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

NCERT solutions for class 7 maths chapter 13 Exponents and Powers: 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 this chapter. 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 CBSE NCERT solutions for class 7 maths chapter 13 exponents and powers. Check NCERT solutions from class 6 to 12 by clicking on the above link. Here you will get solutions to three exercises of this chapter.

Exercise:13.1

Exercise:13.2

Exercise:13.3

## 13.1 Introduction

13.2 Exponents

13.3 Laws Of Exponents

13.3.1 Multiplying Powers With The Same Base

13.3.2 Dividing Powers With The Same Base

13.3.3 Taking Power Of A Power

13.3.4 Multiplying Powers With The Same Exponents

13.3.5 Dividing Powers With The Same Exponents

13.4 Miscellaneous Examples Using The Laws Of Exponents

13.5 Decimal Number System

13.6 Expressing Large Numbers In The Standard Form

## Solutions of NCERT for class 7 maths chapter 13 exponents and powers topic 13.2 exponents

Question:(i) Express:

729 as a power of 3

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

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

343 as a power of 7 can be givena as

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

CBSE NCERT solutions for class 7 maths chapter 13 exponents and powers-Exercise: 13.1

Question:1(i) Find the value of:

$(i)2^{6}$

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:

$(ii)9^{3}$

the value of $(ii)9^{3}$ is given by

$9\times 9\times 9=729$

Question:1(iii) Find the value of:

$(iii)11^{2}$

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

$11\times 11=121$

Question:1(iv) Find the value of:

$(iv)5^{4}$

the value of $(iv)5^{4}$ is given by

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

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

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

$(i)6\times 6\times 6\times 6$ 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

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

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

(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

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

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

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

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

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

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 $8^2$

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

$100^2=100\times 100=10000$

$2^1^0^0=2\times 2\times 2\times 2\times 2...........till\, 100\, times\, 2$

since  $2^1^0^0> 100^2$

$2^1^0^0$ 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 $10^2$

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

(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:

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

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

can be simplified as

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

Question:6(ii) Simplify:

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

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

can be simplified as

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

Question:6(iii) Simplify:

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

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

can be simplified as

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

Question:6(iv) Simplify:

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

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

can be simplified as

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

Question:6(v) Simplify:

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

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

can be simplified as

$0\times 10\times 10=0$

Question:6(vi) Simplify:

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

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

can be simplified as

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

Question:6(vii) Simplify:

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

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

can be simplified as

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

Question:6(viii) Simplify:

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

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

can be simplified as

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

Question:7(i) Simplify:

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

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

can be simplified as

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

Question:7(ii) Simplify:

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

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

can be simplified as

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

Question:7(iii) Simplify:

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

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

can be simplified as

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

Question:7(iv) Simplify:

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

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

can be simplified as

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

Question:8 Compare the following numbers:

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

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

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.

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

## Solutions for NCERT class 7 maths chapter 13 exponents and powers topic 13.3.1 multiplying powers with the same base

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

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

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

can be simplified as

$2^{5+3}=2^8$

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

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

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

can be simplified as

$p(3+2)=p^5$

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

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

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

can be simplified as

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

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

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

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

can be simplified as

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

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

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

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

can be simplified as

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

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

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

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

can be simplified as

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

Solutions of NCERT for class 7 maths chapter 13 exponents and powers topic 13.3.2 dividing powers with the same base

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

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 $7^{(13-10)}=7^3$

Solutions for NCERT class 7 maths chapter 13 exponents and powers topic 13.3.3 taking powers of a powers

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

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

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 $5^{(3\times 7)}=5^{21}$

Solutions of NCERT for class 7 maths chapter 13 exponents and powers topic 13.3.4 multiplying powers with the same exponents

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

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

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 $((-2)\times (-3))^4= 6^4$

Solutions of NCERT for class 7 maths chapter 13 exponents and powers topic 13.3.5 dividing powers with the same exponents

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

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

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

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

## CBSE NCERT solutions for class 7 maths chapter 13 exponents and powers-Exercise: 13.2

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

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

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

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

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 $8^{(t-2)}$

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

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

can be simplified as

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

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

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

$= 2^{(5-5)} \times 3^{(4-1)}$

$=3^3$

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

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

can be simplified as

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

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

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

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

$=5^{10-7}$

$=5^3$

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

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

can be simplified as

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

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

$=5^8\div 5^3$

$=5^{(8-3)}$

$=5^{5}$

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

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

can be simplified as

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

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

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

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

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

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

can be simplified as

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

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

$=3^{(7-7)}$

$=3^0$

$=1$

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

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

can be simplified as

$=1+1+1$

$=3$

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

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

can be simplified as

$=1\times 1\times 1=1$

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

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

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

can be simplified as

$=(1+1)\times 1$

$=2\times 1$

$=2$

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

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

can be simplified as

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

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

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

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

$=(2a)^2$

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

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

can be simplified as

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

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

$=a^{(2+8)}$

$=a^{(10)}$

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

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

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:

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

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

can be simplified as

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

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

$=2^{(4 \times 2)}$

$=2^8$

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

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

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.

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

$(i)\: 108\times 192$

 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:

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

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

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:

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

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

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:

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

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

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 class 7 maths chapter 13 exponents and powers topic 13.6 expressing large numbers in the standard form

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

(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

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

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

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

Solutions of NCERT for class 7 maths chapter 13 exponents and powers-Exercise: 13.3

279404, 3006194, 2806196, 120719, 20068

(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

$20068=20000+0+0+60+8$

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

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

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

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

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

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

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

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

$=80000+6000+000+40+5$

$=86045$

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

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

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

$=405302$

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

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

$=30000+0000+700+00+5$

$=30705$

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

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

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

$=900230$

Thus, the above problems are simplified in simpler forms.

(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

(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

$=3.90878\times 1000$

$=3.90878\times 10^3$

(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.

(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.

$=1.027\times 1000000000$

$=1.027\times 10^9$

## NCERT Solutions for Class 7 Maths- Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

## 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 solutions for class 7 maths chapter 13 exponents and powers if you are facing difficulties while solving the problems.