Haryana State Board HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.2 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Exercise 7.2

Question 1.

Find the cube root of each of the following numbers by prime factorisation method s

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125.

Solution:

(i) 64 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)

= 2^{3} × 2^{3}

= 2 × 2

∴ \(\sqrt[3]{64}\) = 4

(ii) 512 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)

= 2^{3} × 2^{3} × 2^{3}

= 2 × 2 × 2

∴ \(\sqrt[3]{512}\) = 8

(iii) 10648 = \(\underline{2 \times 2 \times 2}\) × \(\underline{11 \times 11 \times 11}\)

= 2^{3} × 11^{3}

= 2 × 11

∴ \(\sqrt[3]{10648}\) = 22

(iv) 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5

= 2^{3} × 3^{3} × 5^{3}

∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5

= 30

(v) 15625 = 5 × 5 × 5 × 5 × 5 × 5

= 5^{3} × 5^{3}

= 5 × 5

∴ \(\sqrt[3]{15625}\) = 25

(vi) 13824 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

= 2^{3} × 2^{3} × 2^{3} × 3^{3}

= 2 × 2 × 2 × 3

∴ \(\sqrt[3]{13824}\) = 24

(vii) 110592 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

= 2^{3} × 2^{3} × 2^{3} × 2^{3} × 3^{3}

= 2 × 2 × 2 × 2 × 3

∴ \(\sqrt[3]{110592}\) = 48

(viii) 46656 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\)

= 2^{3} × 2^{3} × 3^{3} × 3^{3}

= 2 × 2 × 3 × 3

∴ \(\sqrt[3]{46656}\) = 36

(ix) 175616 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{7 \times 7 \times 7}\)

= 2^{3} × 2^{3} × 2^{3} × 7^{3}

= 2 × 2 × 2 × 7

∴ \(\sqrt[3]{175616}\) = 56

(x) 91125 = \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)

= 3^{3} × 3^{3} × 5^{3}

= 3 × 3 × 5

∴ \(\sqrt[3]{91125}\) = 45

Question 2.

State true or false.

(i) Cube of any odd number is even

(ii) A perfect cube does not end with two zeros

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

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

Solution:

(i) False,

(ii) True,

(iii) True,

(iv) False,

(v) False,

(vi) False,

(vii) True

Question 3.

You ar told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

Solution:

The given number is 1331

Step 1. From groups of three starting from the rightmost digit of 1331 is 1331.In this case one group 331 has three digits whereas 1 has only two digits.

Step 2. Take 331.

The digit 1 is at its one’s place.

We take the one’s place of the required cube root as 1.

Step 3. Take the other group is 1.

Cube of 1 is 1 and cube of 2 is 8.1 lies between 0 and 8.

The smaller number among 1 and 2 is 1. The one’s place of 1 is 1 itself. Take 1 as ten’s place of the cube root of 1331.

Thus, \(\sqrt[3]{1331}\) = 11

Similarly, 4913

Step 1. 4913

Step 2. Take 913

The digit 3 is at its one’s place we take the one’s place of the required cube root as (3 × 3 × 3 = 27)7.

Step 3. Take the other group is 4. Cube of 1 is 1 and cube of 2 is 8. 4 lies between 1 and 8.

The smaller number among 1 and 2 is 1. The one’s place of 1 is 1 itself. Take 1 as ten’s place of the cube root of 4913.

Thus, \(\sqrt[3]{4913}\) = 17

Similarly, 12167

Step 1. 12167

Step 2. 7 × 7 × 7 = 343 i.e. one’s place is 3

Step 3. 12, 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27 8 < 12 < 27

The smaller number among 2 and 3 is 2. The one’s place of 2 is 2 itself. Take 2 as tens place of the cube root of 12167.

Thus, \(\sqrt[3]{12167}\) = 23

Similarly, 32768

Step 1. 32768

Step 2. 8 × 8 × 8 = 512

i.e. one’s place is 2

Step 3. 32, 3 × 3 × 3 = 27

4 × 4 × 4 = 64

The smaller number along 3 and 4 is 3. The one’s place of 3 is 3 itself. Take 3 as ten’s place of the cube root 32768.

Thus \(\sqrt[3]{32768}\) = 32