NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

# NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

## NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1.

### Ex 7.1 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) Let us find the prime factorization of 216:
216 = 2 × 2 × 2 × 3 × 3 × 3
Here, 2 and 3 form a group of three.
Thus, 216 is a perfect cube.

(ii) Let us find the prime factorization of 128:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, 2 is left without making a group of three.
Thus, 128 is not a perfect cube.
(iii) Let us find the prime factorization of 1000:
1000 = 2 × 2 × 2 × 5 × 5 × 5
Here, 2 and 5 form a group of three.
Thus, 1000 is a perfect cube.
(iv) Let us find the prime factorization of 100:
100 = 2 × 2 × 5 × 5
Here, 2 and 5 do not form a group of three.
Thus, 100 is not a perfect cube.
(v) Let us find the prime factorization of 46656:
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, 2 and 3 form the groups of three.
Thus, 46656 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) Let us find the prime factorization of 243:
243 = 3 × 3 × 3 × 3 × 3 = 33 × 3 × 3
Here, one more 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.

(ii) Let us find the prime factorization of 256:
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 2 × 2
Here, one more 2 is required to make 2 × 2 a group of three, i.e., 2 × 2 × 2.

Thus, the required smallest number to be multiplied is 2.

(iii) Let us find the prime factorization of 72:
72 = 2 × 2 × 2 × 3 × 3 = 23 × 3 × 3
Here, one more 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.
(iv) Let us find the prime factorization of 675:
675 = 3 × 3 × 3 × 5 × 5 = 33 × 5 × 5
Here, one more 5 is required to make 5 × 5 a group of three, i.e., 5 × 5 × 5.
Thus, the required smallest number to be multiplied is 5.
(v) Let us find the prime factorization of 100:
100 = 2 × 2 × 5 × 5
Here, the numbers 2 and 5 are required to make 2 × 2 × 5 × 5 a group of three, i.e., 2 × 2 × 2 × 5 × 5 × 5.
Thus, the required smallest number to be multiplied is 2 × 5 = 10.

### Ex 7.1 Class 8 Maths Question 3.

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 92
(v) 704

Solution:
(i) Let us find the prime factorization of 81:
81 = 3 × 3 × 3 × 3 = 33 × 3
Here, the number 3 is not in group of three. Thus, 3 is the number by which 81 is divided to make it a perfect cube, i.e., 81 ÷ 3 = 27, which is a perfect cube.
Thus, 3 is the required smallest number.

(ii) Let us find the prime factorization of 128:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 2
Here, the number 2 is not in group of three. Thus, 2 is the smallest number by which 128 is divided to make it a perfect cube, i.e., 128 ÷ 2 = 64, which is a perfect cube.
Thus, 2 is the required smallest number.
(iii) Let us find the prime factorization of 135:
135 = 3 × 3 × 3 × 5 = 33 × 5
Here, the number 5 is not in group of three. Thus, 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.
(iv) Let us find the prime factorization of 192:
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 23 × 23 × 3
Here, the number 3 is not in group of three. Thus, 3 is the smallest number by which 192 is divided to make it a perfect cube, i.e., 192 ÷ 3 = 64, which is a perfect cube.
Thus, 3 is the required smallest number.
(v) Let us find the prime factorization of 704:
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 23 × 23 × 11
Here, the number 11 is not in group of three. Thus, 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 of plasticine are 5 cm, 2 cm and 5 cm.
Volume of the cuboid = 5 cm × 2 cm × 5 cm = 50 cm3
Let us find the prime factorization of 50:
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 is 20.

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