Rs Aggarwal 2019 Solutions for Class 8 Math Chapter 4 Cubes And Cube Roots are provided here with simple step-by-step explanations. These solutions for Cubes And Cube Roots are extremely popular among Class 8 students for Math Cubes And Cube Roots Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 8 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 64:

Question 1:

Evaluate:
(i) (8)3
(ii) (15)3
(iii) (21)3
(iv) (60)3

Answer:

(i) (8)3 = 8×8×8= 512.
Thus, the cube of 8 is 512.
(ii) (15)3 = 15×15×15= 3375.
Thus, the cube of 15 is 3375.
(iii) (21)3 = 21×21×21= 9261.
Thus, the cube of 21 is 9261.
(iv) (60)3 = 60×60×60= 216000.
Thus, the cube of 60 is 216000.

Page No 64:

Question 2:

Evaluate:
(i) (1.2)3
(ii) (3.5)3
(iii) (0.8)3
(iv) (0.05)3

Answer:

(i) (1.2)3 = 1.2×1.2×1.2= 1.728
Thus, the cube of 1.2 is 1.728.
(ii) (3.5)3= 3.5×3.5×3.5= 42.875
Thus, the cube of 3.5 is 42.875.
(iii) (0.8)3= 0.8×0.8×0.8= 0.512
Thus, the cube of 0.8 is 0.512.
(iv) (0.05)3= 0.05×0.05×0.05= 0.000125
Thus, the cube of 0.05 is 0.000125.



Page No 65:

Question 3:

Evaluate:
(i) 473
(ii) 10113
(iii) 1153
(iv) 13103

Answer:

(i) 473= 47×47×47= 64343
Thus, the cube of 47 is 64343.

(ii)
 10113= 1011×1011×1011= 10001331
Thus, the cube of 1011 is 10001331.(47)3
(iii)
 1153= 115×115×115= 13375
Thus, the cube of 115 is 13375
(iv)
1310313103= 1310×1310×1310= 21971000
Thus, the cube of 1310 is 21971000. (47)3

Page No 65:

Question 4:

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
(i) 125
(ii) 243
(iii) 343
(iv) 256
(v) 8000
(vi) 9261
(vii) 5324
(viii) 3375

Answer:

(i) 125
Resolving 125 into prime factors:
125 = 5×5×5
Here, one triplet is formed, which is 53. Hence, 125 can be expressed as the product of the triplets of 5.
Therefore, 125 is a perfect cube.

(ii) 243 is not a perfect cube.

(iii) 343
Resolving 125 into prime factors:
343 = 7×7×7
Here, one triplet is formed, which is 73. Hence, 343 can be expressed as the product of the triplets of 7.
Therefore, 343 is a perfect cube.

(iv) 256 is not a perfect cube.

(v) 8000
Resolving 8000 into prime factors:
8000 = 2×2×2×2×2×2×5×5×5
Here, three triplets are formed, which are 23, 23 and 53. Hence, 8000 can be expressed as the product of the triplets of 2, 2 and 5, i.e. 23 ×23 ×53  = 203 .
Therefore, 8000 is a perfect cube.

(vi) 9261
Resolving 9261 into prime factors:
9261 = 3×3×3×7×7×7
Here, two triplets are formed, which are 33  and 73. Hence, 9261 can be expressed as the product of the triplets of 3 and 7, i.e. 33 × 73= 213 .
Therefore, 9261 is a perfect cube.

(vii) 5324 is not a perfect cube.

(viii) 3375 .
Resolving 3375 into prime factors:
3375 = 3×3×3×5×5×5.
Here, two triplets are formed, which are 33  and 53. Hence, 3375 can be expressed as the product of the triplets of 3 and 5, i.e. 33 × 53= 153 .
Therefore, 3375 is a perfect cube.

Page No 65:

Question 5:

Which of the following are the cubes of even numbers?
(i) 216
(ii) 729
(iii) 512
(iv) 3375
(v) 1000

Answer:

The cubes of even numbers are always even. Therefore, 216, 512 and 1000 are the cubes of even numbers.

 216 = 2×2×2×3×3×3 = 23×33 = 63
 512 = 2×2×2×2×2×2×2×2×2 = 23×23×23 = 83
 1000 = 2×2×2×5×5×5 = 23×53 = 103

Page No 65:

Question 6:

Which of the following are the cubes of odd numbers?
(i) 125
(ii) 343
(iii) 1728
(iv) 4096
(v) 9261

Answer:

The cube of an odd number is an odd number. Therefore, 125, 343 and 9261 are the cubes of odd numbers.

 125 = 5×5×5 = 53

343 = 7×7×7 = 73

9261 = 3×3×3×7×7×7 = 33 × 73= 213 

Page No 65:

Question 7:

Find the smallest number by which 1323 must be multiplied so that the product is a perfect cube.

Answer:

1323

1323 = 3×3×3×7×7.
To make it a perfect cube, it has to be multiplied by 7.

Page No 65:

Question 8:

Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.

Answer:

2560

2560 can be expressed as the product of prime factors in the following manner:


2560 = 2×2×2×2×2×2×2×2×2×5

To make this a perfect square, we have to multiply it by 5×5.
Therefore, 2560 should be multiplied by 25 so that the product is a perfect cube.

Page No 65:

Question 9:

What is the smallest number by which 1600 must be divided so that the quotient is a perfect cube?

Answer:

1600

1600 can be expressed as the product of prime factors in the following manner:


1600 = 2×2×2×2×2×2×5×5

Therefore, to make the quotient a perfect cube, we have to divide 1600 by:
5×5=25

Page No 65:

Question 10:

Find the smallest number by which 8788 must be divided so that the quotient is a perfect cube.

Answer:


8788
8788 can be expressed as the product of prime factors as 2×2×13×13×13.
Therefore, 8788 should be divided by 4, i.e. (2×2), so that the quotient is a perfect cube.



Page No 66:

Question 1:

Find the value of using the short-cut method:
(25)3

Answer:

253
Here, a = 2 and b = 5
 
Using the formula a3+3a2b+3ab2+b3:

     4
× 2
      4
×  15
     25
×  6
   25
×5
    8
 +7
      60
  + 16
   150
+  12
  125
    15       76    162  

253 = 15625

Page No 66:

Question 2:

Find the value of using the short-cut method:
(47)3

Answer:

473
Here, a = 4 and b = 7

Using the formula a3+3a2b+3ab2+b3:

     16
× 4
      16
×  21
     49
×  12
   49
×7
    64
 +39
      336
  +   62
   588
+  34
  343
    103        398    622  

473 = 103823

Page No 66:

Question 3:

Find the value of using the short-cut method:
(68)3

Answer:

683
Here, a = 6 and b = 8

Using the formula a3+3a2b+3ab2+b3:

     36
× 6
      36
×  24
     64
×  18
   64
×8
    216
 +  98
      864
  + 120
   1152
+    51
  512
     314        984    1203  

683 = 314432

Page No 66:

Question 4:

Find the value of using the short-cut method:
(84)3

Answer:

843

Here, a = 8 and b = 4

Using the formula a3+3a2b+3ab2+b3:

     64
×   8
      64
×  12
     16
×  24
   16
× 4
    512
 +  80
    768
  + 39
   384
+    6
  64
    592      807    390  

843 = 592704



Page No 67:

Question 1:

Evaluate:
643

Answer:

643
By prime factorisation:
64 = 2×2×2×2×2×2
     = 2×2×2×2×2×2

643=(2)3×(2)33=2×2=4

Page No 67:

Question 2:

Evaluate:
3433

Answer:

3433
By prime factorisation:
343 = 7×7×7
      = ( 7×7×7 )

3433=733=7

Page No 67:

Question 3:

Evaluate:
7293

Answer:

7293
By prime factorisation:

729 = 3×3×3×3×3×3
       = 3×3×3×3×3×3

7293 = 3×3= 9

Page No 67:

Question 4:

Evaluate:
17283

Answer:

17283
By prime factorisation:

1728 = 2×2×2×2×2×2×3×3×3
         = 2×2×2×2×2×2×3×3×3=23×23×33

17283 = 2×2×3= 12

Page No 67:

Question 5:

Evaluate:
92613

Answer:

92613
By prime factorisation:

9261 = 3×3×3×7×7×7
         = 3×3×3×7×7×7=33×73

92613 = 3×7 = 21

Page No 67:

Question 6:

Evaluate:
40963

Answer:

40963
By prime factorisation:

4096 = 2×2×2×2×2×2×2×2×2×2×2×2
         = 2×2×2×2×2×2×2×2×2×2×2×2=23×23×23×23

40963 = 2×2×2×2 = 16

Page No 67:

Question 7:

Evaluate:
80003

Answer:

80003
By prime factorisation:

8000 = 2×2×2×2×2×2×5×5×5
         = 2×2×2×2×2×2×5×5×5

80003 = 2×2×5 = 20

Page No 67:

Question 8:

Evaluate:
33753

Answer:

33753
By prime factorisation:

3375 = 3×3×3×5×5×5
         = 3×3×3×5×5×5

33753 = 3×5 = 15



Page No 68:

Question 9:

Evaluate:
-2163

Answer:

-2163
By prime factorisation:

216 = 2×2×2×3×3×3
        = 2×2×2×3×3×3
-2163 = -2×3 = -6

-2163 =  -2163=-6

Page No 68:

Question 10:

Evaluate:
-5123

Answer:

-5123
By prime factorisation:

5123 = 2×2×2×2×2×2×2×2×2
             = 2×2×2×2×2×2×2×2×2
-5123 = -2×2×23=- 8

-5123-5123=-8

Page No 68:

Question 11:

Evaluate:
-13313

Answer:

-13313
By prime factorisation:
13313 = 11×11×113

             
-13313= -11×11×1113 = -11
-13313=-13313=-11 

Page No 68:

Question 12:

Evaluate:
27643

Answer:

27643
By prime factorisation:
                 

27643 = 273643= 3×3×332×2×2×2×2×23= 3×3×334×4×43=34
27643= 34

Page No 68:

Question 13:

Evaluate:
1252163

Answer:

1252163
By prime factorisation:

                

1252163  = 5×5×532×2×2×3×3×33 = 5×5×536×6×63=56

1252163 = 56

Page No 68:

Question 14:

Evaluate:
-271253

Answer:

-271253
            
By factorisation:
271253= 3×3×35×5×53

-271253= -35

Page No 68:

Question 15:

Evaluate:
-643433

Answer:

-643433
On factorisation:
                

643433= 2×2×2×2×2×27×7×73
-643433= -47

Page No 68:

Question 16:

Evaluate:
64×7293

Answer:

64×7293
64×7293 = 643×7293
                 = 4×4×43 ×3×3×3×3×3×33
                 = 4×4×43 ×9×9×93
64×7293 = 4×9= 36

Page No 68:

Question 17:

Evaluate:
72910003

Answer:

72910003
             
On factorisation:
729100033×3×3×3×3×332×2×2×5×5×53= 9×9×9310×10×103
72910003= 910

Page No 68:

Question 18:

Evaluate:
-5123433

Answer:

-5123433
By factorisation:

                        

5123433= 8×8×837×7×73
-5123433= -87

Page No 68:

Question 1:

Tick (✓) the correct answer
Which of the following numbers is a perfect cube?
(a) 141
(b) 294
(c) 216
(d) 496

Answer:

(a)
141 is not a perfect cube.

(b)
294 is not a perfect cube.

(c) (✓)
216 is a perfect cube.
216 = 2×2×2×3×3×3=23×33= 63

(d)
496 is not a perfect cube.

Page No 68:

Question 2:

Tick (✓) the correct answer
Which of the following numbers is a perfect cube?
(a) 1152
(b) 1331
(c) 2016
(d) 739

Answer:

(a)
1152 = 2×2×2×2×2×2×2×3×3 = 23×23×2×3×3.
Hence, 1152 is not a perfect cube.

(b) (✓)
1331 = 11×11×11 = 113
Hence, 1331 is a perfect cube.

(c)
2016 = 2×2×2×2×2×3×3×7 = 23×2×2×3×3×7
Hence, 2016 is not a perfect cube.

(d)
739 is not a perfect cube.

Page No 68:

Question 3:

Tick (✓) the correct answer
5123=?
(a) 6
(b) 7
(c) 8
(d) 9

Answer:

(c) 8

5123 = 2×2×2×2×2×2×2×2×2 3 =2×2×2×2×2×2×2×2×23
5123 = 23×23×23 3= 8

Hence, the cube root of 512 is 8.

Page No 68:

Question 4:

Tick (✓) the correct answer
125×643=?
(a) 100
(b) 40
(c) 20
(d) 30

Answer:

(c) 20

125×643 = 1253×643 =5×5×53×2×2×2×2×2×23
125×643 = 533×23×233 = 533×433
125×643 = 5×4 = 20

Hence, the cube root of 125×643 is 20.

Page No 68:

Question 5:

Tick (✓) the correct answer
643433=?
(a) 49
(b) 47
(c) 87
(d) 821

Answer:

(b) 47
643433 = 6433433 = 4×4×437×7×73= 433733
643433 = 47
643433 = 47

Page No 68:

Question 6:

Tick (✓) the correct answer
-5127293=?
(a) -79
(b) -89
(c) 79
(d) 89

Answer:

(b) -89
-5127293= -51237293 = -8×-8×-839×9×93 = -833933
-5127293 = -89
-5127293 = -89

Page No 68:

Question 7:

Tick (✓) the correct answer
By what least number should 648 be multiplied to get a perfect cube?
(a) 3
(b) 6
(c) 9
(d) 8

Answer:

(c) 9

648 = 2×2×2×3×3×3×3 = 23×33×3
Therefore, to get a perfect cube, we need to multiply 648 by 9, i.e. 3×3.

Page No 68:

Question 8:

Tick (✓) the correct answer
By what least number should 1536 be divided to get a perfect cube?
(a) 3
(b) 4
(c) 6
(d) 8

Answer:

(a) 3


1536 = 2×2×2×2×2×2×2×2×2×3 = 23×23×23×3
Therefore, to get a perfect cube, we need to divide 1536 by 3.

Page No 68:

Question 9:

Tick (✓) the correct answer
13103=?
(a) 1271000
(b) 2271000
(c) 21971000
(d) none of these

Answer:

(c) 21971000
13103 = 13103 = 133103 = 13×13×1310×10×1013103 = 21971000 = 21971000

13103 = 21971000

Page No 68:

Question 10:

Tick (✓) the correct answer
(0.8)3 = ?
(a) 51.2
(b) 5.12
(c) 0.512
(d) none of these

Answer:

(c) 0.512

0.83  = 0.8×0.8×0.8 = 0.512
0.83  = 0.512



Page No 70:

Question 1:

Evaluate 1253.

Answer:

1253
1253 = 753 = 7353 = 7×7×75×5×5 =343125
1253 = 343125

Page No 70:

Question 2:

Evaluate 40963.

Answer:

40963

By prime factorisation method:

40963 = 2×2×2×2×2×2×2×2×2×2×2×23 = 23×23×23×233
40963 = 2×2×2×2 =16.

40963 = 16

Page No 70:

Question 3:

Evaluate 216×3433.

Answer:

216×3433
By prime factorisation:

216×3433 = 2163×3433 = 2×2×2×3×3×33×7×7×73 = 23×333×733
216×3433 = 2×3×7 = 42

216×3433 = 42

Page No 70:

Question 4:

Evaluate -641253.

Answer:

-641253
By prime factorisation method:

-641253 = -6431253= -4×-4×-435×5×53 = -433533
-641253 = -45
-641253 = -45

Page No 70:

Question 5:

Mark (✓) against the correct answer
1343=?
(a) 12764
(b) 22764
(c) 52364
(d) none of these

Answer:

(c) 52364
1343 = 743 =7343 =7×7×74×4×4= 34364
1343 = 34364 = 52364
1343 = 52364

Page No 70:

Question 6:

Mark (✓) against the correct answer
Which of the following numbers is a perfect cube?
(a) 121
(b) 169
(c) 196
(d) 216

Answer:

(d) 216

121=11×11169=13×13196=7×7×2×2

216 = 2×2×2×3×3×3 = 23×33 = 63

216 = 63
Hence, 216 is a perfect cube.

Page No 70:

Question 7:

Mark (✓) against the correct answer
216×643=?
(a) 64
(b) 32
(c) 24
(d) 36

Answer:

(c) 24

216×643  = 2163×643 = 2×2×2×3×3×33×2×2×2×2×2×23
216×643  = 23×333×23×233 = 633×433
216×643  = 6×4 = 24
216×643  =  24

Page No 70:

Question 8:

Mark (✓) against the correct answer
-3437293=?
(a) 79
(b) -79
(c) -97
(d) 97

Answer:

(b) -79
By prime factorisation:
-3437293 = -34337293 = -7×-7×-733×3×3×3×3×33 = -73333×333 
-3437293 = -733933 = -79

-3437293 = -79

Page No 70:

Question 9:

Mark (✓) against the correct answer
By what least number should 324 be multiplied to get a perfect cube?
(a) 12
(b) 14
(c) 16
(d) 18

Answer:

(d) 18



324 = 2×2×3×3×3×3 = 2×2×3×33

Therefore, to show that the given number is the product of three triplets, we need to multiply 324 by 2×3×3.
In other words, we need to multiply 324 by 18 to make it a perfect cube.

Page No 70:

Question 10:

Mark (✓) against the correct answer
12832503=?
(a) 35
(b) 45
(c) 25
(d) none of these

Answer:

(b) 45

Resolving the numerator and the denominator into prime factors:

12832503=1282503=2×8×82×5×5×53=2×8×82×5×5×53=8×85×5×53=23×23533=2×25=45

Page No 70:

Question 11:

Mark (✓) against the correct answer
Which of the following is a cube of an odd number?
(a) 216
(b) 512
(c) 343
(d) 1000

Answer:

(c) 343
The cube of an odd number will always be an odd number.
Therefore, 343 is the cube of an odd number.

Page No 70:

Question 12:

Fill in the blanks.
(i) ab3=a3×..........
(ii) ab3=.........
(iii) -x3=.........
(iv) (0.5)3 = .........

Answer:

(i) b3

ab3 = a3×b3

(ii) a3b3

ab3 = a3b3

(iii) -x3

-x3 = -x3

(iv) 0.125

0.53 = 0.5×0.5×0.5 = 0.125



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