# 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$\sqrt[3]{64}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\\\sqrt[3]{64}=2\times 2$

= 4

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

(B) 512

5123=2×2×2×2×2×23$\sqrt[3]{512}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}$

= 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$\sqrt[3]{27000}=\sqrt[3]{2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5}$

=>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$\sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}$

=> 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$\sqrt[3]{13824}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3}$

=> 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$\sqrt[3]{110592}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \times3\times 3\times 3}$

=> 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$\sqrt[3]{46656}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3}$

=> 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$\sqrt[3]{175616}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 7\times 7}$

=> 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$\sqrt[3]{91125}=\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3\times 5\times 5\times 5}$

=> 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