# Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.3

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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.3

Question 1.
Fill in the blanks:
(i) The ones digits in the cube of 73 is……….
(ii) The maximum number of digits in the cube of a two digit number is……….
(iii) The cube root of 540 × 50 is………..
(iv) The cube root of 0.000004913 is………..
(v) The number to be added to 3333 to make it a perfect cube is………….
Solution:
(i) 7
(ii) 6
(iii) 30
(iv) 0.017
(v) 42

Question 2.
Say True or False:
(i) The cube of 0.0012 is 0.000001728.
(ii) The cube root of 250047 is 63.
(iii) 79570 is not a perfect cube.
Solution:
(i) false
(ii) true
(iii) true

Question 3.
Show that 1944 is not a perfect cube.
Solution:

1994 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2³ × 3³ × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.

Question 4.
Find the cube root of 24 × 36 × 80 × 25.
Solution:

Question 5.
Find the smallest number by which 10985 should be divided so that the quotient is a perfect cube.
Solution:

We have 10985 = 5 × 13 × 13 × 13
= 5 × 13 × 13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.

Question 6.
Find two smallest perfect square numbers which when multiplied together gives a perfect cube number.
Solution:
Consider the numbers 22 and 42
The numbers are 4 and 16.
Their product 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16

Question 7.
If the cube of a squared number is 729, find the square root of that number.
Solution:
729 = 3 × 3 × 3 × 3 × 3 × 3
(729)1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]

we have to find out √3, √3 = 1.732

Question 8.
What is the square root of cube root of 46656?
Solution:
We have to find out $$\sqrt{(\sqrt[3]{46656})}$$
First we will find $$\sqrt[3]{46656}$$

$$\sqrt[3]{46656}$$ = (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3)1/3
$$\sqrt[3]{46656}$$ = 2 × 2 × 3 × 3
$$\sqrt[3]{46656}$$ = 2² × 3² = 36
Now $$\sqrt{(\sqrt[3]{46656})}$$ = $$\sqrt{36}$$ = $$\sqrt{2^2×3^2 }$$ = 2 × 3 = 6
∴ The required is 6.

Question 9.
Find the cube root of 729 and 6859 by prime factorisation.
Solution:

(i) $$\sqrt[3]{729}$$ = $$\sqrt[3]{3 × 3 × 3 × 3 × 3 × 3}$$
= 3 × 3
$$\sqrt[3]{729}$$ = 9

(ii) $$\sqrt[3]{6859}$$ = $$\sqrt[3]{19 × 19 × 19 }$$
$$\sqrt[3]{6859}$$ = 19

Question 10.
Find the smallest number by which 200 should be multiplied to make it a perfect cube.
Solution:
We find 200 = 2 × 2 × 2 × 5 × 5
Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.

200 × 5 = (2 × 2 × 2 × 5 × 5) × 5
1000 = 2 × 2 × 2 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.

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