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Class 8 RD Sharma Solutions- Chapter 4 Cubes and Cube Roots – Exercise 4.1 | Set 2
  • Last Updated : 06 Apr, 2021

Chapter 4 Cubes and Cube Roots – Exercise 4.1 | Set 1

Question 12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube?

(i) 675 (ii) 8640
(iii) 1600 (iv) 8788
(v) 7803 (vi) 107811
(vii) 35721 (viii) 243000

Solution:

(i) 675

Finding the prime factors of 675

675 = 3 × 3 × 3 × 5 × 5



= 33 × 52

Hence, 675 is not a perfect cube.

We divide it by 52 = 25 to make the quotient a perfect cube, that gives 27 as quotient which is a perfect cube.

Hence, 25 is the required smallest number.

(ii) 8640

Finding the prime factors of 8640

8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

= 23 × 23 × 33 × 5



Hence, 8640 is not a perfect cube.

We divide it by 5 to make the quotient a perfect cube, which gives 1728 as quotient which is a perfect cube.

Hence, 5 is the required smallest number.

(iii) 1600

Finding the prime factors of 1600

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 23 × 23 × 52

Hence, 1600 is not a perfect cube.

We divide it by 52 = 25 to make the quotient a perfect cube, which gives 64 as quotient which is a perfect cube

Hence, 25 is the required smallest number.

(iv) 8788

Finding the prime factors of 8788

8788 = 2 × 2 × 13 × 13 × 13

= 22 × 133

Hence, 8788 is not a perfect cube.

We divide it by 4 to make the quotient a perfect cube, which gives 2197 as quotient which is a perfect cube

Hence, 4 is the required smallest number.

(v) 7803

Finding the prime factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 33 × 172

Hence, 7803 is not a perfect cube.

We divide it by 172 = 289 to make the quotient a perfect cube, which gives 27 as quotient which is a perfect cube.

Hence, 289 is the required smallest number.

(vi) 107811

Finding the prime factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 33 × 113 × 3

Hence, 107811 is not a perfect cube.

We divide it by 3 to make the quotient a perfect cube , which gives 35937 as quotient which is a perfect cube.

Hence, 3 is the required smallest number.

(vii) 35721

Finding the prime factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 33 × 33 × 72

Hence, 35721 is not a perfect cube.

We divide it by 72 = 49 to make the quotient a perfect cube, which gives 729 as quotient which is a perfect cube.

Hence, 49 is the required smallest number.

(viii) 243000

Finding the prime factors of 243000

243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5

= 23 × 33 × 53 × 32

Hence, 243000 is not a perfect cube.

We divide it by 32 = 9 to make the quotient a perfect cube, which gives 27000 as quotient which is a perfect cube

Hence, 9 is the required smallest number.

Question 13. Prove that if a number is trebled then its cube is 27 time the cube of the given number.

Solution:

Suppose the number is a

Therefore, its cube is = a3

Trebling the number = 3 × a = 3a

Therefore, the cube of new number = (3a) 3 = 27a3

This implies, the new cube is 27 times the original cube.

Hence, proved.

Question 14. What happens to the cube of a number if the number is multiplied by

(i) 3?
(ii) 4?
(iii) 5?

Solution:

(i) 3?

Suppose the number is a

Therefore, its cube is = a3

Now, when the number is multiplied by 3

The new number becomes = 3a

Hence, the cube of new number is = (3a) 3 = 27a3

This implies, number will become 27 times the cube of the number.

(ii) 4?

Suppose the number is a

Therefore, its cube is = a3

Now, when the number is multiplied by 4

The new number becomes = 4a

Hence, the cube of new number is = (4a) 3 = 64a3

This implies, number will become 64 times the cube of the number.

(iii) 5?

Suppose the number is a

Therefore, its cube is = a3

Now, when the number is multiplied by 5

The new number becomes = 5a

Hence, the cube of new number is = (5a) 3 = 125a3

This implies number will become 125 times the cube of the number.

Question 15. Find the volume of a cube, one face of which has an area of 64m2.

Solution:

It is given that the area of one face of the cube is = 64 m2

Suppose the length of edge of cube be ‘a’ metres

a2 = 64

a = √ 64

= 8m

Now, volume of cube = a3

a3 = 83 = 8 × 8 × 8

= 512m3

Hence, Volume of a cube is 512m3

Question 16. Find the volume of a cube whose surface area is 384m2.

Solution:

It is given that the surface area of cube is = 384 m2

Suppose the length of each edge of cube be ‘a’ meters

6a2 = 384

a2 = 384/6

= 64

a = √64

= 8m

Now, volume of cube = a3

a3 = 83 = 8 × 8 × 8

= 512m3

Hence, Volume of a cube is 512m3

Question 17. Evaluate the following:

(i) {(52 + 122)1/2}3
(ii) {(62 + 82)1/2}3

Solution:

(i) {(52 + 122)1/2}3

From the above equation we get,

{(25 + 144)1/2}3

{(169)1/2}3

{(132)1/2}3

(13)3

2197

(ii) {(62 + 82)1/2}3

From the above equation we get,

{(36 + 64)1/2}3

{(100)1/2}3

{(102)1/2}3

(10)3

1000

Question 18. Write the units digit of the cube of each of the following numbers:
31, 109, 388, 4276, 5922, 77774, 44447, 125125125

Solution:

31

We will find the cube of unit digit only to find unit digit of cube of a number 

Unit digit of 31 is 1

Cube of 1 = 13 = 1

Hence, the unit digit of cube of 31 is always 1

109

We will find the cube of unit digit only to find unit digit of cube of a number

Unit digit of 109 is = 9

Cube of 9 = 93 = 729

Hence, the unit digit of cube of 109 is always 9

388

We will find the cube of unit digit only to find unit digit of cube of a number.

Unit digit of 388 is = 8

Cube of 8 = 83 = 512

Hence, the unit digit of cube of 388 is always 2

4276

We will find the cube of unit digit only to find unit digit of cube of a number.

Unit digit of 4276 is = 6

Cube of 6 = 63 = 216

Hence, the unit digit of cube of 4276 is always 6

5922

We will find the cube of unit digit only to find unit digit of cube of a number.

Unit digit of 5922 is = 2

Cube of 2 = 23 = 8

Hence, the unit digit of cube of 5922 is always 8

77774

We will find the cube of unit digit only to find unit digit of cube of a number

Unit digit of 77774 is = 4

Cube of 4 = 43 = 64

Hence, the unit digit of cube of 77774 is always 4

44447

We will find the cube of unit digit only to find unit digit of cube of a number.

Unit digit of 44447 is = 7

Cube of 7 = 73 = 343

Hence, the unit digit of cube of 44447 is always 3

125125125

We will find the cube of unit digit only to find unit digit of cube of a number.

Unit digit of 125125125 is = 5

Cube of 5 = 53 = 125

Hence, the unit digit of cube of 125125125 is always 5

Question 19. Find the cubes of the following numbers by column method:

(i) 35
(ii) 56
(iii) 72

Solution:

(i) 35

We have, a = 3 and b = 5

Column I

a3

Column II

3×a2×b

Column III

3×a×b2

Column IV

b3

33 = 273×9×5 = 1353×3×25 = 22553 = 125
+15+23+12125
42158237 
42875

Hence, the cube of 35 is 42875

(ii) 56

We have, a = 5 and b = 6

Column I

a3

Column II

3×a2×b

Column III

3×a×b2

Column IV

b3

53 = 1253×25×6 = 4503×5×36 = 54063 = 216
+50+56+21126
175506561 
175616

Hence, the cube of 56 is 175616

(iii) 72

Column I

a3

Column II

3×a2×b

Column III

3×a×b2

Column IV

b3

73 = 3433×49×2 = 2943×7×4 = 8423 = 8
+30+8+08
37330284 
373248

Hence, the cube of 72 is 373248

Question 20. Which of the following numbers are not perfect cubes?

(i) 64
(ii) 216
(iii) 243
(iv) 1728

Solution:

(i) 64

Finding the prime factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2

= 23 × 23

= 43

Therefore, it’s a perfect cube.

(ii) 216

Finding the prime factors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3

= 23 × 33

= 63

Therefore, it’s a perfect cube.

(iii) 243

Finding the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3

= 33 × 32

Therefore, it’s not a perfect cube.

(iv) 1728

Finding the prime factors of 1728

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 23 × 23 × 33

= 123

Therefore, it’s a perfect cube.

Question 21. For each of the non-perfect cubes in Q. No 20 find the smallest number by which it must be

(a) Multiplied so that the product is a perfect cube.
(b) Divided so that the quotient is a perfect cube.

Solution:

In the previous question the only non-perfect cube was = 243

(a) Multiplied so that the product is a perfect cube.

Finding the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3 = 33 × 32

Therefore, we should multiply it by 3 to make it a perfect cube.

(b) Divided so that the quotient is a perfect cube.

Finding the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3 = 33 × 32

Therefore, we have to divide it by 9 to make it a perfect cube.

Question 22. By taking three different, values of n verify the truth of the following statements:

(i) If n is even, then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(ii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as the remainder when divided by 3.
(iv) If a natural number n is of the form 3p+2 then n3 also a number of the same type.

Solution:

(i) If n is even, then n3 is also even.

Consider three even natural numbers 2, 4, 6

Hence, cubes of 2, 4 and 6 are

23 = 8

43 = 64

63 = 216

Therefore, we can see that all cubes are even in nature.

Hence proved.

(ii) If n is odd, then n3 is also odd.

Consider three odd natural numbers 3, 5, 7

Hence, cubes of 3, 5 and 7 are

33 = 27

53 = 125

73 = 343

Therefore, we can see that all cubes are odd in nature.

Hence proved.

(iii) If n is divided by 3 leaves remainder of 1, then when n3 is divided by 3 also leaves 1 as remainder.

Consider 4, 7 and 10 as three natural numbers of the form (3n+1)

Hence, cube of 4, 7, 10 are

43 = 64

73 = 343

103 = 1000

We get 1 as remainder in each case if we divide these numbers by 3.

Hence proved.

(iv) If a natural number n is of the form 3p+2 then n3 also a number of the same type.

Consider 5, 8 and 11 as three natural numbers of the form (3p+2)

Hence, cube of 5, 8 and 10 are

53 = 125

83 = 512

113 = 1331

Let’s write these cubes in form of (3p + 2)

125 = 3 × 41 + 2

512 = 3 × 170 + 2

1331 = 3 × 443 + 2

Hence proved.

Question 23. Write true (T) or false (F) for the following statements:

(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a3 is always greater than a2.
(vi) If a and b are integers such that a2>b2, then a3>b3.
(vii) If a divides b, then a3 divides b3.
(viii) If a2 ends in 9, then a3 ends in 7.
(ix) If a2 ends in an even number of zeros, then a3 ends in 25.
(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.

Solution:

(i) 392 is a perfect cube.

Finding the prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 23 × 72

Therefore, the statement is False.

(ii) 8640 is not a perfect cube.

Finding the prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 23 × 23 × 33 × 5

Therefore, the statement is True

(iii) No cube can end with exactly two zeros.

Statement is True.

As a perfect cube always have zeros in multiple of 3.

(iv) There is no perfect cube which ends in 4.

It is known that 64 is a perfect cube = 4 × 4 × 4 and ends with 4.

Therefore, the statement is False.

(v) For an integer a, a3 is always greater than a2.

Statement is False in the case of negative integers ,

(-2)2 = 4 and (-2)3 = -8

(vi) If a and b are integers such that a2>b2, then a3>b3.

Statement is False.

Because, in the case of the negative integers,

(-5)2 > (-4)2 = 25 > 16

But, (-5)3 > (-4)3 = -125 > -64 is not true.

(vii) If a divides b, then a3 divides b3.

Statement is True.

If a divides b

b/a = k, so b=ak

b3/a3 = (ak)3/a3 = a3k3/a3 = k3,

For each value of b and a its true.

(viii) If a2 ends in 9, then a3 ends in 7.

Statement is False.

Let a = 7

72 = 49 and 73 = 343

(ix) If a2 ends in an even number of zeros, then a3 ends in 25.

Statement is False.

Since, when a = 20

a2 = 202 = 400 and a3 = 8000 (a3 doesn’t end with 25)

(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.

Statement is False.

Since, when a = 100

a2 = 1002 = 10000 and a3 = 1003 = 1000000 (a3 doesn’t end with odd number of zeros)

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