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Ncert Solutions for cube roots Class 8 CBSE Part 1



In this page we have NCERT book Solutions for Class 8th Maths:cube roots for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

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

=a3 where a is the side of the cube

So that means we need  2×2×5=20 cuboid to make the cube


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