RS Aggarwal solution class 8 chapter 4 Cubes and Cube roots Exercise 4A

Exercise 4A

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Q1 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 1:

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

Answer 1:

(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.


Q2 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 2:

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

Answer 2:

(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.


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Q3 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 3:

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

Answer 3:

(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


Q4 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

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 4:

(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.


Q5 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 5:

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

Answer 5:

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


Q6 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 6:

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

Answer 6:

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 


Q7 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 7:

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

Answer 7:

1323

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


Q8 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 8:

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

Answer 8:

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.


Q9 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 9:

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

Answer 9:

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


Q10 | Ex-4A | Class 8 | RS AGGARWAL | Cubes and Cube roots | Chapter 4 | myhelper

Question 10:

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

Answer 10:


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.

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