Ncert Solutions For Class 7 Maths Ex 13.2

Ncert Solutions For Class 7 Maths Chapter 13 Ex 13.2

Q.1.Using laws of exponents, simplify and write the answer in exponential form:

(i)32×34×38

(ii)615÷610

(iii)a3×a2

(iv)7x×72

(v)(52)2÷53

(vi)25×55

(vii)a4×b4

(viii)(34)3

(ix)(220÷215)×23

(x)8t÷82

Solution:

(i)32×34×38=3(2+4+8)=314

(ii)615÷610=61510=65

(iii)a3×a2=a3+2=a5

(iv)7x×72=7x+2

(v)(52)2÷53=52×3÷53=56÷53=563=53

(vi)25×55=(2×5)5=105

(vii)a4×b4=(a×b)4

(viii)(34)3=34×3=312

(ix)(220÷215)×23=(22015)×23=25×23=25+3=28

(x) 8t÷82=8t2

Formulae:

(i)am×an=am+n

(ii)am÷an=amn

(iii)(am)n=am×n

(iv)am×bm=(a×b)m

 

Q.2.(i)23×34×43×32

(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

Solution:

(i)23×34×43×32=23×34×223×25=23+2×343×25=25×343×25=255×343=20×33=1×33=33

 

(ii)[(52)3×54]÷57=[56×54]÷57=[56+4]÷57=510÷57=5107=53

 

(iii)254÷53=(52)4÷53=58÷53=583=55

 

(iv) 3×72×11821×113=3×72×1183×7×113=311×721×1183=30×71×115=7×115

 

(v)3734×33=3734+3=3737=377=30=1

 

(vi)20+30+40=1+1+1=3

 

(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=286×a52=22×a2=(2a)2

 

(x)(a5a3)×a8=(a53)×a8=a2×a8=a2+8=a10

 

(xi)45×a8b345×a5b2=455×a85×b32=40×a3×b=1×a3×b=a3b

 

(xii)(23×2)2=(23+1)2=(24)2=24×2=28

 

Q.3.Say true or false and justify your answer:

(i)10×1011=10011

(ii)23>52

(iii)23×32=65

(iv)30=(1000)0

Solution:

(i)10×1011=10011

L.H.S.  101+11=1012

And

R.H.S. (102)11=1022

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

Therefore, it is false.

(ii)23>52

L.H.S. 23=8

And

R.H.S. 52=25

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

Therefore, it is false.

(iii)23×32=65

L.H.S. 23×32=8×9=72

And

R.H.S. 65=7,776

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

Therefore, it is false.

(iv)30=(1000)0

L.H.S. 30=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)108×192

(ii)270

(iii)729×64

(iv)768

Solution:

(i)108×192

108×192=(22×33)×(26×3)=22+6×33+1=28×34

 

(ii)270

270=2×35×5

 

(iii)729×64

729×64=36×26\

 

(iv)768

768=28×3

 

Q.5. Simplify:

(i)(25)2×7383×7

(ii)25×52×t8103×t4

(iii)35×105×2557×65

Solution:

(i)(25)2×7383×7

(25)2×7383×7=(25×2)×73(23)3×7=210×7329×7=2109×731=2×72=2×49=98

 

(ii)25×52×t8103×t4

25×52×t8103×t4=52×52×t8(5×2)3×t4=52+2×t8423×33=54×t423×53=543×t423=5t48

 

(iii)35×105×2557×65

35×105×2557×65=35×(2×5)5×5257×(2×3)5=35×25×55×5257×25×35=35×25×55+257×25×35=35×25×5757×25×35=255×355×555=20×30×50=1×1×1=1