You can Download KSEEB Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1 Questions and Answers helps you to revise the complete syllabus.
KSEEB Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1
Question 1.
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Solution:
(i) 216 = 2 × 2 × 2 × 3 × 3 × 3
= 23 × 33
= (2 × 3)3
= (6)3
Hence, 216 is a perfect cube.
(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 33 × 2
Here, 2 is not in a group of three.
Hence, 128 is not a perfect cube.
(iii) 1000 = 2 × 2 × 2 × 5 × 5 × 5
= 23 × 53
= (10)3
Hence, 1000 is a perfect cube.
(iv) 100 = 2 × 2 × 5 × 5
Here 2 and 5 are not in a group of three.
Hence, 100 is not a perfect cube.
(v) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 9 × 9 × 9
= 23 × 23 × 93
= (2 × 2 × 9)3
= (36)3
Hence 46656 is a perfect cube.
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) 243 = 3 × 3 × 3 × 3 × 3
The prime factorisation shows that 3 × 3 is not in a pair of three.
Hence to make it a perfect cube, it should be multiplied by 3.
∴ 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3
Hence, 729 is a perfect cube.
(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
The prime factorisation shows that 2 × 2 is not in a pair of three.
Hence to make it a perfect cube, it should be multiplied by 2.
∴ 256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Hence, 512 is a perfect cube.
(iii) 72 = 2 × 2 × 2 × 3 × 3
The prime factorisation shows that 3 × 3 is not in a pair of three.
Hence to make it a perfect cube, it should be multiplied by 3.
∴ 72 × 3 = 2 × 2 × 2 × 3 × 3 × 3
Hence, 216 is a perfect cube.
(iv) 675 = 3 × 3 × 3 × 5 × 5
The prime factorisation shows that 5 × 5 is not in a pair of three.
Hence to make it a perfect cube, it should be multiplied by 5.
∴ 675 × 5 = 3 × 3 × 3 × 5 × 5 × 5
Hence, 3375 is a perfect cube.
(v) 100 = 2 × 2 × 5 × 5
The prime factorisation shows that 2 and 5 are not in a pair of three.
Hence to make it a perfect cube, it should be multiplied by 2 and 5.
∴ 100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5
Hence, 1000 is a perfect cube.
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) 192
(v) 704
Solution:
(i) 81 = 3 × 3 × 3 × 3
3 does not form a pair of three.
Hence, 81 should be divided by 3 to make it a perfect cube.
(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
2 does not form a pair of three.
So, 128 should be divided by 2 to make it a perfect cube.
(iii) 135 = 3 × 3 × 3 × 5
5 does not form a pair of three.
So, 135 should be divided by 5 to make it a perfect cube.
(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
3 does not form a pair of three.
So, 192 should be divided by 3 to make it a perfect cube.
(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
11 does not form a pair of three.
So, 704 should be divided by 11 to make it a perfect cube.
Question 4.
Parikshit makes a cuboid of plasticine on sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution:
Length of cuboid = 5 cm
Breadth of cuboid = 2 cm
Height of cuboid = 5 cm
Volume of cuboid = 5 × 5 × 2 = 50 cm3
Since 50 is not a perfect cube. To make it a perfect cube, it should be multiplied by 5 (is not in pair of three) and 2 × 2.
Hence, 5 × 2 × 2 = 20 cubes needed.