NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

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Written by Team Trustudies
Updated at 2021-05-07


NCERT solutions for class 7 Maths Chapter 13 Exponents And Powers Exercise 13.1

Q.1 Find the value of
(i) 26
(ii)93
(iii) 112
(iv) 54



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 26=2×2×2×2×2×2=64
(ii) 93=9×9×9=729
(iii) 112=11×11=121
(iv)54=5×5×5×5=625

Q.2 Exress the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c× c × d



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 6×6×6×6=63
(ii)t×t=t2
(iii) b×b×b×b=b4
(iv) 5×5×7×7×7=52×73=52·73
(v) 2×2×a×a=22×a2=22·a2
(vi)a×a×a×c×c×c×c×d=a3×c4×d=a3·c4·d

Q.3 Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 2    5122    2162    1282    642    322    162    82    42    2    1 512= 2×2×2×2×2×2×2×2×2 = 29
(ii) 7    3437    497    7    1 343 = 7×7×7 = 73
(iii) 3    7293    2433    813    273    9    3 729 = 3×3×3×3×3×3 = 36
(iv) 5    31255    6255    1255    255    5    1 3125 = 5×5×5×5×5= 55

Q.4 Identify the greater number, wherever possible, in each of the following?
(i) 43or34
(ii)53or35
(iii)28or82
(iv)1002or2100
(v) 210or102



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 43or34
43=4×4×4=64,
34=3×3×3×3=81
Since 81 > 64
?34>43.
(ii)53or35
53=5×5×5=125
35=3×3×3×3×3=243
Since 243 > 125
35>53.
(iii)28or82
28=2×2×2×2×2×2×2×2=256
82=8×8=64
Since 256 > 64
28>82.
(iv) 1002or2100 1002=100×100=10000
2100=2×2×2×100times=214=16384
Since 16384 > 10,000
2100>1002.
(v)210or102
210=2×2×2×2×2×2×2×2×2×2=1024
102=10×10=100
?210>102.

Q.5 Express each of the following as the product of powers of their prime
(i) 648
(ii) 405
(iii) 540
(iv) 3600



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 2    6482    3242    1623    813    273    93    31 648= 23×34
(ii) 3    4053    1353    453    155    5    1 405= 34×51
(iii) 2    5402    2703    1353    453    155    5    1 540=22×33×51
(iv) 2    36002    18002    9002    4503    2253    755    255    51 3600=24×32×52

Q.6 Simplify:
(i) 2×103
(ii) 72×22
(iii) 23×5
(iv) 3×44
(v) 0×102
(vi)52×33
(vii) 24×32
(viii)32×104



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 2×103=2×10×10×10==2000
(ii)72×22==7×7×2×2=196
(iii)23×5=2×2×2×5=40
(iv) 3×44=3×4×4×4×4=768
(v)0×102=0×10×10==0
(vi) 52×33=5×5×3×3×3=675
(vii) 24×32=2×2×2×2×3×3=144
(viii) 32×104=3×3×10×10×10×10=90000

Q.7 Simplify:
(i)(?4)3
(ii)(?3)×(?2)3
(iii)(?3)2×(?5)2
(iv)(?2)3×(?10)3



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i)(?4)3=(?4)×(?4)×(?4)=?64 [?(-a)odd number = (-a)odd number]
(ii) (?3)×(?2)3=(?3)×(?2)×(?2)×(?2)
=(?3)×(?8)= 24 [?(-a)odd number = (-a)odd number]
(iii)(?3)2×(?5)2=[(?3)×(?5)]2
=152=225[?am×bm=(ab)m]
(iv)(?2)3×(?10)3=[(?2)×(?10)]3
=202=8000[?am×bm=(ab)m]

Q.8 Compare the following:
(i)2.7×1012;1.5×108
(ii) 4×1014;3×1017



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 2.7×1012=2.7×104×108=27000×108
27000×108>1.5×108
2.7×1012>1.5×108
(ii) 3×1017=3×103×1014=3000×1014
3000×1014>4×1014
4×1014<3×1017

NCERT solutions for class 7 Maths Chapter 13 Exponents And Powers Exercise 13.2

Q.1 Using laws of e×ponents, simplify and write the answer in e×ponential form:
(i) 32×34×38
(ii)615÷610
(iii)a3×a2
(iv)7x×72
(v)(52)3÷53
(vi) 25×55
(vii) a4×b4
(viii)(34)3
(ix)(220÷215)×23
(x)8t÷82



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 32×34×38=32+4+8=314[am÷an=am+n]
(ii) 615÷610=615?10=65[am÷an=am?n]
(iii) a3×a2=a3+2=a5[am×an=am+n]
(iv)7x×72=7x+2[am×an=am+n]
(v)(52)3÷53=52×3÷53=56÷53=56?3
=53[(am)n=amn,am÷an=am?n]
(vi) 25×55=(2×5)5=105[am×bm=(ab)m]
(vii) a4×b4=(ab)4[am×bm=(ab)4]
(ix)(220÷215)×23=220?15×23=25×23=25+3=28
(x) 8t÷82=8t?2

Q.2 Simplify and express each of the following in exponential form:
(i) (23×34×4)(3×32)
(ii) [(52)3×54]÷57
(iii) 254÷53
(iv) 3×72×11821×113
(v) 3734×33
(vi) 20+30+40
(vii) 20×30×40
(viii) (30+20)×50
(ix)28×a543×a3
(x) (a5a3)×a8
(xi) 45×a8b345×a5b2
(xii) (23×2)2



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 23×34×43×32 = 23×34×223×25
= 23+2×343×25=25×343×25
= 25?5×34?1 = 20×33= 33
(ii)[(52)3×54]÷57 = (56×54)÷57
= (56+4÷57 = 510?7 = 53
(iii) 254÷53 = (52)4÷53= 58÷53 = 58?3= 55
(iv) 3×72×11821×113 = 3×72×1187×3×113
= 72?1×118?3 = 7×115 \) (v) 3734×33 = 3734+3 = 3737 = 1
(vi) 20+30+40 = 1+1+1 = 0[?a0=1]
(vii) 20×30×40 = 1×1×1 = 1
(viii) (30+20)×50 = (1+1)×1=2×1=2
(ix) 28×a543×a3 = 28×a5(22)3×a3
= 28×a526×a3 = \( 2^{8-6}{×a^{5-3} \) = 22×a2 = (2a)2
(x) (a5a3)×a8 = a5+8a3 = a13a3
= a13?3 = a10
(xi) 45×a8b345×a5b2 = a8b3a5b2
= a8?5b3?2 = a3b
(xii) (23×2)2 = 26×22 = 26+2 = 28

Q.3 Say true or false and justify your answer: (i)10×1011=10011
(ii) 23>52
(iii)23×32=65
(iv) 320=(1000)0



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i)LHS = 10×1011=101+11=1012
RHS=10011=(102)11=1022
1012?1022
? Statement is false.
(ii)23>52
LHS = 23=8
RHS =52=25
8 < 25
?23<52
Thus, the statement is false.
(iii) 23×32=65
LHS =23×32=8×9=72
RHS =65=6×6×6×6×6=7776
72?7776
?The statement is false.
(iv)30=(1000)0
LHS = 30=1
RHS = (1000)0=1
LHS = RHS

Q.4 Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) 108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
=28×34
(ii) 270=2×3×3×3×5=2×33×5
(iii) 729 × 64 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2
=36×26
(iv) 768 = 2×2×2×2×2×2×2×2×3 = 28×3

Q.5 Simplify:
(i) (25)2×7383×7
(ii) 25×52×t8103×t4
(iii) 35×105×2557×65



NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers


Answer :

(i) (25)2×7383×7 = (210×73(23)3×7
= \(\frac{(2^{10}× 7^3}{{2^9 × 7} \) = 210?9×73?1 = 21×72 =2× 49 = 98
(ii) 25×52×t8103×t4 = 52×52×t823×53×t4
= 54×t823×53×t4 = 54?3×t8?423 = 5t423
(iii) 35×105×2557×65= 35×25×55×5257×25×35
=