Class 8 RD Sharma Solutions Chapter 4 Cubes and Cube Roots Exercise 4.1 | Set 2 (original) (raw)
Last Updated : 11 Sep, 2024
In this section, students continue exploring the concepts of cubes and cube roots, focusing on calculating cube roots of different numbers and identifying perfect cubes. This exercise emphasizes the prime factorization method for finding cube roots and solving problems related to cubes of both positive and negative integers.
**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 64m 2 .
**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 384m 2 .
**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) {(5** 2 + 12 2 ) 1/2 } 3
****(ii) {(6** 2 + 8 2 ) 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 Ia3 **Column II3×a2×b **Column III3×a×b2 **Column IVb3 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 Ia3 **Column II3×a2×b **Column III3×a×b2 **Column IVb3 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 Ia3 **Column II3×a2×b **Column III3×a×b2 **Column IVb3 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 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
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 n 3 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 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
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 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
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, 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 = 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, 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 > 16
But, (-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
b3/a3 = (ak)3/a3 = a3k3/a3 = k3,
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
72 = 49 and 73 = 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
a2 = 202 = 400 and a3 = 8000 (a3 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
a2 = 1002 = 10000 and a3 = 1003 = 1000000 (a3 doesn’t end with odd number of zeros)
Summary
In this exercise, students further strengthen their understanding of cubes and cube roots by working on problems that require finding cubes of integers and determining cube roots through the prime factorization method. The ability to recognize perfect cubes and calculate cube roots is developed through step-by-step practice with both positive and negative integers.