RS Aggarwal Class 8 Math Fourth Chapter Cubes and Cube Roots Exercise 4A Solution

RS Aggarwal Class 8 Math Fourth Chapter Cubes and Cube Roots Exercise 4A Solution

EXERCISE 4A

(1) Evaluate:

(i) (8)3 = 8 × 8 × 8 = 512

(ii) (15)3 = 15 × 15 × 15 = 3375

(iii) (21)3 = 21 × 21 × 21 = 9261

(iv) (60)3 = 60 × 60 × 60 = 216000

(2) Evaluate:

(i) (1.2)3 = 1.2 × 1.2 × 1.2 = 1.728

(ii) (3.5)3 = 3.5 × 3.5 × 3.5 = 42.875

(iii) (0.8)3 = 0.8 × 0.8 × 0.8 = 0.512

(iv) (0.05)3 = 0.05 × 0.05 × 0.05 = 0.000125

(3) Evaluate:

(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 = 5 × 5 × 5

(ii) 243 = 3 × 3 × 3 × 3 × 3

(iii) 343 = 7 × 7 × 7 = (7)3; Thus, 343 is a perfect cube.

(iv) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2; Thus, 256 is not a perfect cube.

(v) 8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5

= (2 × 2 × 5)3 = (20)3; Thus, 8000 is a perfect cube.

(vi) 9261 = 3 × 3 × 3 × 7 × 7 × 7

= (3 × 7)3 = 213; Thus, 9261 is a perfect cube.

(vii) 5324 = 2 × 2 × 11 × 11 × 11; Thus, 5324 is not a perfect cube.

(viii) 3375 = 3 × 3 × 3 × 5 × 5 × 5

= (3 × 5)3 = (15)3; Thus, 3375 is a perfect cube.

(5) Which of the following are the cubes of even numbers?

Properties: The cubes of every even number are even.

(i) 216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33 = 63

(iii) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 23 = (8)3

(v) 1000 = 2 × 2 × 2 × 5 × 5 × 5 = 23 × 53 = (10)3

(6) Which of the following are the cubes of odd numbers?

Properties: The cube of every odd number is odd.

(i) 125 = 5 × 5 × 5 = 53

(ii) 343 = 7 × 7 × 7 = 73

(v) 9261 = 3 × 3 × 3 × 7 × 7 × 7 = 33 × 73 = (21)3

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

Solution: Writing 1323 as a perfect of prime factors, we have:

1323 = 3 × 3 × 3 × 7 × 7

Clearly, to make it a perfect cube, it must be multiplied by 7.

(8) Find the smallest number by which 2560 must be divided so that the quotient is a perfect cube.

Solution: Writing 2560 as a perfect of prime factors, we have:

2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

Clearly, to make it a perfect cube, it must be multiplied by 5 × 5 = 25.

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

Solution: Writing 1600 as a product of prime factors, we have:

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

Clearly, to make it a perfect cube, it must be divided by (5 × 5) = 25.

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

Solution: Writing 1600 as a product of prime factors, we have:

8788 = 2 × 2 × 13 × 13 × 13

Clearly, to make it a perfect cube, it must be divided by (2 × 2) = 4.


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  1. Thank you so much sir

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