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:

1^{3} = 1

2^{3} = 8 = 3 + 5

3^{3} = 27 = 7 + 9 + 11

4^{3} = 64 = 13 + 15 + 17 + 19

5^{3} = 125 = 21 + 23 + 25 + 27 + 29

6^{3} = 216 = 31 + 33 + 35 + 37 + 39 + 41

7^{3} = 343 = 43 + 45 + 47 + 49 + 51 + 53 + 55

8^{3} = 512 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

9^{3} = 729 = 73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89

10^{3} = 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

= 2^{3} × 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 1^{3} < 6 < 2^{3}

∴ 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 2^{3} < 12 < 3^{3}

∴ 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 3^{3} < 32 < 4^{3}

∴ 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

⇒ 99x^{3}= 33957

⇒ x^{3} = \(\frac{33957}{99}\) = 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

8^{3} = 512 < 632..

16^{3} = 4096 < 4741 < 4913 = 17^{3}

i.e., 16^{3} < 4741 < 17^{3}.

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 m^{2}.

Solution:

Surface area of a cube = 150 m^{2}

⇒ 6 × side^{2} = 150

⇒ side^{2} = \(\frac{150}{6}\) = 25

∴ side = \(\sqrt{25}\) = 5 m

Thus, the volume of the cube = (side)^{3} = (5)^{3} = 125 m^{3}.

Additional Questions:

Question 1.

Find the cube root of – 3375.

Solution:

3375 = 3 × 3 × 3 × 5 × 5 × 5

= 3^{3} × 5^{3}

∴ \(\sqrt[3]{3375}\) = 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 × 7^{3}

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