NCERT Solutions for Class 8 Maths Chapter 6 Cube and Cube roots Exercise 6.1
In this page we have NCERT book Solutions for Class 8 Maths Chapter 6 Cube and cube roots for EXERCISE 6.1 . This exercise has questions about cubes of the nunber, How to find if number is perfect cube or not, Which number we can multiply or divide to make it perfect cube. Hope you like them and do not forget to like , social share
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NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1
Question 1
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656 Answer: (i) Converting the number into common factors
216 = 2 x 2 x 2 x 27
= 2 x 2 x 2 x 3 x 3 x 3
As number of 2s and 3s is 3 in the factorization so it is a perfect cube
Therefore 3√216 = 2×3=6 (ii)128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Number of 2s is 7 and 7 is not divisible by three so 128 is not a perfect cube (iii) 1000 = 2 x 2 x 2 x 5 x 5 x 5
Number of 2s and 5s is 3 each so 1000 is a perfect cube.
Therefore 3√1000 = 2×5=10 (iv) 100 = 2 x 2 x 5 x 5
Number of 2s and 5s is 2 each and not 3 so 100 is not a perfect cube. (v) 46656
= 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Number of 2s and 3s is 6 each and 6 is divisible by 3 so 46656 is a perfect cube
Therefore 3√46656= 2×2×3×3=36
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 Answer Converting the number into common factors (i) 243 = 3 x 3 x 3 x 3 x 3
Number of 3s is 5, so we need to another 3 in the factorization to make 243 a perfect cube. 243 multiplied by 3 will be a perfect cube. (ii) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Number of 2s is 8 so 256 needs to be multiplied by 2 to become a perfect cube. (iii) 72 = 2 x 2 x 2 x 3 x 3
Number of 2s is 3 and that of 3s is 2, so 72 needs to be multiplied by 3 to become a perfect cube. (iv) 675 = 5 x 5 x 27 = 5 x 5 x 3 x 3 x 3
675 needs to be multiplied by 5 to become a perfect cube. (v) 100 = 10 x 10
100 needs to be multiplied by 10 to become 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 Answer Converting the number into common factors (i) 81 = 3 x 3 x 3 x 3
So we have 4 3’s,Therefore
81 needs to be divided by 3 to become a perfect cube. (ii) 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
So we have 7 2’s,Therefore
128 needs to be divided by 2 to become a perfect cube. (iii) 135 = 5 x 3 x 3 x 3
So we have 1 5’s and 3 3’s Therefore
135 needs to be divide by 5 to become a perfect cube. (iv) 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
So we have 6 2’s and 1 3’s Therefore
192 needs to be divided by 3 to become a perfect cube. (v) 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
So we have 6 2’s and 1 11’s Therefore
704 needs to be divided by 11 to become a perfect cube.
Question 4
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? Answer:
The volume of the cuboid =2×5×5 cm3
If we multiply this expression by 2×2×5, The it become perfect cube.
Now Volume of cube is given by
V=a3 where a is the side of the cube
So that means we need 2×2×5=20 cuboid to make the cube
Summary
NCERT Solutions for Class 8 Maths Chapter 6 Cube and Cube roots Exercise 6.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download this assignment as pdf
This chapter 6 has total 2 Exercise 6.1 and 6.2. This is the first exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below
