Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Cubes and Cube Roots

    Exercise 6.1

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

8th class > Cubes and Cube Roots > Exercise 6.1

Exercise 6.1

1.Which of the following numbers are not perfect cubes?

Instructions

216
128
1000
100
46656
Perfect Cube
Not Perfect Cube

2.Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (Enter the factors in increasing order)

Instructions

(i) 243
Prime factorisation of 243 = × × × × = × ×
Here, number is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3. Thus, the required smallest number to be multiplied is 3.
(ii) 256
Prime factorisation 256 = × × × × × × × = × × ×
Here, a number is needed to make 2 × 2 a group of three, i.e., 2 × 2 × 2
(iii) 72
Prime factorisation of 72 = × × × × = × ×
Here, a number is required to make 3 × 3 a group of three, i.e. 3 × 3 × 3. Thus, the required smallest number to be multiplied is 3.
(iv) 675
Prime factorisation of 675 = × × × × = × ×
Here, a number is required to make 5 × 5 a group of three to make it a perfect cube, i.e. 5 × 5 × 5. Thus, the required smallest number is 5.
(v) 100
Prime factorisation of 100 = × × ×
Here, number and are needed to multiplied 2 × 2 × 5 × 5 to make it a perfect cube i.e., 2 × 2 × 2 × 5 × 5 × 5.
Thus, the required smallest number to be multiplied is 2 × 5 = .

3.Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (Enter the factors in increasing order)

Instructions

(i) 81
Prime factorisation of 81 = × × × = ×
Here, a number is the number by which 81 is divided to make it a perfect cube, i.e., 81 ÷ 3 = which is a perfect cube. Thus, the required smallest number to be divided is .
(ii) 128
Prime factorisation of 128 = × × × × × × = × ×
Here, a number is the smallest number by which 128 is divided to make it a perfect cube, i.e., 128 ÷ 2 = which is a perfect cube. Thus, is the required smallest number.
(iii) 135
Prime factorisation of 135 = × × × = ×
Here, is the smallest number by which 135 is divided to make a perfect cube, i.e., 135 ÷ 5 = which is a perfect cube. Thus, is the required smallest number.
(iv) 192
Prime factorisation of 192 = × × × × × × = × ×
Here, is the smallest number by which 192 is divided to make it a perfect cube i.e., 192 ÷ 3 = which is a perfect cube. Thus, is the required smallest number.
(v) 704
Prime factorisation 704 = × × × × × × = × ×
Here, is the smallest number by which 704 is divided to make it a perfect cube, i.e., 704 ÷ 11 = which is a perfect cube. Thus, is the required smallest number.
  1. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube? (Enter the factors in increasing order for prime factorisation)

Instructions

The sides of the cuboid are given as 5 cm, 2 cm and 5 cm. Volume of the cuboid = 5 cm × 2 cm × 5 cm = cm3.
For the prime factorisation of 50, we have: 50 = 2 × 5 × 5
To make it a perfect cube, we must have the prime factorisation: 2 × 2 × 2 × 5 × 5 × 5 i.e. × (2 × 5 × 5) = × volume of the given cuboid
Thus, the required number of cuboids = .