Simplify and express each of the following in exponential form:

(vi)\: 2^{0}+3^{0}+4^{0}

Answer:

(vi)\: 2^{0}+3^{0}+4^{0}

can be simplified as

=1+1+1

=3

Question: 2(vii) Simplify and express each of the following in exponential form:

(vii)\: 2^{0}\times 3^{0}\times 4^{0}

Answer:

(vii)\: 2^{0}\times 3^{0}\times 4^{0}

can be simplified as

=1\times 1\times 1=1

Question: 2(viii) Simplify and express each of the following in exponential form:

(viii)\: (3^{0}+2^{0})\times 5^0

Answer:

(viii)\: (3^{0}+2^{0})\times 5^0

can be simplified as

=(1+1)\times 1

=2\times 1

=2

Question: 2(ix) Simplify and express each of the following in exponential form:

(ix)\: \frac{2^{8}\times a^{5}}{4^{3}\times a^{3}}

Answer:

(ix)\: \frac{2^{8}\times a^{5}}{4^{3}\times a^{3}}

can be simplified as

=\frac{2^{8}\times a^{5}}{(2^2)^{3}\times a^{3}}

=\frac{2^{8}\times a^{5}}{2^{6}\times a^{3}}

=2^{(8-6)}\times a^{(5-3)}

=2^{(2)}\times a^{(2)}

=(2a)^2

Question: 2(x) Simplify and express each of the following in exponential form:

(x)\: (\frac{a^{5}}{a^{3}})\times a^{8}

Answer:

(x)\: (\frac{a^{5}}{a^{3}})\times a^{8}

can be simplified as

=a^{(5-3)}\times a^8

=a^{(2)}\times a^8

=a^{(2+8)}

=a^{(10)}

Question: 2(xi) Simplify and express each of the following in exponential form:

(xi)\: \frac{4^{5}\times a^{8}b^{3}}{4^{5}\times a^{5}b^{2}}

Answer:

(xi)\: \frac{4^{5}\times a^{8}b^{3}}{4^{5}\times a^{5}b^{2}}

can be simplified as

=4^{(5-5)}\times a^{(8-5)}\times b^{(3-2)}

=4^{0}\times a^{3}\times b^{1}

= a^{3}\times b

Question:2(xii) Simplify and express each of the following in exponential form:

(xii)\: (2^{3}\times 2)^{2}

Answer:

(xii)\: (2^{3}\times 2)^{2}

can be simplified as

=(2^{3+1})^{2}

=(2^{(4)})^2

=2^{(4 \times 2)}

=2^8

Question: 3 Say true or false and justify your answer:

(i)\: 10\times 10^{11}=100^{11} (ii)\: 2^{3}> 5^{2} (iii)\: 2^{3}\times 3^{2}= 6^{5} (iv)\: 3^{0}=(1000)^{0}

Answer:

(i)\: 10\times 10^{11}=100^{11}

can be simplified as

LHS:10^{(1+11)}

=10^{(12)}

Since , LHS \neq RHS

Thus, it is false

(ii)\: 2^{3}> 5^{2}

can be simplified as

LHS=2^3=8

RHS=5^2=25

Since, LHS\ngtr RHS

Thus, it is false

(iii)\: 2^{3}\times 3^{2}= 6^{5}

can be simplified as

LHS:2^3\times 3^2=8\times 9=72

RHS: 6^5=7776

Since , LHS \neq RHS

Thus, it is false

(iv)\: 3^{0}=(1000)^{0}

can be simplified as

LHS: 3^0=1

RHS:1000^0=1

Since, LHS = RHS

Thus, it is true.

Question: 4 Express each of the following as a product of prime factors only in exponential form:

(i)\: 108\times 192 (ii)\: 270 (iii)\: 729\times 64 (iv)\: 768

Answer:

(i)\: 108\times 192

2
108
2
54
3
27
3
9
3
3

1


2
192
2
96
2
48
2
24
2
12
2
6
3
3

1


108\times 192=(2^2\times 3^3)\times (2^6\times 3)

=2^{(2+6)}\times 3^{(3+1)}

=2^8\times 3^4


(ii)\: 270

2
270
3
135
3
45
3
15
5
5

1

270=2\times 3^3\times 5


(iii)\: 729\times 64

3
729
3
243
3
81
3
27
3
9
3
3

1


2
64
2
32
2
16
2
8
2
4
2
2

1

729\times 64=3^6\times 2^6


(iv)\: 768

2
768
2
384
2
192
2
96
2
48
2
24
2
12
2
6
3
3

1


768=2^8\times 3

Question: 5(i) Simplify:

(i)\: \frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}

Answer:

(i)\: \frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}

can be simplified as

= \frac{2^{(5\times 2)}\times 7^{3}}{(2^3)^{3}\times 7}

= \frac{2^{10}\times 7^{3}}{(2^{(3\times 3)})\times 7}

= \frac{2^{10}\times 7^{3}}{(2^{9})\times 7}

=2^{(10-9)}\times 7^{(3-1)}

=2\times 7^{2}

=2\times49=98

Question: 5(ii) Simplify:

(ii)\frac{25\times 5^{2}\times t^{8}}{10^{3}\times t^{4}}

Answer:

(ii)\frac{25\times 5^{2}\times t^{8}}{10^{3}\times t^{4}}

can be simplified as

=\frac{5^2\times 5^{2}\times t^{8}}{(2\times 5)^{3}\times t^{4}}

=\frac{5^{(2+2)}\times t^{8}}{2^3\times 5^{3}\times t^{4}}

=\frac{5^{4}\times t^{8}}{2^3\times 5^{3}\times t^{4}}

=\frac{5^{4-3}\times t^{(8-4)}}{2^3}

=\frac{5\times t^{4}}{2^3}

=\frac{5 t^{4}}{8}

Question: 5(iii) Simplify:

(iii)\: \frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}

Answer:

(iii)\: \frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}

can be simplified as

= \frac{3^{5}\times (2\times 5)^{5}\times 5^2}{5^{7}\times (2\times 3)^{5}}

= \frac{3^{5}\times 2^5 \times 5^{5}\times 5^2}{5^{7}\times 2^5 \times 3^{5}}

= 3^{(5-5)} \times 2^{(5-5)} \times 5^{(5+2-7)}

= 3^{0} \times 2^{0} \times 5^{0}

= 1 \times 1 \times 1=1

NCERT Solutions for Chapter 13 Maths Class 7 Exponents and Powers Topic 13.6

Question: Expand by expressing powers of 10 in the exponential form:

(i) 172 (ii) 5,643 (iii) 56,439 (iv) 1,76,428

Answer:

(i) 172

172=100+70+2

=100+(7\times 10)+2

=1\times 10^2+(7\times 10^1)+2\times 10^0

(ii) 5,643

5643=5000+600+40+3

=5\times 1000+6\times 100+4\times 10+3

=5\times 10^3+6\times 10^2+4\times 10^1+3\times 10^0

(iii) 56,439

56439=50000+6000+400+30+9

=5\times 10000+6\times 1000+4\times 100+3\times 10+9

=5\times 10^4+6\times 10^3+4\times 10^2+3\times 10^1+9\times 10^0

(iv) 1,76,428

176428=100000+70000+6000+400+20+8

=1\times 100000+7\times 10000+6\times 1000+4\times 100+2\times 10+8

=1\times 10^5+7\times 10^4+6\times 10^3+4\times 10^2+2\times 10^1+8\times 10^0

NCERT Solutions for Chapter 13 Maths Class 7 Exponents and Powers Exercise 13.3

Question: 1 Write the following numbers in the expanded forms:

279404, 3006194, 2806196, 120719, 20068

Answer:

(i) 279404

279404=200000+70000+9000+400+00+4

=2\times 100000+7\times 10000+9\times 1000+4\times 100+0\times 10+4\times 1

=2\times 10^5+7\times 10^4+9\times 10^3+4\times 10^2+0\times 10^1+4\times 10^0


(ii) 3006194

3006194=3000000+0+0+6000+100+90+4

=3\times 10^6+0\times 10^5+0\times 10^4+6\times 10^3+1\times 10^2+9\times 10^1+4\times 10^0

(iii) 2806196

2806196=2000000+800000+0+6000+100+90+6

=2\times 1000000+8\times 100000+0\times 10000+6\times 1000+1\times 100+9\times 10+6\times 1

=2\times 10^6+8\times 10^5+0\times 10^4+6\times 10^3+1\times 10^2+9\times 10^1+6\times 10^0


(IV) 120719

120719=100000+20000+0+700+10+9

=1\times 100000+2\times 10000+0\times 1000+7\times 100+1\times 10+9\times 1 =1\times 10^5+2\times 10^4+0\times 10^3+7\times 10^2+1\times 10^1+9\times 10^0

(V) 20068

20068=20000+0+0+60+8

=2\times 10000+0\times 1000+0\times 100+6\times 10+8\times 1

=2\times 10^4+0\times 10^3+0\times 10^2+6\times 10^1+8\times 10^0

Question: 2 Find the number from each of the following expanded forms:

(a)\: 8\times 10^{4}+6\times 10^{3}+0\times 10^{2}+4\times 10^{1}+5\times 10^{0}

(b)\: 4\times 10^{5}+5\times 10^{3}+3\times 10^{2}+2\times 10^{0}

(c)\: 3\times 10^{4}+7\times 10^{2}+5\times 10^{0}

(d)\: 9\times 10^{5}+2\times 10^{2}+3\times 10^{1}

Answer:

(a)\: 8\times 10^{4}+6\times 10^{3}+0\times 10^{2}+4\times 10^{1}+5\times 10^{0}

=8\times 10000+6\times 1000+0\times 100+4\times 10+5\times 1

=80000+6000+000+40+5

=86045

(b)\: 4\times 10^{5}+5\times 10^{3}+3\times 10^{2}+2\times 10^{0}

=4\times 100000+0\times 10000+5\times 1000+3\times 100+0\times 10+2\times 1

=400000+00000+5000+300+00+2

=405302

(c)\: 3\times 10^{4}+7\times 10^{2}+5\times 10^{0}

=3\times 10000+0\times 1000+7\times 100+0\times 10+5\times 1

=30000+0000+700+00+5

=30705

(d)\: 9\times 10^{5}+2\times 10^{2}+3\times 10^{1}

=9\times 100000+0\times 10000+0\times 1000+2\times 100+3\times 10+0\times 1

=900000+00000+0000+200+30+0

=900230

Thus, the above problems are simplified in simpler forms.

Question: 3 Express the following numbers in standard form:

(i) 5,00,00,000 (ii) 70,00,000 (iii ) 3,18,65,00,000 (iv) 3,90,878 (v) 39087.8 (vi) 3908.78

Answer:

(i) 5,00,00,000

50000000=5\times 10000000=5\times 10^7

(ii) 70,00,000

7000000=7\times 1000000=7\times 10^6

(iii ) 3,18,65,00,000

3186500000=31865\times 100000

=3.1865\times 10000 \times 100000

=3.1865\times 10^9

(iv) 3,90,878

=3.90878\times 100000

=3.90878\times 10^5

(v) 39087.8

=3.90878\times 10000

=3.90878\times 10^4

(vi) 3908.78

=3.90878\times 1000

=3.90878\times 10^3

Question: 4 Express the number appearing in the following statements in standard form.

(a) The distance between Earth and Moon is 384,000,000 m.

(b) Speed of light in vacuum is 300,000,000 m/s.

(c) Diameter of the Earth is 1,27,56,000 m.

(d) Diameter of the Sun is 1,400,000,000 m.

(e) In a galaxy there are on an average 100,000,000,000 stars.

(f) The universe is estimated to be about 12,000,000,000 years old.

(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

(i) The earth has 1,353,000,000 cubic km of sea water.

(j) The population of India was about 1,027,000,000 in March, 2001.

Answer:

(a) The distance between Earth and Moon = 384,000,000 m

=384\times 1000000

=3.84 \times 100\times 1000000

=3.84 \times 10^8m

(b) Speed of light in vacuum =300,000,000 m/s.

=3\times 100000000

=3\times 10^8

(c) Diameter of the Earth = 1,27,56,000 m.

=12756\times 1000

=1.2756\times 10000\times 1000

=1.2756\times 10^7m

(d) Diameter of the Sun = 1,400,000,000 m.

=14\times 100000000

=14\times 10^8m

=1.4\times 10^9m

(e) In a galaxy there are on an average = 100,000,000,000 stars.

=1\times 100000000000

=1\times 10^{11}

(f) The universe is estimated to be about= 12,000,000,000 years old.

=1.2\times 10000000000

=1.2\times 10^{10}years


(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated = 300,000,000,000,000,000,000 m.

=3\times 100000000000000000000000000000

=3 \times 10^{19}

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

60,230,000,000,000,000,000,000

= 6023\times 10000,000,000,000,000,000

= 6023\times 10^{19}

(i) The earth has 1,353,000,000 cubic km of seawater.

=1.353\times 1000000000

=1.353\times 10^9 km^3

(j) The population of India was about 1,027,000,000 in March 2001.

=1.027\times 1000000000

=1.027\times 10^9

Exponents and Powers Class 7 Maths Chapter 13- Topics

  • Exponents
  • Laws Of Exponents
  • Multiplying Powers With The Same Base
  • Dividing Powers With The Same Base
  • Taking Power Of A Power
  • Multiplying Powers With The Same Exponents
  • Dividing Powers With The Same Exponents
  • Miscellaneous Examples Using The Laws Of Exponents
  • Decimal Number System
  • Expressing Large Numbers In The Standard Form

NCERT Solutions for Class 7 Maths Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Integers

Chapter 2

Fractions and Decimals

Chapter 3

Data Handling

Chapter 4

Simple Equations

Chapter 5

Lines and Angles

Chapter 6

The Triangle and its Properties

Chapter 7

Congruence of Triangles

Chapter 8

Chapter 9

Rational Numbers

Chapter 10

Practical Geometry

Chapter 11

Perimeter and Area

Chapter 12

Algebraic Expressions

Chapter 13

Exponents and Powers

Chapter 14

Symmetry

Chapter 15

NCERT Solutions for Class 7 Subject Wise

Some Important Properties and Formulas from NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

For any non-zero integers, a and b and whole numbers m and n it obey certain properties given below

  • \ a^m\times a^n=a^{m+n}
  • \frac{a^m}{a^n}=a^{m-n}
  • \ (a^m)^n=a^{mn}
  • a^{m} \times b^{m}=(a b)^{m}
  • a^{m} \div b^{m}=(\frac{a}{b})^m
  • a^{0}=1
  • (-1)^{\text {even number }}=1
  • (-1)^{\operatorname{add} \operatorname{number}}=-1

Tip- If you understood the fundamental of this chapter, you don't need to remember above these findings. You can simply derive these properties by the strong fundamentals of this chapter. You should try to solve all the NCERT questions including the practice questions given at the end of every topic. You can take help from the CBSE NCERT solution for class 7 maths chapter 13 exponents and powers if you are facing difficulties while solving the problems.

Happy Reading!!!

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What is exponents?

Exponent is a method expressing large numbers in terms of powers.

For example; 2 multiplied 3 times i.e, 2×2×2 can be expressed as 2^3 so where 3 exponents of 2. To get deeper understanding of these concepts students can download exponents and powers class 7 pdf and study both onlline and offline mode. For Hindi students we are working to create class 7 hindi chapter 13 pdf which will be available soon. 

2. What are the topics covered in NCERT Class 7 Maths Chapter13?

Here are the topics covered in NCERT Class 7 Maths chapter13

  • Exponents   
  • Laws Of Exponents   
  • Multiplying Powers With The Same Base   
  • Dividing Powers With The Same Base   
  • Taking Power Of A Power   
  • Multiplying Powers With The Same Exponents   
  • Dividing Powers With The Same Exponents   
  • Miscellaneous Examples Using The Laws Of Exponents   
  • Decimal Number System   
  • Expressing Large Numbers In The Standard Form  

Hindi students can download class 7th hindi chapter 13 question answer pdf soon. 

3. How many exercises in NCERT Class 7 Maths chapter 13?

There are 3 exercises in NCERT Class 7 Maths chapter 13

NCERT Exponents and Powers class 7 Exercise 13.1 - 8 Questions

NCERT Exponents and Powers class 7 Exercise 13.2 - 5 Questions

NCERT Exponents and Powers class 7 Exercise 13.3 - 4 Questions

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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