# 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:**

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(i)675Finding the prime factors of 675

675 = 3 × 3 × 3 × 5 × 5

= 3

^{3}× 5^{2}Hence, 675 is not a perfect cube.

We divide it by 5

^{2}= 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)8640Finding the prime factors of 8640

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

= 2

^{3}× 2^{3}× 3^{3}× 5Hence, 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)1600Finding the prime factors of 1600

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 2

^{3}× 2^{3}× 5^{2}Hence, 1600 is not a perfect cube.

We divide it by 5

^{2}= 25 to make the quotient a perfect cube, which gives 64 as quotient which is a perfect cubeHence, 25 is the required smallest number.

(iv) 8788Finding the prime factors of 8788

8788 = 2 × 2 × 13 × 13 × 13

= 2

^{2}× 13^{3}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) 7803Finding the prime factors of 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3

^{3}× 17^{2}Hence, 7803 is not a perfect cube.

We divide it by 17

^{2}= 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)107811Finding the prime factors of 107811

107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 3

^{3}× 11^{3}× 3Hence, 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) 35721Finding the prime factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 3

^{3}× 3^{3}× 7^{2}Hence, 35721 is not a perfect cube.

We divide it by 7

^{2}= 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)243000Finding the prime factors of 243000

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

= 2

^{3}× 3^{3}× 5^{3}× 3^{2}Hence, 243000 is not a perfect cube.

We divide it by 3

^{2}= 9 to make the quotient a perfect cube, which gives 27000 as quotient which is a perfect cubeHence, 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 = a

^{3}Trebling the number = 3 × a = 3a

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

^{ 3}= 27a^{3}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 = a

^{3}Now, when the number is multiplied by 3

The new number becomes = 3a

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

^{ 3}= 27a^{3}This implies, number will become 27 times the cube of the number.

(ii)4?Suppose the number is a

Therefore, its cube is = a

^{3}Now, when the number is multiplied by 4

The new number becomes = 4a

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

^{ 3}= 64a^{3}This implies, number will become 64 times the cube of the number.

(iii) 5?Suppose the number is a

Therefore, its cube is = a

^{3}Now, when the number is multiplied by 5

The new number becomes = 5a

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

^{ 3}= 125a^{3}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 64m**^{2}.

^{2}.

**Solution:**

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

^{2}Suppose the length of edge of cube be ‘a’ metres

a

^{2}= 64a = √ 64

= 8m

Now, volume of cube = a

^{3}a

^{3}= 8^{3}= 8 × 8 × 8= 512m

^{3}Hence, Volume of a cube is 512m

^{3}

**Question 16. Find the volume of a cube whose surface area is 384m**^{2}.

^{2}.

**Solution:**

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

^{2}Suppose the length of each edge of cube be ‘a’ meters

6a

^{2}= 384a

^{2}= 384/6= 64

a = √64

= 8m

Now, volume of cube = a

^{3}a

^{3}= 8^{3}= 8 × 8 × 8= 512m

^{3}Hence, Volume of a cube is 512m

^{3}

**Question 17. Evaluate the following:**

**(i) {(5 ^{2} + 12^{2})^{1/2}}^{3}**

**(ii) {(6**

^{2}+ 8^{2})^{1/2}}^{3}**Solution:**

(i){(5^{2}+ 12^{2})^{1/2}}^{3}From the above equation we get,

{(25 + 144)

^{1/2}}^{3}{(169)

^{1/2}}^{3}{(13

^{2})^{1/2}}^{3}(13)

^{3}2197

(ii){(6^{2}+ 8^{2})^{1/2}}^{3}From the above equation we get,

{(36 + 64)

^{1/2}}^{3}{(100)

^{1/2}}^{3}{(10

^{2})^{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:**

31We 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 = 1

^{3}= 1Hence, the unit digit of cube of 31 is always 1

109We 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 = 9

^{3}= 729Hence, the unit digit of cube of 109 is always 9

388We 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 = 8

^{3}= 512Hence, the unit digit of cube of 388 is always 2

4276We 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 = 6

^{3}= 216Hence, the unit digit of cube of 4276 is always 6

5922We 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 = 2

^{3}= 8Hence, the unit digit of cube of 5922 is always 8

77774We 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 = 4

^{3}= 64Hence, the unit digit of cube of 77774 is always 4

44447We 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 = 7

^{3}= 343Hence, the unit digit of cube of 44447 is always 3

125125125We 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 = 5

^{3}= 125Hence, 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)35We have, a = 3 and b = 5

Column Ia

^{3}

Column II3×a

^{2}×b

Column III3×a×b

^{2}

Column IVb

^{3}3 ^{3}= 273×9×5 = 135 3×3×25 = 225 5 ^{3}= 125+15 +23 +12 125 42 158 237 42 8 7 5 Hence, the cube of 35 is 42875

(ii) 56We have, a = 5 and b = 6

Column Ia

^{3}

Column II3×a

^{2}×b

Column III3×a×b

^{2}

Column IVb

^{3}5 ^{3}= 1253×25×6 = 450 3×5×36 = 540 6 ^{3}= 216+50 +56 +21 126 175 506 561 175 6 1 6 Hence, the cube of 56 is 175616

(iii) 72

Column Ia

^{3}

Column II3×a

^{2}×b

Column III3×a×b

^{2}

Column IVb

^{3}7 ^{3}= 3433×49×2 = 294 3×7×4 = 84 2 ^{3}= 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) 64Finding the prime factors of 64

64 = 2 × 2 × 2 × 2 × 2 × 2

= 2

^{3}× 2^{3}= 4

^{3}Therefore, it’s a perfect cube.

(ii) 216Finding the prime factors of 216

216 = 2 × 2 × 2 × 3 × 3 × 3

= 2

^{3}× 3^{3}= 6

^{3}Therefore, it’s a perfect cube.

(iii) 243Finding the prime factors of 243

243 = 3 × 3 × 3 × 3 × 3

= 3

^{3}× 3^{2}Therefore, it’s not a perfect cube.

(iv) 1728Finding the prime factors of 1728

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

= 2

^{3}× 2^{3}× 3^{3}= 12

^{3}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 = 3

^{3}× 3^{2}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 = 3

^{3}× 3^{2}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 n ^{3} is also even.**

**(ii) If n is odd, then n**

^{3}is also odd.**(ii) If n leaves remainder 1 when divided by 3, then n**

^{3}also leaves 1 as the**remainder when divided by 3.**

**(iv) If a natural number n is of the form 3p+2 then n**

^{3}also a number of the same type.**Solution:**

(i) If n is even, then n^{3}is also even.Consider three even natural numbers 2, 4, 6

Hence, cubes of 2, 4 and 6 are

2

^{3}= 84

^{3}= 646

^{3}= 216Therefore, we can see that all cubes are even in nature.

Hence proved.

(ii)If n is odd, then n^{3}is also odd.Consider three odd natural numbers 3, 5, 7

Hence, cubes of 3, 5 and 7 are

3

^{3}= 275

^{3}= 1257

^{3}= 343Therefore, 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 n^{3}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

4

^{3}= 647

^{3}= 34310

^{3}= 1000We 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 n^{3}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

5

^{3}= 1258

^{3}= 51211

^{3}= 1331Let’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, a ^{3} is always greater than a^{2}.**

**(vi) If a and b are integers such that a**

^{2}>b^{2}, then a^{3}>b^{3}.**(vii) If a divides b, then a**

^{3}divides b^{3}.**(viii) If a**

^{2}ends in 9, then a^{3}ends in 7.**(ix) If a**

^{2}ends in an even number of zeros, then a^{3}ends in 25.**(x) If a**

^{2}ends in an even number of zeros, then a^{3}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 = 2

^{3}× 7^{2}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 = 2

^{3}× 2^{3}× 3^{3}× 5Therefore, 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, a^{3}is always greater than a^{2}.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 a^{2}>b^{2}, then a^{3}>b^{3}.Statement is False.

Because, in the case of the negative integers,

(-5)

^{2}> (-4)^{2}= 25 > 16But, (-5)

^{3}> (-4)^{3}= -125 > -64 is not true.

(vii) If a divides b, then a^{3}divides b^{3}.Statement is True.

If a divides b

b/a = k, so b=ak

b

^{3}/a^{3}= (ak)^{3}/a^{3}= a^{3}k^{3}/a^{3}= k^{3},For each value of b and a its true.

(viii) If a^{2}ends in 9, then a^{3}ends in 7.Statement is False.

Let a = 7

7

^{2}= 49 and 7^{3}= 343

(ix) If a^{2}ends in an even number of zeros, then a^{3}ends in 25.Statement is False.

Since, when a = 20

a

^{2}= 20^{2}= 400 and a^{3}= 8000 (a^{3}doesn’t end with 25)

(x) If a^{2}ends in an even number of zeros, then a^{3}ends in an odd number of zeros.Statement is False.

Since, when a = 100

a

^{2}= 100^{2}= 10000 and a^{3}= 100^{3}= 1000000 (a^{3}doesn’t end with odd number of zeros)