RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

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

Answer 15: Area of a face of cube is given by:
 As2A=s2, where s = Side of the cube
Further, volume of a cube is given by:
V=s3V=s3, where s = Side of the cube
It is given that the area of one face of the cube is 64 m2. Therefore we have:
s2=64s=RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev=8 ms2=64⇒s=64=8 m
Now, volume is given by:
V=s3=83V= 8×8×8 =512 m3V=s3=83⇒V= 8×8×8 =512 m3
Thus, the volume of the cube is 512 m3.

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

Answer 16: Surface area of a cube is given by:
SA=6s2SA=6s2, where s = Side of the cube
Further, volume of a cube is given by:
V=s3V=s3, where s = Side of the cube
It is given that the surface area of the cube is 384 m2. Therefore, we have:
6s2=384s=RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev 8 m6s2=384⇒s=3846=64=8 m

Now, volume is given by:
V=s3=83V= 8×8×8 =512 m3V=s3=83⇒V= 8×8×8 =512 m3
Thus, the required volume is 512 m3. 

Question 17: Evaluate the following:
(i) 
{(52+122)1/2}3(52+122)1/23
(ii)
 {(62+82)1/2}3(62+82)1/23

Answer 17: (i)
To evaluate the value of the given expression, we can proceed as follows:

(52+122)1/2}3

={(25+144)1/2}3

={(169)1/2}3 

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev

={13}3=13×13×13=2197

(ii) To evaluate the value of the given expression, we can proceed as follows: 

{(62+82)1/2}3={(36+64)1/2}3={(100)1/2}3 RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev

={10}3=10×10×10=1000 

Question 18: Write the units digit of the cube of each of the following numbers:
31, 109, 388, 833, 4276, 5922, 77774, 44447, 125125125

Answer 18: Properties:
If a numbers ends with digits 1, 4, 5, 6 or 9, its cube will have the same ending digit.
If a number ends with 2, its cube will end with 8.
If a number ends with 8, its cube will end with 2.
If a number ends with 3, its cube will end with 7.
If a number ends with 7, its cube will end with 3.
From the above properties, we get:
Cube of the number 31 will end with 1.
Cube of the number 109 will end with 9.
Cube of the number 388 will end with 2.
Cube of the number 833 will end with 7.
Cube of the number 4276 will end with 6.
Cube of the number 5922 will end with 8.
Cube of the number 77774 will end with 4.
Cube of the number 44447 will end with 3.
Cube of the number 125125125 will end with 5. 

Question 19: Find the cubes of the following numbers by column method:
(i) 35
(ii) 56
(iii) 72

Answer 19: (i) We have to find the cube of 35 using column method. We have: a=3 and b=5a=3 and b=5

Column I
a3 
Column II
3×a2×b3×a2×b 
Column III
3×a×b23×a×b2 
Column IV
b3 
33=27 
3×a2×b=3×32×5=135 
3×a×b2=3×3×52=225 
53=125 
+15 
+23 
+ 12 
125
42158237
42875

Thus, cube of 35 is 42875.
(ii) We have to find the cube of 56 using column method. We have: a=5 and b=6a=5 and b=6

Column I
a3 
Column II
3×a2×b3×a2×b 
Column III
3×a×b2 
Column IV
b3 
53=125 
3×a2×b=3×52×6=450 
3×a×b2=3×5×62=540 
63=216 
+50 
+56 
  + 21 
216 
175 506 
  561 

175616

Thus, cube of 56 is 175616.
(iii) We have to find the cube of 72 using column method. We have: a=7 and b=2a=7 and b=2

Column I
a3 
Column II
3×a2×b 
Column III
3×a×b23×a×b2 
Column IV
b3 
73=343 
3×a2×b=3×72×2=294 
3×a×b2=3×7×22=84 
23=8 
+30 
+8+08
37330284
373248

Thus, cube of 72 is 373248. 

Question 20: Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728

Answer 20: (i) On factorising 64 into prime factors, we get:
64=2×2×2×2×2×264=2×2×2×2×2×2
On grouping the factors in triples of equal factors, we get:
64={2×2×2}×{2×2×2}64=2×2×2×2×2×2
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.
(ii) On factorising 216 into prime factors, we get:
216=2×2×2×3×3×3216=2×2×2×3×3×3
On grouping the factors in triples of equal factors, we get:
216={2×2×2}×{3×3×3}216=2×2×2×3×3×3
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.
(iii) On factorising 243 into prime factors, we get:
243=3×3×3×3×3243=3×3×3×3×3
On grouping the factors in triples of equal factors, we get:
243={3×3×3}×3×3243=3×3×3×3×3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.
(iv) On factorising 1728 into prime factors, we get:
1728=2×2×2×2×2×2×3×3×31728=2×2×2×2×2×2×3×3×3
On grouping the factors in triples of equal factors, we get:
1728={2×2×2}×{2×2×2}×{3×3×3}1728=2×2×2×2×2×2×3×3×3
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.
Thus, (iii) 243 is the required number, which is not 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.

Answer 21: The only non-perfect cube in question number 20 is 243.
(a) On factorising 243 into prime factors, we get:
243=3×3×3×3×3243=3×3×3×3×3
On grouping the factors in triples of equal factors, we get:
243={3×3×3}×3×3243=3×3×3×3×3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is multiplied by 3, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 243 should be multiplied by 3 to make it a perfect cube.
(b) On factorising 243 into prime factors, we get:
243=3×3×3×3×3243=3×3×3×3×3
On grouping the factors in triples of equal factors, we get:
243={3×3×3}×3×3243=3×3×3×3×3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is divided by (3×3=93×3=9), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 243 should be divided 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.
(iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as 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.

Answer 22: (i) Let the three even natural numbers be 2, 4 and 8.
Cubes of these numbers are: 
23=8, 43=64, 83=51223=8, 43=64, 83=512 
By divisibility test, it is evident that 8, 64 and 5128, 64 and 512 are divisible by 2.
Thus, they are even.
This verifies the statement.
(ii) Let the three odd natural numbers be 3, 9 and 27.
Cubes of these numbers are: 
33=27, 93=729, 273=1968333=27, 93=729, 273=19683 
By divisibility test, it is evident that 27, 729 and 1968327, 729 and 19683 are divisible by 3.
Thus, they are odd.
This verifies the statement.
(iii) Three natural numbers of the form (3n + 1) can be written by choosing