Class 8 RD Sharma Solutions- Chapter 4 Cubes and Cube Roots – Exercise 4.1 | Set 2

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 = 27	3×9×5 = 135	3×3×25 = 225	53 = 125
+15	+23	+12	125
42	158	237
42	8	7	5

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 = 125	3×25×6 = 450	3×5×36 = 540	63 = 216
+50	+56	+21	126
175	506	561
175	6	1	6

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 = 343	3×49×2 = 294	3×7×4 = 84	23 = 8
+30	+8	+0	8
373	302	84
373	2	4	8

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)