Ncert Solutions For Class 8 Maths Ex 7.2

Ncert Solutions For Class 8 Maths Chapter 7 Ex 7.2

Q1 :

By the method of prime factorization find the cube root for the following.

(A) 64

(B) 512                                                                                                                    

(C) 10648

(D) 27000

(E) 15625

(F) 13824

(G) 110592

(H) 46656

(I) 175616

(J) 91125

Solution:

(A) 64

643=2×2×2×2×2×23643=2×2

= 4

02 064
02 032
02 016
02 08
02 04
02 02
01

 

(B) 512

5123=2×2×2×2×2×23

= 2 x 2 x 2

= 8

02 0512
02 0256
02 0128
02 064
02 032
02 016
02 08
02 04
02 02
01

 

(C) 10648

\sqrt[3]{10648}=\sqrt[3]{2\times 2\times 2\times 11\times 11\times 11}

= 2 x 11

=22

02 010648
02 05324
02 02662
011 01331
011 0121
011 011
01

 

(D) 27000

270003=2×2×2×3×3×3×5×5×53

=>2 x 3 x 5

=>30

02 027000
02 013500
02 06750
03 03375
03 01125
03 0375
05 0125
05 025
05 05
01

(E) 15625

156253=5×5×5×5×5×53

=> 5 x 5

=> 25

05 015625
05 03125
05 0625
05 0125
05 025
05 05
01

 

(F) 13824

138243=2×2×2×2×2×2×2×2×2×3×3×33

=> 2 x 2 x 2 x 3

=> 24

02 13824
02 06912
02 03456
02 01728
02 0864
02 0432
02 0216
02 0108
02 054
03 27
03 09
03 03
01

 

(G) 110592

1105923=2×2×2×2×2×2×2×2×2×2×2×2×3×3×33

=> 2 x 2 x 2 x 2 x 3

=> 48

02 0110592
02 055296
02 027648
02 013824
02 06912
02 03456
02 01728
02 0864
02 0432
02 0216
02 0108
02 054
03 027
03 09
03 03
01

 

(H) 46656

466563=2×2×2×2×2×2×3×3×3×3×3×33

=> 2 x 2 x 2 x 3 x 3 x 3

=> 36

(I) 175616

1756163=2×2×2×2×2×2×2×2×2×7×7×73

=> 2 x 2 x 2 x 7

=> 56

02 0175616
02 087808
02 043904
02 021952
02 010976
02 05488
02 02744
02 01372
02 0686
07 0343
07 049
07 07
01

 

(J) 91125

911253=3×3×3×3×3×3×5×5×53

=> 3 x 3 x 5

=> 45

03 091125
03 010125
03 03375
03 01125
03 0375
05 0125
05 025
05 05
01

 

Q2:

State whether the following is true of false:

(A) Any off number of a cube is even.

(B) When a number end with two zeros, it is never a perfect cube.

(C) If the square of a given number ends with 5 then its cube will end with 25.

(D) There is no number that ends with 8 which is a perfect cube.

(E) The cube of a given two digit number will always be a three digit number.

(F) The cube of a two digit number will have either seven or more digits.

(G) The cube of single digit number may also be a single digit number.

Solution:

(A) The statement given is false.

Since, 13  = 1, 33 = 27, 53 = 125, . . . . . . . . . . are all odd.

(B) The given statement is true.

Since, a perfect cube ends with three zeroes.

Eg. 103 = 1000, 20= 8000, 303  = 27,000 , . . . . . . . . . . . . so on.

(C) The given statement is false

Since, 52 = 25, 53 = 125 , 152 = 225, 153 = 3375 ( Did not end with 25)

(D)  the given statement is false.

Since 123 = 1728 [the number ends with 8]

223 = 10648 [ the number ends with 8]

(E) The given statement is false

Since, 103 = 1000 [Four digit number]

And 113= 1331 [four digit number]

(F) The statement is False.

Since 993 = 970299 [Six digit number]

(G) the given statement is true

13 = 1 [single digit]

23 = 8 [single digit]

Q3 :

1331 is told to be a perfect cube. What are the factorization methods in which you can find its cube root? Similarly, find the cube roots for

(i)4913

(ii)12167

(iii)32768.

Solution:

We know that 103 = 1000 and possible cute of 113  = 1331

Since, the cube of units digit is 13  = 1

Then, cube root of 1331 is 11

(i) 4913

We know that 73 is 343

Next number that comes with 7 as the units place is 173  = 4913

Therefore the cube root of 4913 is 17

(ii) 12167

Since we know that 33 = 27

Here in cube, the ones digit is 7

Now the next number with 3 In the ones digit is 133 = 2197

And the next number with 3 in the ones digit is 233  = 12167

Hence the cube root of 12167 is 23

(iii) 32768

We know that 2= 8

Here in the cube, the ones digit is 8

Now the next number with 2 in the ones digit is 123  = 1728

And the next number with 2 as the ones digit 223 = 10648

And the next number with 2 as the ones digit 323 = 32768

Hence the cube root of 32768 is 32