Exercise R3 | Class 10 | Mathematics | Revision Exercise

1. Which of the following is not a perfect cube?
(a) 3757
(b) 3375
(c) 3332
(d) 4096
Solution:
(a) 3757
From prime factorisation, we have
$\ 3757 = 13×17^2$
⸫ 3757 is not a perfect cube.

(b) 3375
From prime factorisation, we have
$\ 3375 = 3^3×5^3$
⸫ 3375 is a perfect cube.

(c) 3332
From prime factorisation, we have
$\ 3332 = 2^2×7^2×17$
⸫ 3332 is not a perfect cube.

(d) 4096
From prime factorisation, we have
$\ 4096 = 2^{12}$
⸫ 34096 is a perfect cube.


2. Find the cubes of the following numbers.
(i) 19
Solution:
The cube of 19 is
$\ 19^3 = 19×19×19 = 6859$

(ii) 21
Solution:
The cube of 21 is
$\ 21^3 = 21×21×21 = 9261$

(i) 23
Solution:
The cube of 23 is
$\ 23^3 = 23×23×23 = 12167$

(i) 27
Solution:
The cube of 27 is
$\ 27^3 = 27×27×27 = 19683$


3. Write the digits in the unit place of the cubes of the following numbers.
(i) 14
Ans. 4

(ii) 18
Ans. 2

(iii) 13
Ans. 7

(iv) 27
Ans. 3

Explanation: If 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are in the unit place of any number then 0, 1, 8, 7, 4, 5, 6, 3, 2, 9 respectively are in the unit place of the cube of those numbers.

4. Find the smallest number with which the following numbers are to be multiplied so that they become perfect cube.
(i) 5324
Solution:
From prime factorisation, we have
$\ 5324 = 2^2×11^3$
⸫ The number 5324 is to be multiplied by 2 so that it becomes a perfect cube.

(ii) 3087
Solution:
From prime factorisation, we have
$\ 3087 = 3^2×7^3$
⸫ The number 3087 is to be multiplied by 3 so that it becomes a perfect cube.

(iii) 3125
Solution:
From prime factorisation, we have
$\ 3125 = 5^5$
⸫ The number 3125 is to be multiplied by 5 so that it becomes a perfect cube.


(iv) 468
Solution:
From prime factorisation, we have
$\ 648 = 2^3×3^4$
⸫ The number 648 is to be multiplied by $\ 3^2=9$ so that it becomes a perfect cube.


5. Find the smallest numbers with which the following numbers are to be divided so that they become perfect cubes.
(i) 10368
Solution:
From prime factorisation, we have
$\ 10368 = 2^4×3^4$
⸫ The number 10368 is to be divided by $\ 2×3=6$ so that it becomes a perfect cubes.

(ii) 2187
Solution:
From prime factorisation, we have
$\ 2187 = 3^7$
⸫ The number 2187 is to be divided by 3 so that it becomes a perfect cubes.

(iii) 5000
Solution:
From prime factorisation, we have
$\ 5000 = 2^3×5^4$
⸫ The number 5000 is to be divided by 5 so that it becomes a perfect cubes.

(iv) 8192
Solution:
From prime factorisation, we have
$\ 8192 = 2^{13}$
⸫ The number 8192 is to be divided by 2 so that it becomes a perfect cubes.



6. Find the cube roots of the following numbers.
(i) 1331
Ans. 11
Explanation: Cube roots by Assumption Method
Step1: To make groups of three digits from unit place of the number, i.e.,
1 331
Step2 Observe the last digit of the number, and put the number which will be the unit digit of the cube or cube root of the number (which is expained in Q.1).
Since, it is 1, the digit in the unit place of the cube root of 1331 will also be 1.
Step3: Observe the group left, in this group the number is 1. Now to find the perfect cube number which very near to this number and to put the cube root of the number.
But, since 1 itself is a perfect cube number, so we take 1.
Now, 1 is the cube root of 1. So, the digit in the tenth place of the cube root of 1331 will also be 1.


(ii) 1728
Ans. 12

(iii) 2197
Ans. 13
Explanation: Cube roots by Assumption Method
Step1: To make groups of three digits from unit place of the number, i.e.,
2 197
Step2 Observe the last digit of the number, and put the number which will be the unit digit of the cube or cube root of the number (which is expained in Q.1).
Since, it is 7, the digit in the unit place of the cube root of 1331 will also be 3.
Step3: Observe the group left, in this group the number is 2. Now to find the perfect cube number which very near to this number and to put the cube root of the number.
Since, 1 is the perfect cube number and just smaller than 2, so we take 1.
Now, 1 is the cube root of 1. So, the digit in the tenth place of the cube root of 2197 will also be 1.


(i) 2744
Ans. 14



7. Find the cube roots of the following by factorisation
(i) 3375
Solution:
From prime factorisation, we have
$\ 3375 = 3^3×5^3$
⸫ $\ \sqrt{3375} = 3×5 = 15$

(ii) 4913
Solution:
From prime factorisation, we have
$\ 4913 = 17^3$
⸫ $\ \sqrt{4913} = 17$

(iii) 9261
Solution:
From prime factorisation, we have
$\ 9261 = 3^3×7^3$
⸫ $\ \sqrt{9261} = 3×7 = 21$

(iv) 13824
Solution:
From prime factorisation, we have
$\ 13824 = 2^9×3^3$
⸫ $\ \sqrt{13824} = 2^3×3 = 24$



8. Find the cube roots of the following without factorisation.
(i) 12167
Ans. 23
Explanation: Cube roots by Assumption Method
Step1: To make groups of three digits from unit place of the number, i.e.,
12 167
Step2 Observe the last digit of the number, and put the number which will be the unit digit of the cube or cube root of the number (which is expained in Q.1).
Since, it is 7, the digit in the unit place of the cube root of 12167 will also be 3.
Step3: Observe the group left, in this group the number is 12. Now to find the perfect cube number which very near to this number and to put the cube root of the number.
Since, 8 is the perfect cube number and smaller than 12, so we take 8
Now, 2 is the cube root of 8. So, the digit in the tenth place of the cube root of 12167 will also be 2.


(ii) 8000
Ans. 20

(iii) 4096
Ans. 16


(i) 5832
Ans. 18


9. The length of the edge of a cube is 1.2 cm. Find its volume.
Solution:
Here, $\ l = 1.2$ cm
⸫ Volume $\ = l^3 = (1.2)^3 = 1.728~ cm^3$


10. The volume of a cube shaped box is 6859 $\ cm^3$. Find its height.
Solution:
Let the height of the box be $\ h$
Given, Volume $\ = 6859~ cm^3$
⇨ $\ h^3 = 6859$
⇨ $\ h^3 = 19^3$
⇨ $\ h = 19$ cm


11. Choose the correct options:
(a) The digit in the unit place in the cube of 23 is
(i) 6
(ii) 7
(iii) 8
(iv) 9
Solution: (ii) 7

(b) Which of the following is a perfect cube?
(i) 652)
(ii) 933
(iii) 343
(iv) 1002
Solution: (iii) 343

(c) The value of $\ \sqrt[3]{1000}$ is
(i) 30
(ii) 100
(iii) 10
(iv) 1000
Solution: (iii) 10

(d) If $\ m$ is the cube root of $\ n$ then the value of $\ n$ is
(i) $\ \sqrt{m}$
(ii) $\ \sqrt[3]{m}$
(iii) $\ m^3$
(iv) $\ m^2$
Solution: (iii) $\ m^3$

(e) The value of $\ \sqrt[3]{8}+\sqrt[3]{27}+\sqrt[3]{64}$ is-
(i) 6
(ii) 7
(iii) 8
(iv) 9
Solution: (iv) 9
Explanation:
$\ \sqrt[3]{8}+\sqrt[3]{27}+\sqrt[3]{64}$
$\ = 2+3+4 = 9$

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