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

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Question 1:

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

Answer 1:

(i) (8)³ = $\left(8\times 8\times 8\right)$= 512.
Thus, the cube of 8 is 512.

(ii) (15)³ = $\left(15\times 15\times 15\right)$= 3375.
Thus, the cube of 15 is 3375.

(iii) (21)³ = $\left(21\times 21\times 21\right)$= 9261.
Thus, the cube of 21 is 9261.

(iv) (60)³ = $\left(60\times 60\times 60\right)$= 216000.
Thus, the cube of 60 is 216000.

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

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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 = $\left(1.2\times 1.2\times 1.2\right)$= 1.728
Thus, the cube of 1.2 is 1.728.

(ii) (3.5)³= $\left(3.5\times 3.5\times 3.5\right)$= 42.875
Thus, the cube of 3.5 is 42.875.

(iii) (0.8)³= $\left(0.8\times 0.8\times 0.8\right)$= 0.512
Thus, the cube of 0.8 is 0.512.

(iv) (0.05)³= $\left(0.05\times 0.05\times 0.05\right)$= 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

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Question 3:

Evaluate:
(i) ${\left(\frac{4}{7}\right)}^3$
(ii) ${\left(\frac{10}{11}\right)}^3$
(iii) ${\left(\frac{1}{15}\right)}^3$
(iv) ${\left(1\frac{3}{10}\right)}^3$

Answer 3:

(i) ${\left(\frac{4}{7}\right)}^3$= $\left(\frac{4}{7}\times \frac{4}{7}\times \frac{4}{7}\right)$= $\left(\frac{64}{343}\right)$
Thus, the cube of $\left(\frac{4}{7}\right)$ is $\left(\frac{64}{343}\right)$.

(ii)  ${\left(\frac{10}{11}\right)}^3$= $\left(\frac{10}{11}\times \frac{10}{11}\times \frac{10}{11}\right)$= $\left(\frac{1000}{1331}\right)$
Thus, the cube of $\left(\frac{10}{11}\right)$ is $\left(\frac{1000}{1331}\right)$.(47)3
(iii)  ${\left(\frac{1}{15}\right)}^3$= $\left(\frac{1}{15}\times \frac{1}{15}\times \frac{1}{15}\right)$= $\left(\frac{1}{3375}\right)$
Thus, the cube of $\left(\frac{1}{15}\right)$ is $\left(\frac{1}{3375}\right)$
(iv) ${\left(1\frac{3}{10}\right)}^3$=  ${\left(\frac{13}{10}\right)}^3$= $\left(\frac{13}{10}\times \frac{13}{10}\times \frac{13}{10}\right)$= $\left(\frac{2197}{1000}\right)$
Thus, the cube of $\left(1\frac{3}{10}\right)$ is $\left(\frac{2197}{1000}\right)$. (47)3

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

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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$\times$5$\times$5
Here, one triplet is formed, which is $5^3$. 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$\times$7$\times$7
Here, one triplet is formed, which is $7^3$. 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$\times$2$\times$2$\times$2$\times$2$\times$2$\times$5$\times$5$\times$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. $2^{3 }$$\times$$2^{3 }$$\times$$5^{3 }$ = ${20}^{3 }$.
Therefore, 8000 is a perfect cube.

(vi) 9261
Resolving 9261 into prime factors:
9261 = 3$\times$3$\times$3$\times$7$\times$7$\times$7
Here, two triplets are formed, which are $3^{3 }$ and $7^3$. Hence, 9261 can be expressed as the product of the triplets of 3 and 7, i.e. $3^{3 }$$\times$ $7^3$= ${21}^{3 }$.
Therefore, 9261 is a perfect cube.

(vii) 5324 is not a perfect cube.

(viii) 3375 .
Resolving 3375 into prime factors:
3375 = 3$\times$3$\times$3$\times$5$\times$5$\times$5.
Here, two triplets are formed, which are $3^{3 }$ and $5^3$. Hence, 3375 can be expressed as the product of the triplets of 3 and 5, i.e. $3^{3 }$$\times$ $5^3$= ${15}^{3 }$.
Therefore, 3375 is a perfect cube.

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

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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$\times$2$\times$2$\times$3$\times$3$\times$3 = $2^3\times 3^{3 }= 6^3$
 512 = $2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 = 2^3\times 2^3\times 2^3 = 8^3$
 1000 = $2\times 2\times 2\times 5\times 5\times 5 = 2^3\times 5^3 = {10}^3$

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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$\times$5$\times$5 = $5^3$

343 = 7$\times$7$\times$7 = $7^3$

9261 = 3$\times$3$\times$3$\times$7$\times$7$\times$7 = $3^{3 }$$\times$ $7^3$= ${21}^{3 }$

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Question 7:

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

Answer 7:

1323

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

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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 = $2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 5$

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

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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 = $2\times 2\times 2\times 2\times 2\times 2\times 5\times 5$

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

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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 $2\times 2\times 13\times 13\times 13$.
Therefore, 8788 should be divided by 4, i.e. ($2\times 2$), so that the quotient is a perfect cube.