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 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
= 23 × 23
= 2 × 2
∴ \(\sqrt[3]{64}\) = 4
(ii) 512 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
= 23 × 23 × 23
= 2 × 2 × 2
∴ \(\sqrt[3]{512}\) = 8
(iii) 10648 = \(\underline{2 \times 2 \times 2}\) × \(\underline{11 \times 11 \times 11}\)
= 23 × 113
= 2 × 11
∴ \(\sqrt[3]{10648}\) = 22
(iv) 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
= 23 × 33 × 53
∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5
= 30
(v) 15625 = 5 × 5 × 5 × 5 × 5 × 5
= 53 × 53
= 5 × 5
∴ \(\sqrt[3]{15625}\) = 25
(vi) 13824 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)
= 23 × 23 × 23 × 33
= 2 × 2 × 2 × 3
∴ \(\sqrt[3]{13824}\) = 24
(vii) 110592 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)
= 23 × 23 × 23 × 23 × 33
= 2 × 2 × 2 × 2 × 3
∴ \(\sqrt[3]{110592}\) = 48
(viii) 46656 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\)
= 23 × 23 × 33 × 33
= 2 × 2 × 3 × 3
∴ \(\sqrt[3]{46656}\) = 36
(ix) 175616 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{7 \times 7 \times 7}\)
= 23 × 23 × 23 × 73
= 2 × 2 × 2 × 7
∴ \(\sqrt[3]{175616}\) = 56
(x) 91125 = \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)
= 33 × 33 × 53
= 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