# NCERT Solutions For Class 7 Maths Chapter 13

## NCERT Solutions Class 7 Maths Exponents and Powers

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers includes solutions for all the questions provided in the NCERT 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.

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

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

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

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

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

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 $$100^{2}$$.

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

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

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

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,

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

Exercise-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.  $$10^{1+11}=10^{12}$$

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.

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

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

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

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

Exercise-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

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

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

$$9\times 100000+2 \times 100+3\times 10\\ =900000+200+30\\ =9,00,230$$

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$$= 1,027,000,000\\ = 1027 \times 1000000\\ = 1.027 \times 1000 \times 1000000\\ =1.027 \times 10 ^{9}$$

In the exponents and powers chapter, some of the most crucial concepts are covered which forms the basis of many mathematics concepts. This chapter is extremely crucial for the class 7 exam and students need to be thorough with all the concepts to be able to comprehend many topics in the later grades easily.

The exponents and powers chapter starts with the introduction of these two terms and why exponents and powers are needed in mathematics. Then, some of the important terminologies and concepts are discussed like laws of exponents, decimal number system, and expressing large numbers in the standard form.

It is important for the students to go through all the example questions and then practice the problems given in the NCERT Book For Class 7 Maths. Students can always check these solutions in case of any doubts.

Keep visiting BYJU’S to get all the NCERT solutions and to get full assistance for the exams. Students are also suggested to download BYJU’S- The Learning App to have a more personalized and effective learning experience.