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.

The main topics of NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers are:

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

The NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers are given below:

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

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

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}

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

       2^1^0^0=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

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(viii) 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} 

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}

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}

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

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

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

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}

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

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

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

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}

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

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

Solutions of NCERT for class 7 maths chapter 13 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

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 

Solutions of NCERT for class 7 maths

CBSE NCERT solutions for class 7 science

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.    

Happy Reading!!!

 

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