RBSE Class 8 Maths Important Questions Chapter 7 Cube and Cube Roots

Rajasthan Board RBSE Class 8 Maths Important Questions Chapter 7 Cube and Cube Roots Important Questions and Answers.

RBSE Class 8 Maths Chapter 7 Important Questions Cube and Cube Roots

Multiple Choice Questions

Question 1.
The smallest Hardi-Ramanujan number is
(a) 729
(b) 1729
(c) 13832
(d) 4104
Answer:
(b) 1729

Question 2.
How many perfect cube numbers are there in between 1 to 1000?
(a) 10
(b) 18
(c) 25
(d) 52
Answer:
(a) 10

Question 3.
How many consecutive odd numbers will be required to get 103?
(a) 5
(b) 8
(c) 10
(d) 100
Answer:
(c) 10

Question 4.
The cube of 8 is
(a) 64
(b) 240
(c) 256
(d) 512
Answer:
(d) 512

Question 5.
What is the cube root of 13824?
(a) 24
(b) 56
(c) 18
(d) 124
Answer:
(a) 24

Question 6.
The least natural number by 392 should be multiplied to get a number which is a perfect cube is
(a) 2
(b) 3
(c) 5
(d) 7
Answer:
(d) 7

Fill in the blanks

Question 1.
Cubes of even numbers are ....................
Answer:
even

Question 2.
The unit’s digit of the cube of 18 is ..................
Answer:
2

Question 3.
The operation of finding the .................. is the opposite operation of finding the cube.
Answer:
cube root

Question 4.
The cube root is denoted by the ................. sign.
Answer:
\(\sqrt[3]{ }\)

Question 5.
The cube root of 857 375 is ..................
Answer:
95

Very Short Answer Type Questions

Question 1.
What do you mean by Hardi- Ramanujan number?
Answer:
The number expressible as the sum of two cubes in two different ways, are called Hardi-Ramanujan numbers. For eg. 1729, 4104, 13832 etc.

Question 2.
What do you mean by the cube of a number?
Answer:
The number obtained after multiplying a number thrice by itself is called cube of that number.

Question 3.
Express 9³ as the sum of consecutive odd numbers?
Answer:
73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89 = 729 = 9³

Question 4.
Find the cube root of 8000 by prime factorization method.
Answer:
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
∴ \(\sqrt[3]{8000}\) = 2 × 2 × 5 = 20.

Question 5.
Find the least number by which 1188 should be divided so as to get a perfect cube number.
Answer:
Prime factors of 1188 = 2 × 2 × 3 × 3 × 3 × 11
Remaining factors after obtaining the triples, are 2 × 2 × 11. So we need to divide 1188 by 2 × 2 × 11 = 44 to get a perfect cube.

Short Answer Type Questions

Question 1.
Suman makes a cuboid of soil of sides 15 cm, 30 cm and 15 cm. How many such cuboids will he need to form a cube?
Answer:
Volume of a cuboid = 15 × 30 × 15
= 3 × 5 × 2 × 3 × 5 × 3 × 5
= 2 × 3 × 3 × 3 × 5 × 5 × 5
In above. By Prime factorisation 2 comes only one time there fore such, cuboid will he need to form a cube = 2 × 2 = 4

Question 2.
Mansi has a cuboidal box whose sides are 5 cm, 3 cm and 5 cm. How many such cuboidal boxes will be required for making one cubical box?
Answer:
Volume of cuboidal box = 5 × 3 × 5 = 5 × 5 × 3
To make it a cube, we require 5 × 3 × 3 = 45 i.e., 45 such cuboids