# NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

## NCERT Solutions For Class 7 Maths Chapter 13 PDF Free Download

NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, includes solutions for all the questions provided in the class 7 maths textbook. It is important for the students to practice various questions from this book as it will help them to develop a thorough understanding of the concepts.

While solving the questions from NCERT books, students often find some doubts and eventually end up piling those. All the NCERT solutions provided here can help the students to clear all their doubts instantly and help them to have an effective preparation.

## Class 7 Maths NCERT Solutions – Chapter 13 Exponents and Powers

In the exponents and powers chapter, some of the most crucial concepts are covered which forms the basis of many mathematical concepts and students need to be thorough with all the concepts to comprehend many related topics in the higher level of education.

The exponents and powers chapter covers topics such as;

• 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
• Numbers with exponent zero
• Decimal Number System
• Expressing Large Numbers In The Standard Form

Students can also do their preparation by solving sample papers and previous year question papers to get an idea of question pattern asked from chapter 13 of class 7. Click here to get question papers of NCERT class 7.

### NCERT Solutions For Class 7 Maths Chapter 13 Exercise 13.1

Q.1.Find the value of:

(i) $2^{6}$

(ii) $9^{3}$

(iii) $11^{2}$

(iv) $5^{4}$

Solution:

(i)$2^{6}=2\times 2\times 2\times 2\times2\times 2=64$

(ii)$9^{3}=9\times 9\times 9=729$

(iii)$11^{2}=11\times 11=121$

(iv) $5^{4}=5\times 5\times 5\times 5=625$

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Q.2. Express the following in exponential form:

(i) $6 \times 6 \times 6 \times 6$

Solution: $6 \times 6 \times 6 \times 6= 6^{4}$

(ii) $t \times t$

Solution: $t \times t= t^{2}$

(iii) $b \times b \times b\times b$

Solution: $b \times b \times b\times b= b^{4}$

(iv) $5 \times 5\times 7 \times 7\times 7$

Solution: $5 \times 5\times 7 \times 7\times 7 = 5^{2}\times 7^{3}$

(v) $2 \times 2\times a \times a$

Solution: $2 \times 2\times a \times a = 2^{2}\times a^{2}$

(vi) $a \times a\times a \times c\times c\times c\times c \times d$

Solution: $a \times a\times a \times c\times c\times c\times c \times d =a^{3}\times c^{4}\times d$

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Q.3. Express each of the following numbers using exponential notation:

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution:

(i) 512

$512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^{9}$

(ii)343

$343=7\times 7\times 7=7^{3}$

(iii)729

$729=3\times 3\times 3\times 3\times 3\times3=3^{6}$

(iv)3125

$3125=5\times 5\times 5\times 5\times 5=5^{5}$

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Q.4. Identify the greater number, wherever possible, in each of the following:

(i) $4^{3} \: and \: 3^{4}$

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

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

(iv) $100^{2} \: and \: 2^{100}$

Solution:

(i)$4^{3} =4\times 4\times 4=64\\ 3^{4}=3\times 3\times 3\times3=81$

Since 64<81

So, $3^{4}$ is greater than $4^{3}$.

(ii)$5^{3} =5\times 5\times 5=125\\ 3^{5}=3\times 3\times 3\times 3\times 3=243$

Since 125<243

So, $3^{5}$ is greater than $5^{3}$.

(iii)$2^{8} =2\times 2\times 2 \times 2\times 2\times 2\times 2\times 2=256\\ 8^{2}=8\times 8=64$

Since 64<256

So, $2^{8}$ is greater than $8^{2}$.

(iv) $100^{2}=100\times 100=10,000\\ 2^{100}=2\times 2\times 2\times 2\times 2 \times……..14\: times\:\times ……….\times 2=16,384\times ………..\times 2$

So, $2^{100}$ is greater than $100^{2}$.

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

(i) 648

(ii) 405

(iii) 540

Solution:

(i)648

$648=2^{3}\times 3^{4}$

(ii)405

$405=5\times 3^{4}$

(iii)540

$540=2^{2}\times 3^{3}\times 5$

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Q.6. Simplify:

(i) $2 \times 10^{3}$

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

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

(iv)$3 \times 4^{4}$

(v)$0 \times 10^{2}$

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

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

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

Solution:

(i)$2 \times 10^{3}=2\times 10\times 10\times 10=2000$

(ii)$7^{2}\times2^{2}=7\times 7\times 2\times 2=196$

(iii)$2^{3}\times5=2\times 2\times 2\times 5=40$

(iv)$3\times4^{4}=3\times 4\times 4\times 4\times 4=768$

(v)$0\times10^{2}=0\times 10\times 10=0$

(vi)$5^{3}\times3^{3}=5\times 5\times 5\times 3\times 3\times 3=675$

(vii)$2^{4}\times3^{2}=2\times 2\times 2\times 2\times 3\times 3=144$

(viii)$3^{2}\times10^{4}=3\times 3\times 10\times 10\times 10\times 10=90,000$

Q.7.Simplify:

(i)$(-4)^{3}$

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

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

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

Solution:

(i)$(-4)^{3}=(-4)\times (-4)\times (-4)=-64$

(ii)$(-3)\times (-2)^{3}=(-3)\times (-2)\times (-2)\times (-2)=24$

(iii)$(-3)^{2}\times (-5)^{2}=(-3)\times (-3)\times (-5)\times (-5)=225$

(iv)$(-2)^{3}\times (-10)^{3}=(-2)\times (-2)\times (-2)\times (-10)\times (-10)\times (-10)=8000$

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Q.8.Compare the following numbers:

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

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

Solution:

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

On comparing the 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 the exponents of base 10,

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

### NCERT Solutions For Class 7 Maths Chapter 13 Exercises 13.2

Q.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)$(5^{2})^{2} \div 5^{3}$

(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)$8^{t}\div 8^{2}$

Solution:

(i)$3^{2}\times 3^{4}\times 3^{8}=3^{(2+4+8)}=3^{14}$

(ii)$6^{15}\div 6^{10}=6^{15-10}=6^{5}$

(iii)$a^{3}\times a^{2}=a^{3+2}=a^{5}$

(iv)$7^{x}\times 7^{2}=7^{x+2}$

(v)$(5^{2})^{2} \div 5^{3}=5^{2\times 3}\div 5^{3}=5^{6}\div 5^{3}=5^{6-3}=5^{3}$

(vi)$2^{5} \times 5^{5}=(2\times 5)^{5}=10^{5}$

(vii)$a^{4}\times b^{4}=(a\times b)^{4}$

(viii)$(3^{4})^{3}=3^{4\times 3}=3^{12}$

(ix)$(2^{20}\div 2^{15})\times 2^{3}=(2^{20-15})\times 2^{3}=2^{5}\times 2^{3}=2^{5+3}=2^{8}$

(x) $8^{t}\div 8^{2}=8^{t-2}$

Formulae:

(i)$a^{m}\times a^{n}=a^{m+n}$

(ii)$a^{m}\div a^{n}=a^{m-n}$

(iii)$(a^{m})^{n}=a^{m\times n}$

(iv)$a^{m}\times b^{m}=(a\times b)^{m}$

Q.2.(i)$\frac{2^{3}\times 3^{4}\times 4}{3\times 32}$

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

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

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

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

(vi)$2^{0}+3^{0}+4^{0}$

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

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

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

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

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

(xii)$(2^{3}\times2)^{2}$

Solution:

(i)$\frac{2^{3}\times 3^{4}\times 4}{3\times 32}=\frac{2^{3}\times 3^{4}\times 2^{2}}{3\times 2^{5}}=\frac{2^{3+2}\times 3^{4}}{3\times 2^{5}}\\ =\frac{2^{5}\times 3^{4}}{3\times 2^{5}}=2^{5-5}\times 3^{4-3}\\ =2^{0}\times 3^{3}=1\times 3^{3}=3^{3}$

(ii)$[(5^{2})^{3}\times 5^{4}]\div 5^{7}=[5^{6}\times 5^{4}]\div 5^{7}\\ =[5^{6+4}]\div 5^{7}\\ =5^{10}\div 5^{7}\\ =5^{10-7}=5^{3}$

(iii)$25^{4}\div 5^{3}=(5^{2})^{4}\div 5^{3}\\ =5^{8}\div 5^{3}\\ =5^{8-3}=5^{5}$

(iv) $\frac{3\times 7^{2}\times 11^{8}}{21\times 11^{3}}=\frac{3\times 7^{2}\times 11^{8}}{3\times 7\times 11^{3}}\\ =3^{1-1}\times 7^{2-1}\times 11^{8-3}\\ =3^{0}\times 7^{1}\times 11^{5}=7\times 11^{5}$

(v)$\frac{3^{7}}{3^{4}\times 3^{3}}=\frac{3^{7}}{3^{4+3}}=\frac{3^{7}}{3^{7}}\\ =3^{7-7}=3^{0}=1$

(vi)$2^{0}+3^{0}+4^{0}=1+1+1=3$

(vii)$2^{0}\times 3^{0}\times 4^{0}=1\times 1\times 1=1$

(viii)$3^{0}+2^{0}\times 5^{0}=(1+1)\times 1=2\times 1=2$

(ix)$\frac{2^{8}\times a^{5}}{4^{3}\times a^{3}}=\frac{2^{8}\times a^{5}}{(2^{2})^{3}\times a^{3}} =\frac{2^{8}\times a^{5}}{2^{6}\times a^{3}}\\ =2^{8-6}\times a^{5-2}=2^{2}\times a^{2}\\ =(2a)^{2}$

(x)$(\frac{a^{5}}{a^{3}})\times a^{8}=(a^{5-3})\times a^{8}=a^{2}\times a^{8}\\ =a^{2+8}=a^{10}$

(xi)$\frac{4^{5}\times a^{8}b^{3}}{4^{5}\times a^{5}b^{2}}=4^{5-5}\times a^{8-5}\times b^{3-2}=4^{0}\times a^{3}\times b\\ =1\times a^{3}\times b=a^{3}b$

(xii)$(2^{3}\times2)^{2}=(2^{3+1})^{2}=(2^{4})^{2}\\ =2^{4\times 2}=2^{8}$

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

(ii)$2^{3} > 5^{2}$

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

(iv)$3^{0} = (1000)^{0}$

Solution:

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

L.H.S. Â $10^{1+11}=10^{12}$

And

R.H.S. $(10^{2})^{11}=10^{22}$

Since, $L.H.S.\neq R.H.S.$

Therefore, it is false.

(ii)$2^{3} > 5^{2}$

L.H.S. $2^{3}=8$

And

R.H.S. $5^{2}=25$

Since, L.H.S. is not greater than R.H.S.

Therefore, it is false.

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

L.H.S. $2^{3}\times 3^{2}=8\times 9=72$

And

R.H.S. $6^{5}=7,776$

Since, $L.H.S.\neq R.H.S.$

Therefore, it is false.

(iv)$3^{0} = (1000)^{0}$

L.H.S. $3^{0}=1$

And

R.H.S. $(1000)^{0}=1$

Since, L.H.S. = R.H.S.

Therefore, it is true.

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Q.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

Solution:

(i)$108 \times 192$ $108 \times 192 =(2^{2}\times 3^{3})\times (2^{6}\times 3)\\ =2^{2+6}\times 3^{3+1}\\ =2^{8}\times 3^{4}$

(ii)270

$270 =2\times 3^{5}\times 5$

(iii)$729 \times 64$ $729 \times 64=3^{6}\times 2^{6}$\

(iv)768

$768=2^{8}\times 3$

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Q.5. Simplify:

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

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

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

Solution:

(i)$\frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}$ $\frac{(2^{5})^{2}\times 7^{3}}{8^{3}\times 7}\\ \\ =\frac{(2^{5\times 2})\times 7^{3}}{(2^{3})^{3}\times 7}\\ \\ =\frac{2^{10}\times 7^{3}}{2^{9}\times 7}\\ \\ =2^{10-9}\times 7^{3-1}=2\times 7^{2}\\ \\ =2\times 49=98$

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

(iii)$\frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}$ $\frac{3^{5}\times 10^{5}\times 25}{5^{7}\times 6^{5}}=\frac{3^{5}\times (2\times 5)^{5}\times 5^{2}}{5^{7}\times (2\times 3)^{5}}\\ \\ =\frac{3^{5}\times 2^{5}\times 5^{5}\times 5^{2}}{5^{7}\times 2^{5}\times 3^{5}}\\ \\ =\frac{3^{5}\times 2^{5}\times 5^{5+2}}{5^{7}\times 2^{5}\times 3^{5}}\\ \\ =\frac{3^{5}\times 2^{5}\times 5^{7}}{5^{7}\times 2^{5}\times 3^{5}}\\ \\ =2^{5-5}\times 3^{5-5}\times 5^{5-5}\\ =2^{0}\times 3^{0}\times 5^{0}=1\times 1\times 1=1$

### NCERT Solutions For Class 7 Maths Chapter 13 Exercises 13.3

Q.1. Write the following numbers in the expanded form:

(i)279404

(ii)3006194

(iii)2806196

(iv)120719

(v)20068

Solution:

(i)279404=2,00,000+70,000+9,000+400+00+4

=$2\times 100000+7\times 10000+9\times 1000+4\times 100+0\times10+4\times 1$

=$2\times 10^{5}+7\times 10^{4}+9\times 10^{3}+4\times 10^{2}+0\times10^{1}+4\times 10^{0}$

(ii)3006194=30,00,000+0+0+6,000+100+90+4

=$3\times 1000000+0 \times 100000+0\times 10000+6\times 1000+1\times 100+9\times10+4\times 1\\ =3\times 10^{6}+0 \times 10^{5}+0\times 10^{4}+6\times 10^{3}+1\times 10^{2}+9\times10^{1}+4\times 10^{0}$

(iii)2806196=20,00,000+8,00,000+0+6,000+100+90+6

$=2\times 1000000+8 \times 100000+0\times 10000+6\times 1000+1\times 100+9\times10+6\times 1\\ =2\times 10^{6}+8 \times 10^{5}+0\times 10^{4}+6\times 10^{3}+1\times 10^{2}+9\times10^{1}+6\times 10^{0}$

(iv)120719=1,00,000+20,000+0+700+10+9

=$=1\times 100000+2 \times 10000+0\times 1000+7\times 100+1\times 10+9\times1\\ =1\times 10^{5}+2 \times 10^{4}+0\times 10^{3}+7\times 10^{2}+1\times 10^{1}+9\times10^{0}$

(v)20068=20,000+00+00+60+8

=$=2\times 10000+0 \times 1000+0\times 100+6\times 10+8\times 1\\ =2\times 10^{4}+0 \times 10^{3}+0\times 10^{2}+6\times 10^{1}+8\times 10^{0}$

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Q.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)$9\times 10^{5}+2 \times 10^{2}+3\times 10^{1}$

Solution:

(a)$8\times 10^{4}+6 \times 10^{3}+0\times 10^{2}+4\times 10^{1}+5\times 10^{0}$ $8\times 10000+6 \times 1000+0\times 100+4\times 10+5\times 1\\ =80000+6000+0+40+5\\ =86,045$

(b)$4\times 10^{5}+5 \times 10^{3}+3\times 10^{2}+2\times 10^{0}$ $4\times 100000+5 \times 1000+3\times 100+2\times 1\\ =400000+5000+300+2\\ =4,05,302$

(c) $3\times 10^{4}+7 \times 10^{2}+5\times 10^{0}$ $3\times 10000+7 \times 100+5\times 1\\ =30000+700+5\\ =30,705$

(d)$9\times 10^{5}+2 \times 10^{2}+3\times 10^{1}$ $9\times 100000+2 \times 100+3\times 10\\ =900000+200+30\\ =9,00,230$

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Q.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

Solution:

(i)5,00,00,000 =$5\times 1,00,00,000=5\times 10^{7}$

(ii)70,00,000=$7\times 10,00,000=7\times 10^{6}$

(iii)3,18,65,00,000= $31865\times 100000=3.1865\times10000\times 100000=3.1865\times 10^{9}$

(iv)3,90,878 =$3.90878\times 100000=3.90878\times10^{5}$

(v)39087.8 =$3.90878\times 10000=3.90878\times10^{4}$

(vi)3908.78=$3.90878\times 1000=3.90878\times10^{3}$

Q.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 Earth id 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 100,000,000,0000 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.

Solution:

(i)The distance between Earth and Moon =$= 384,000,000 \, m\\ = 384 \times 1000000 \, m\\ = 3.84 \times 100 \times 1000000\\ =3.84 \times 10^{8} \, m$

(ii)Speed of light in vacuum=$300,000,000\, m/s\\ = 3 \times 100000000 \, m/s\\ =3\times 10^{8} \, m/s$

(iii)Diameter of the Earth=$1,27,56,000\, m\\ = 12756 \times 1000 \, m\\ = 1.2756 \times 10000 \times 1000\, m\\ =1.2756 \times 10^{7}\, m$

(iv)Diameter of the Sun=$1,400,000,000 m\\ = 14 \times 100,000,000 m\\ = 1.4 \times 10 \times 100,000,000 m\\ =1.4\times 10^{9}\, m$

(v)Average of Stars$= 100,000,000,000\\ = 1 \times 100,000,000,000\\ = 1 \times 10^{11 }$

(vi)Years of Universe$= 12,000,000,000\, years\\ = 12 \times 1000,000,000\, years\\ = 1.2 \times 10 \times 1000,000,000\, years\\ = 1.2\times 10^{10}\, years$

(vii)Distance of the Sun from the centre of the Milky Way Galaxy$= 300,000,000,000,000,000,000 m\\ = 3 \times 100,000,000,000,000,000,000 m\\ = 3 \times 10^{20} \, m$

(viii)Number of molecules in a drop of water weighing 1.8 gm$= 60,230,000,000,000,000,000,000\\ = 6023 \times 10,000,000,000,000,000,000\\ = 6.023 \times 1000 \times 10,000,000,000,000,000,000\\ = 6.023 \times 10^{22}$

(ix)The Earth has Sea water$= 1,353,000,000\, km3^{3}\\ = 1,353 \times 1000000 \, km^{3}\\ = 1.353 \times 1000 \times 1000,000\, km^{3}\\ =1.353 \times 10 ^{9}\, km^{3}$

(x) The population of India$= 1,027,000,000\\ = 1027 \times 1000000\\ = 1.027 \times 1000 \times 1000000\\ =1.027 \times 10 ^{9}$

The Exponents and Power solutions available here are extremely easy to understand and any students can access these solutions for free. Here, NCERT Solutions For Chapter 13 (Exponents and Powers) of Class 7 Maths PDF is also provided for students which they can either view online or download the PDF to access the solutions in offline mode. These solutions are prepared by the subject experts of BYJUâ€™S with respect to CBSE syllabus (2018-2019) and NCERT curriculum, prescribed by the board.

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