DAV Class 8 Maths Chapter 2 Brain Teasers Solutions

The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 2 Brain Teasers Solutions of Cubes and Cube Roots offer comprehensive answers to textbook questions.

DAV Class 8 Maths Ch 2 Brain Teasers Solutions

Question 1A.
Tick (✓) the correct option.

(i) Cube of 0.1 is equal to ______
(a) 1.11
(b) 0.001
(c) 0.101
(d) 0.01
Solution:
(b) 0.001
cube of 0.1 = (0.1)3 = 0.1 × 0.1 × 0.1 = 0.001.

(ii) The smallest number by which 1944 should be multiplied so that it becomes a perfect cube is ______
(a) 3
(b) 2
(c) 5
(d) 4
Solution:
(a) 3
1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
Thus, the number 1944 should be multiplied by 3 to make it a perfect cube.

(iii) Value of $$\sqrt[3]{1000000}$$ is ______
(a) 10
(b) 1000
(c) 100
(d) none of these
Solution:
(c) 100
1000000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5 × 5
= 10 × 10 × 10 × 10 × 10 × 10
∴ $$\sqrt[3]{1000000}=\sqrt[3]{10 \times 10 \times 10 \times 10 \times 10 \times 10}$$
= 10 × 10 = 100

(iv) $$\sqrt[3]{0.027}-\sqrt[3]{0.008}$$ is equal to ______
(a) 1
(b) 0.1
(c) 0.11
(d) 0.09
Solution:
(b) 0.1
$$\sqrt[3]{0.027}-\sqrt[3]{0.008}$$ = 0.3 – 0.2 = 0.1

(v) Cube of $$\left(\frac{-1}{3}\right)$$ is equal to ______
(a) $$\frac{1}{27}$$
(b) $$\frac{-1}{9}$$
(c) $$-\frac{1}{27}$$
(d) $$\frac{1}{9}$$
Solution:
(c) $$-\frac{1}{27}$$
cube of $$\left(-\frac{1}{3}\right)=\left(-\frac{1}{3}\right)^3$$
= $$\left(-\frac{1}{3}\right) \times\left(-\frac{1}{3}\right) \times\left(-\frac{1}{3}\right)=-\frac{1}{27}$$

Question 1B.
Answer the following questions.

(i) Find the number whose cube is 1728.
Solution:
Cube root of 1728 = $$\sqrt[3]{1728}$$
= $$\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}$$
= 2 × 2 × 3 = 12

(ii) Find the value of $$\sqrt[3]{216 \times(-125)}$$.
Solution:
$$\sqrt[3]{216 \times(-125)}$$ = $$\sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3 \times(-5) \times(-5) \times(-5)}$$
= 2 × 3 × (- 5) = – 30

(iii) Find the cube root of 0.000001.
Solution:
$$\sqrt[3]{0.000001}$$
= $$\sqrt[3]{\underline{0.1 \times 0.1 \times 0.1} \times \underline{0.1 \times 0.1 \times 0.1}}$$
= 0.1 × 0.1 = 0.01

(iv) What is the smallest number by which 1715 should be divided so that the quotient is a perfect cube?
Solution:
1715 = 5 × 7 × 7 × 7
Thus, the smellest number by which 1715 should be divided to make it a perfect cube is 5.

(v) Evaluate: $$\sqrt[3]{\frac{0.512}{0.343}}$$
Solution:
$$\sqrt[3]{\frac{0.512}{0.343}}=\sqrt[3]{\frac{512}{343}}$$

= $$\sqrt{\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{7 \times 7 \times 7}}$$

= $$\frac{2 \times 2 \times 2}{7}=\frac{8}{7}$$.

Question 2.
Prove that if a number is tripled, then its cube is 27 times the cube of the given number.
Solution:
Let x be any number.
When it is tripled, it becomes 3x.
∴ (3x)3 = 3x × 3x × 3x
= 27x which is 27 times of the given number.

Question 3.
Write cubes of all natural numbers from 1 to 10 and observe the pattern.
Solution:
13 = 1
23 = 8 = 3 + 5
33 = 27 = 7 + 9 + 11
43 = 64 = 13 + 15 + 17 + 19
53 = 125 = 21 + 23 + 25 + 27 + 29
63 = 216 = 31 + 33 + 35 + 37 + 39 + 41
73 = 343 = 43 + 45 + 47 + 49 + 51 + 53 + 55
83 = 512 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71
93 = 729 = 73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89
103 = 1000 = 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109

Question 4.
Find the cubes of:
(i) 0.6
(ii) – 3.1
(iii) – 0.01
Solution:
(i) (0.6)3 = 0.6 × 0.6 × 0.6 = 0.216
(ii) (- 3.1)3 = (- 3.1) × (- 3.1) × (- 3.1) = – 29.791
(iii) (- 0.01)3 = (- 0.01) × (- 0.01) × (- 0.01) = – 0.000001

Question 5.
Find the value of the following cube roots:
(i) $$\sqrt[3]{0.008}$$
(ii) $$\sqrt[3]{\frac{-64}{1331}}$$
(iii) $$\sqrt[3]{27 \times 2744}$$
Solution:
(i) $$\sqrt[3]{0.008}=\sqrt[3]{0.2 \times 0.2 \times 0.2}$$
= $$\sqrt[3]{(0.2)^3}$$ = 0.2

(ii) $$\sqrt[3]{\frac{-64}{1331}}=\sqrt[3]{\frac{(-4) \times(-4) \times(-4)}{11 \times 11 \times 11}}$$
= $$\sqrt[3]{\frac{(-4)^3}{(11)^3}}=\frac{-4}{11}$$

(iii) $$\sqrt[3]{27 \times 2744}=\sqrt[3]{27} \times \sqrt[3]{2744}$$ [∵ $$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$$]

= $$\sqrt[3]{3^3} \times \sqrt[3]{2 \times 2 \times 2 \times 7 \times 7 \times 7}$$
= 3 × $$\sqrt[3]{2^3 \times 7^3}$$
= 3 × 2 × 7 = 42

Question 6.
Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.
Solution:

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 23 × 2 × 3 × 3 × 5 × 5
Here 2, 3 and 5 do not form their triplet.
∴ The required smallest number is 2 × 2 × 3 × 5, i.e. 60.
Now the new number becomes 3600 × 60 = 216000
∴ $$\sqrt[3]{216000}=\sqrt[3]{216} \times \sqrt[3]{1000}$$
= $$\sqrt[3]{6^3} \times \sqrt[3]{10^3}$$
= 6 × 10 = 60

Question 7.
Evaluate: $$\sqrt[3]{\frac{0.027}{0.008}}+\sqrt{\frac{0.09}{0.04}}$$ – 1
Solution:
$$\sqrt[3]{\frac{0.027}{0.008}}+\sqrt{\frac{0.09}{0.04}}$$ – 1 = $$\sqrt[3]{\frac{27}{8}}+\sqrt{\frac{9}{4}}$$ – 1
= $$\frac{3}{2}+\frac{3}{2}$$ – 1
= 1 – 1 = 0

Question 8.
Guess the cube root of the following numbers
(i) 6859
(ii) 12167
(iii) 32768.
Solution:
(i) 6859
The two groups are 6 and 859
The unit place in the cube root of 6859 must be 9
Now for tens place 13 < 6 < 23
∴ ten’s place is 1
Hence $$\sqrt[3]{6859}$$ = 19

(ii) 1216712167
The two groups are 12 and 167
Here unit place in 167 is 7
∴ the unit place in the $$\sqrt{12167}$$ must be 3
Now 23 < 12 < 33
∴ Tens place must be 2
Hence $$\sqrt[3]{12167}$$ = 23

(iii) 32768
The two groups are 32 and 768
Here unit place in 768 is 8
∴ the unit place in the $$\sqrt{32768}$$ must be 2
Now 33 < 32 < 43
∴ Ten’s place must be 3
Hence $$\sqrt[3]{32768}$$ = 32

DAV Class 8 Maths Chapter 2 HOTS

Question 1.
Evaluate: $$\sqrt[3]{288 \sqrt[3]{72 \sqrt[3]{27}}}$$
Solution:
$$\sqrt[3]{288 \sqrt[3]{72 \sqrt[3]{27}}}$$

Question 2.
Three numbers are in the ratio 2 : 3 : 4. The sum of their cubes is 33957. Find the numbers.
Solution:
Let the numbers be 2x, 3x and 4x such that
(2x)3 + (3x)3 + (4x)3 = 33957
⇒ 99x3= 33957
⇒ x3 = $$\frac{33957}{99}$$ = 343

∴ x = $$\sqrt[3]{343}=\sqrt[3]{7 \times 7 \times 7}$$ = 7
Thus, the numbers are 2 × 7, 3 × 7 and 4 × 7, i.e. 14, 21 and 28.

DAV Class 8 Maths Chapter 2 Enrichment Questions

Question 1.
Find the cube root of 4741632 by estimation.
Solution:
We have to find cube root of 4741632.
Group the given number into 3 and 4 digits from right.
4741,632 – Here, ones digit is 2.
So, ones digit of $$\sqrt[3]{4741632}$$ will be 8 as
83 = 512 < 632..
163 = 4096 < 4741 < 4913 = 173
i.e., 163 < 4741 < 173.
So, we take 16 as the first two digits, i.e. hundreds and tens digits.
Thus, $$\sqrt[3]{4741632}$$ = 168.

Question 2.
Find the volume of a cube whose surface area is 150 m2.
Solution:
Surface area of a cube = 150 m2
⇒ 6 × side2 = 150
⇒ side2 = $$\frac{150}{6}$$ = 25
∴ side = $$\sqrt{25}$$ = 5 m
Thus, the volume of the cube = (side)3 = (5)3 = 125 m3.

Question 1.
Find the cube root of – 3375.
Solution:

3375 = 3 × 3 × 3 × 5 × 5 × 5
= 33 × 53
∴ $$\sqrt[3]{3375}$$ = 3 × 5 = 15
Hence, $$\sqrt[3]{-3375}$$ = – 15
(∴ The cube root of a negative number is always negative.)

Question 2.
Evaluate: $$\sqrt[3]{100} \times \sqrt[3]{270}$$
Solution:
$$\sqrt[3]{100} \times \sqrt[3]{270}=\sqrt[3]{100 \times 270}$$ [∵ $$\sqrt[3]{a} \times \sqrt[3]{b}=\sqrt[3]{a b}$$]
= $$\sqrt[3]{27000}=\sqrt[3]{27} \times \sqrt[3]{1000}$$
= $$\sqrt[3]{3^3} \times \sqrt[3]{10^3}$$
= 3 × 10 = 30

Question 3.
What is the smallest number by which 3087 may be multiplied so that the product is a perfect cube?
Solution:

3087 = 3 × 3 × 7 × 7 × 7
= 3 × 3 × 73
Here 3 is needed to make the triplet of 3
∴ 3 is the smallest number required to make 3087 a perfect cube.

Question 4.
Evaluate: $$\sqrt[3]{(-1728) \times(-2197)}$$.
Solution:
$$\sqrt[3]{(-1728) \times(-2197)}$$

= 2 × 2 × 3 × 13
= 12 × 13 = 156

Multiple Choice Questions:

Question 1.
The cube of an even number is always an _________
(a) odd
(b) even
(c) any natural number
(d) none of these
Solution:
(b) even

Question 2.
The cube of an odd number is always an _________
(a) even
(b) any real number
(c) odd
(d) none of these
Solution:
(c) odd

Question 3.
The cube of any multiple of 2 is always divisible by _________
(a) 8
(b) 6
(c) 12
(d) 4
Solution:
(a) 8

Question 4.
The cube of any negative number is always _________
(a) positive
(b) negative
(c) reciprocal
(d) none of these
Solution:
(b) negative

Question 5.
Any nuumber ending with two zeros may have its _________
(a) square root
(b) cube root
(c) both
(d) none of these
Solution:
(a) square root