Ex 6.1 Class 8 Maths Question 1.

Which of the following numbers are not perfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Solution:

(i) Prime factorisation of 216 is:

216 = 2 × 2 × 2 × 3 × 3 × 3

In the above factorisation, 2 and 3 have formed a group of three.

Thus, 216 is a perfect cube.

(iii) Prime factorisation of 1000, is:

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, no number is left for making a group of three.

Thus, 1000 is a perfect cube.

Ex 7.1 Class 8 Maths Question 2.

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Solution:

(i) Prime factorisation of 243, is:

243 = 3 × 3 × 3 × 3 × 3 = 3^{3} × 3 × 3

Here, number 3 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.

(iii) Prime factorisation of 72, is:

72 = 2 × 2 × 2 × 3 × 3 = 2^{3} × 3 × 3

Here, a number 3 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.

(v) Prime factorisation of 100, is:

100 = 2 × 2 × 5 × 5

Here, number 2 and 5 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 = 10.

(iii) Prime factorisation of 135 is:

135 = 3 × 3 × 3 × 5 = 3^{3} × 5

Here, 5 is the smallest number by which 135 is divided to make a perfect cube,

i.e., 135 ÷ 5 = 27 which is a perfect cube.

Thus, 5 is the required smallest number.

(v) Prime factorisation of 704 is:

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 2^{3} × 2^{3} × 11

Here, 11 is the smallest number by which 704 is divided to make it a perfect cube,

i.e., 704 ÷ 11 = 64 which is a perfect cube.

Thus, 11 is the required smallest number.

Ex 7.1 Class 8 Maths Question 4.

Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?

Solution:

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 = 50 cm^{3}

For the prime factorisation of 50, we have

50 = 2 × 5 × 5

To make it a perfect cube, we must have

2 × 2 × 2 × 5 × 5 × 5

= 20 × (2 × 5 × 5)

= 20 × volume of the given cuboid

Thus, the required number of cuboids = 20