Grade 8 : Mathematics (Ganita Prakash) - Square and Cube (Set 2) | Concept of Maths

Question 1

1. Find the smallest natural number by which $1008$ must be multiplied so that the product is a perfect square.

A) 2

B) 7

C) 3

D) 14

Answer

Correct Answer: B

Solution: First, find the prime factorization of $1008$: $1008 = 2 \times 504 = 2^2 \times 252 = 2^3 \times 126 = 2^4 \times 63 = 2^4 \times 3^2 \times 7$. For a number to be a perfect square, all prime factors must have even exponents. Here, the factor $7$ has an exponent of $1$ (odd). Therefore, we must multiply by $7$ to make the exponent even ($7^2$).

Question 2

2. If one member of a Pythagorean triplet is $32$, find the sum of all three members of the triplet (using the form $2m, m^2-1, m^2+1$).

A) 512

B) 480

C) 544

D) 1024

Answer

Correct Answer: C

Solution: Let $2m = 32 \implies m = 16$. The other two members are: $m^2 - 1 = 16^2 - 1 = 256 - 1 = 255$ $m^2 + 1 = 16^2 + 1 = 256 + 1 = 257$ Sum $= 32 + 255 + 257 = 544$.

Question 3

3. Evaluate the square root: $\sqrt{0.000529}$

A) 0.023

B) 0.23

C) 0.0023

D) 2.3

Answer

Correct Answer: A

Solution: We can write $0.000529$ as $\frac{529}{1,000,000}$. $\sqrt{\frac{529}{1,000,000}} = \frac{\sqrt{529}}{\sqrt{1,000,000}}$ Since $23^2 = 529$ and $1000^2 = 1,000,000$: $\frac{23}{1000} = 0.023$.

Question 4

4. Calculate the value of: $\sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}}$

A) 4

B) 5

C) 6

D) 7

Answer

Correct Answer: C

Solution: Work from the innermost root outwards: 1. $\sqrt{9} = 3$ 2. $\sqrt{19 - 3} = \sqrt{16} = 4$ 3. $\sqrt{21 + 4} = \sqrt{25} = 5$ 4. $\sqrt{41 - 5} = \sqrt{36} = 6$.

Question 5

5. What is the smallest number by which $54000$ must be divided to make it a perfect cube?

A) 2

B) 3

C) 5

D) 4

Answer

Correct Answer: A

Solution: Prime factorization of $54000$: $54000 = 54 \times 1000 = (2 \times 3^3) \times (10^3) = 2 \times 3^3 \times (2^3 \times 5^3)$ $54000 = 2^4 \times 3^3 \times 5^3$ To be a perfect cube, all exponents must be multiples of 3. The factor $2^4$ needs to become $2^3$. So we divide by $2^1 = 2$.

Question 6

6. How many non-square natural numbers lie between $99^2$ and $100^2$?

A) 100

B) 199

C) 200

D) 198

Answer

Correct Answer: D

Solution: The number of non-square numbers between the squares of two consecutive numbers $n$ and $n+1$ is given by $2n$. Here, $n = 99$. Number of non-squares $= 2 \times 99 = 198$.

Question 7

7. If $\sqrt[3]{x} = 1.2$, find the value of $x$.

A) 1.44

B) 1.728

C) 17.28

D) 0.1728

Answer

Correct Answer: B

Solution: $\sqrt[3]{x} = 1.2 \implies x = (1.2)^3$ $x = 1.2 \times 1.2 \times 1.2 = 1.728$.

Question 8

8. A square garden has an area of $1444 \text{ m}^2$. A path of width $2 \text{ m}$ is built inside the garden along its boundary. Find the area of the path.

A) $280 \text{ m}^2$

B) $288 \text{ m}^2$

C) $300 \text{ m}^2$

D) $272 \text{ m}^2$

Answer

Correct Answer: B

Solution: Side of square garden $= \sqrt{1444} = 38 \text{ m}$. The path is inside, so the side of the inner square $= 38 - (2 + 2) = 34 \text{ m}$. Area of path = Outer Area - Inner Area $= 1444 - (34)^2 = 1444 - 1156 = 288 \text{ m}^2$.

Question 9

9. Which of the following numbers cannot be a perfect square?

A) 1024

B) 1296

C) 4096

D) 1234567

Answer

Correct Answer: D

Solution: A property of perfect squares is that they never end in the digits $2, 3, 7,$ or $8$ at the unit's place. The number $1234567$ ends in $7$, so it cannot be a perfect square.

Question 10

10. Evaluate: $\sqrt[3]{\frac{0.000064}{0.008}}$

A) 0.2

B) 0.02

C) 2

D) 0.4

Answer

Correct Answer: A

Solution: First, simplify the fraction inside the root: $\frac{0.000064}{0.008} = \frac{64 \times 10^{-6}}{8 \times 10^{-3}} = 8 \times 10^{-3} = 0.008$. Now, find the cube root: $\sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} = \frac{2}{10} = 0.2$.

Grade 8 : Mathematics (Ganita Prakash) - Square and Cube (Set 2) | Concept of Maths