Squares and Square Roots, Cubes and Cube Roots

The square of a number is that number raised to the power 2.

⇒ (14)²

⇒ 14 × 14

⇒ 196

Hence, (14)² = 196.

The square of a number is that number raised to the power 2.

⇒ (137)²

⇒ 137 × 137

⇒ 18769

Hence, (137)² = 18769.

The square of a number is that number raised to the power 2.

$$\begin{aligned}\Rightarrow \left(\frac{4}{17}\right)^2\\[1em] \Rightarrow \frac{4}{17} \times \frac{4}{17}\\[1em] \Rightarrow \frac{16}{289}\end{aligned}$$

Hence, $\mathbf{\left(\frac{4}{17}\right)^2 = \frac{16}{289}}$.

Converting mixed fraction into improper fraction:

$2\frac{3}{4} = \frac{11}{4}$

The square of a number is that number raised to the power 2.

$$\begin{aligned}\Rightarrow \left(2\frac{3}{4}\right)^2\\[1em] \Rightarrow \left(\frac{11}{4}\right)^2\\[1em] \Rightarrow \frac{11}{4} \times \frac{11}{4}\\[1em] \Rightarrow \frac{121}{16}\\[1em] \Rightarrow 7\frac{9}{16}\end{aligned}$$

Hence, $\mathbf{\left(2\frac{3}{4}\right)^2 = 7\frac{9}{16}}$.

The square of a number is that number raised to the power 2.

⇒ (0.01)²

⇒ 0.01 × 0.01

⇒ 0.0001

Hence, (0.01)² = 0.0001.

The square of a number is that number raised to the power 2.

⇒ (1.2)²

⇒ 1.2 × 1.2

⇒ 1.44

Hence, (1.2)² = 1.44.

The square of a number is that number raised to the power 2.

⇒ (0.17)²

⇒ 0.17 × 0.17

⇒ 0.0289

Hence, (0.17)² = 0.0289.

The square of a number is that number raised to the power 2.

⇒ (4.6)²

⇒ 4.6 × 4.6

⇒ 21.16

Hence, (4.6)² = 21.16.

Resolving 252 into prime factors, we get :

$$\begin{array}{l|r} 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

252 = (2 × 2) × (3 × 3) × 7

Since the prime factor 7 is left unpaired, so 252 cannot be expressed as the product of pairs of equal prime factors.

Hence, 252 is not a perfect square number.

Resolving 324 into prime factors, we get :

$$\begin{array}{l|r} 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

324 = (2 × 2) × (3 × 3) × (3 × 3)

Since 324 can be expressed as the product of pairs of equal prime factors.

Hence, 324 is a perfect square number.

Resolving 676 into prime factors, we get :

$$\begin{array}{l|r} 2 & 676 \\ \hline 2 & 338 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

676 = (2 × 2) × (13 × 13)

Since 676 can be expressed as the product of pairs of equal prime factors.

Hence, 676 is a perfect square number.

Resolving 1225 into prime factors, we get :

$$\begin{array}{l|r} 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

1225 = (5 × 5) × (7 × 7)

Since 1225 can be expressed as the product of pairs of equal prime factors.

Hence, 1225 is a perfect square number.

Resolving 2916 into prime factors, we get :

$$\begin{array}{l|r} 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

2916 = (2 × 2) × (3 × 3) × (3 × 3) × (3 × 3)

Since 2916 can be expressed as the product of pairs of equal prime factors.

Hence, 2916 is a perfect square number.

Resolving 3528 into prime factors, we get :

$$\begin{array}{l|r} 2 & 3528 \\ \hline 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

3528 = (2 × 2) × (3 × 3) × (7 × 7) × 2

Since the prime factor 2 is left unpaired, so 3528 cannot be expressed as the product of pairs of equal prime factors.

Hence, 3528 is not a perfect square number.

Resolving 5625 into prime factors, we get :

$$\begin{array}{l|r} 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

5625 = (3 × 3) × (5 × 5) × (5 × 5)

Since 5625 can be expressed as the product of pairs of equal prime factors.

Hence, 5625 is a perfect square number.

Resolving 9075 into prime factors, we get :

$$\begin{array}{l|r} 3 & 9075 \\ \hline 5 & 3025 \\ \hline 5 & 605 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

9075 = (5 × 5) × (11 × 11) × 3

Since the prime factor 3 is left unpaired, so 9075 cannot be expressed as the product of pairs of equal prime factors.

Hence, 9075 is not a perfect square number.

Resolving 441 into prime factors, we get :

$$\begin{array}{l|r} 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

441 = (3 × 3) × (7 × 7)

Since 441 can be expressed as the product of pairs of equal prime factors, so 441 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{441} = \sqrt{(3 × 3) × (7 × 7)} \\[1em] \Rightarrow \sqrt{441} = 3 \times 7\\[1em] \Rightarrow \sqrt{441} = 21\end{aligned}$$

Hence, 441 is a perfect square and $\mathbf{\sqrt{441} = 21}$.

Resolving 784 into prime factors, we get :

$$\begin{array}{l|r} 2 & 784 \\ \hline 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

784 = (2 × 2) × (2 × 2) × (7 × 7)

Since 784 can be expressed as the product of pairs of equal prime factors, so 784 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{784} = \sqrt{(2 \times 2) \times (2 \times 2) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{784} = 2 \times 2 \times 7\\[1em] \Rightarrow \sqrt{784} = 28\end{aligned}$$

Hence, 784 is a perfect square and $\mathbf{\sqrt{784} = 28}$.

Resolving 1225 into prime factors, we get :

$$\begin{array}{l|r} 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

1225 = (5 × 5) × (7 × 7)

Since 1225 can be expressed as the product of pairs of equal prime factors, so 1225 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{1225} = \sqrt{(5 \times 5) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{1225} = 5 \times 7\\[1em] \Rightarrow \sqrt{1225} = 35\end{aligned}$$

Hence, 1225 is a perfect square and $\mathbf{\sqrt{1225} = 35}$.

Resolving 3969 into prime factors, we get :

$$\begin{array}{l|r} 3 & 3969 \\ \hline 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

3969 = (3 × 3) × (3 × 3) × (7 × 7)

Since 3969 can be expressed as the product of pairs of equal prime factors, so 3969 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{3969} = \sqrt{(3 \times 3) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{3969} = 3 \times 3 \times 7\\[1em] \Rightarrow \sqrt{3969} = 63\end{aligned}$$

Hence, 3969 is a perfect square and $\mathbf{\sqrt{3969} = 63}$.

Resolving 2601 into prime factors, we get :

$$\begin{array}{l|r} 3 & 2601 \\ \hline 3 & 867 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$$

2601 = (3 × 3) × (17 × 17)

Since 2601 can be expressed as the product of pairs of equal prime factors, so 2601 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{2601} = \sqrt{(3 \times 3) \times (17 \times 17)} \\[1em] \Rightarrow \sqrt{2601} = 3 \times 17\\[1em] \Rightarrow \sqrt{2601} = 51\end{aligned}$$

Hence, 2601 is a perfect square and $\mathbf{\sqrt{2601} = 51}$.

Resolving 5929 into prime factors, we get :

$$\begin{array}{l|r} 7 & 5929 \\ \hline 7 & 847 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

5929 = (7 × 7) × (11 × 11)

Since 5929 can be expressed as the product of pairs of equal prime factors, so 5929 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{5929} = \sqrt{(7 \times 7) \times (11 \times 11)} \\[1em] \Rightarrow \sqrt{5929} = 7 \times 11\\[1em] \Rightarrow \sqrt{5929} = 77\end{aligned}$$

Hence, 5929 is a perfect square and $\mathbf{\sqrt{5929} = 77}$.

Resolving 7056 into prime factors, we get :

$$\begin{array}{l|r} 2 & 7056 \\ \hline 2 & 3528 \\ \hline 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)

Since 7056 can be expressed as the product of pairs of equal prime factors, so 7056 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{7056} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{7056} = 2 \times 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{7056} = 84\end{aligned}$$

Hence, 7056 is a perfect square and $\mathbf{\sqrt{7056} = 84}$.

Resolving 8281 into prime factors, we get :

$$\begin{array}{l|r} 7 & 8281 \\ \hline 7 & 1183 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

8281 = (7 × 7) × (13 × 13)

Since 8281 can be expressed as the product of pairs of equal prime factors, so 8281 is a perfect square.

$$\begin{aligned}\Rightarrow \sqrt{8281} = \sqrt{(7 \times 7) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{8281} = 7 \times 13\\[1em] \Rightarrow \sqrt{8281} = 91\end{aligned}$$

Hence, 8281 is a perfect square and $\mathbf{\sqrt{8281} = 91}$.

Resolving 588 into prime factors, we get :

$$\begin{array}{l|r} 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

588 = (2 × 2) × (7 × 7) × 3

Thus, 2, 2 and 7, 7 exist in pairs while 3 is alone.

So, we should multiply the given number by 3 to get a perfect square number.

$$\begin{aligned}\Rightarrow 588 \times 3 = 1764\\[1em] \Rightarrow 1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{1764} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{1764} = 42\end{aligned}$$

Hence, 588 should be multiplied by 3 and $\mathbf{\sqrt{1764} = 42}$.

Resolving 2592 into prime factors, we get :

$$\begin{array}{l|r} 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

2592 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2

Thus, 2, 2 and 3, 3 exist in pairs while one 2 is alone.

So, we should multiply the given number by 2 to get a perfect square number.

$$\begin{aligned}\Rightarrow 2592 \times 2 = 5184\\[1em] \Rightarrow 5184 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{5184} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{5184} = 2 \times 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{5184} = 72\end{aligned}$$

Hence, 2592 should be multiplied by 2 and $\mathbf{\sqrt{5184} = 72}$.

Resolving 3332 into prime factors, we get :

$$\begin{array}{l|r} 2 & 3332 \\ \hline 2 & 1666 \\ \hline 7 & 833 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$$

3332 = (2 × 2) × (7 × 7) × 17

Thus, 2, 2 and 7, 7 exist in pairs while 17 is alone.

So, we should multiply the given number by 17 to get a perfect square number.

$$\begin{aligned}\Rightarrow 3332 \times 17 = 56644\\[1em] \Rightarrow 56644 = (2 \times 2) \times (7 \times 7) \times (17 \times 17)\\[1em] \Rightarrow \sqrt{56644} = \sqrt{(2 \times 2) \times (7 \times 7) \times (17 \times 17)} \\[1em] \Rightarrow \sqrt{56644} = 2 \times 7 \times 17\\[1em] \Rightarrow \sqrt{56644} = 238\end{aligned}$$

Hence, 3332 should be multiplied by 17 and $\mathbf{\sqrt{56644} = 238}$.

Resolving 3380 into prime factors, we get :

$$\begin{array}{l|r} 2 & 3380 \\ \hline 2 & 1690 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

3380 = (2 × 2) × (13 × 13) × 5

Thus, 2, 2 and 13, 13 exist in pairs while 5 is alone.

So, we should multiply the given number by 5 to get a perfect square number.

$$\begin{aligned}\Rightarrow 3380 \times 5 = 16900\\[1em] \Rightarrow 16900 = (2 \times 2) \times (5 \times 5) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{16900} = \sqrt{(2 \times 2) \times (5 \times 5) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{16900} = 2 \times 5 \times 13\\[1em] \Rightarrow \sqrt{16900} = 130\end{aligned}$$

Hence, 3380 should be multiplied by 5 and $\mathbf{\sqrt{16900} = 130}$.

Resolving 1728 into prime factors, we get :

$$\begin{array}{l|r} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

1728 = (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3) × 3

Clearly, in the product of prime factors, the factor 3 does not exist in pair.

So, we must divide the given number by 3.

$$\begin{aligned}\Rightarrow 1728 \div 3 = 576\\[1em] \Rightarrow 576 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{576} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{576} = 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{576} = 24\end{aligned}$$

Hence, 1728 should be divided by 3, the resulting perfect square number is 576 and $\mathbf{\sqrt{576} = 24}$.

Resolving 4500 into prime factors, we get :

$$\begin{array}{l|r} 2 & 4500 \\ \hline 2 & 2250 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

4500 = (2 × 2) × (3 × 3) × (5 × 5) × 5

Clearly, in the product of prime factors, the factor 5 does not exist in pair.

So, we must divide the given number by 5.

$$\begin{aligned}\Rightarrow 4500 \div 5 = 900\\[1em] \Rightarrow 900 = (2 \times 2) \times (3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{900} = \sqrt{(2 \times 2) \times (3 \times 3) \times (5 \times 5)} \\[1em] \Rightarrow \sqrt{900} = 2 \times 3 \times 5\\[1em] \Rightarrow \sqrt{900} = 30\end{aligned}$$

Hence, 4500 should be divided by 5, the resulting perfect square number is 900 and $\mathbf{\sqrt{900} = 30}$.

Resolving 7776 into prime factors, we get :

$$\begin{array}{l|r} 2 & 7776 \\ \hline 2 & 3888 \\ \hline 2 & 1944 \\ \hline 2 & 972 \\ \hline 2 & 486 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

7776 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2 × 3

Clearly, in the product of prime factors, the factors 2 and 3 do not exist in pairs.

So, we must divide the given number by 2 × 3, i.e., 6.

$$\begin{aligned}\Rightarrow 7776 \div 6 = 1296\\[1em] \Rightarrow 1296 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{1296} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{1296} = 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{1296} = 36\end{aligned}$$

Hence, 7776 should be divided by 6, the resulting perfect square number is 1296 and $\mathbf{\sqrt{1296} = 36}$.

Resolving 8820 into prime factors, we get :

$$\begin{array}{l|r} 2 & 8820 \\ \hline 2 & 4410 \\ \hline 3 & 2205 \\ \hline 3 & 735 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

8820 = (2 × 2) × (3 × 3) × (7 × 7) × 5

Clearly, in the product of prime factors, the factor 5 does not exist in pair.

So, we must divide the given number by 5.

$$\begin{aligned}\Rightarrow 8820 \div 5 = 1764\\[1em] \Rightarrow 1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{1764} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{1764} = 42\end{aligned}$$

Hence, 8820 should be divided by 5, the resulting perfect square number is 1764 and $\mathbf{\sqrt{1764} = 42}$.

Let the number of students in the class be $x$.

Then, the amount contributed by each student = number of students = ₹$x$.

∴ Total contribution = $x \times x = x^2$.

But, total contribution = ₹1,156 (given).

Resolving 1156 into prime factors, we get :

$$\begin{array}{l|r} 2 & 1156 \\ \hline 2 & 578 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$$

1156 = (2 × 2) × (17 × 17)

$$\begin{aligned}\Rightarrow x^2 = 1156\\[1em] \Rightarrow x = \sqrt{1156}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (17 \times 17)}\\[1em] \Rightarrow x = 2 \times 17\\[1em] \Rightarrow x = 34\end{aligned}$$

Hence, the strength of the class is 34.

Let the number of rows be $x$.

Then, the number of chairs in each row = number of rows = $x$.

∴ Total number of chairs = $x \times x = x^2$.

But, total number of chairs = 2025 (given).

Resolving 2025 into prime factors, we get :

$$\begin{array}{l|r} 3 & 2025 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

2025 = (3 × 3) × (3 × 3) × (5 × 5)

$$\begin{aligned}\Rightarrow x^2 = 2025\\[1em] \Rightarrow x = \sqrt{2025}\\[1em] \Rightarrow x = \sqrt{(3 \times 3) \times (3 \times 3) \times (5 \times 5)}\\[1em] \Rightarrow x = 3 \times 3 \times 5\\[1em] \Rightarrow x = 45\end{aligned}$$

Hence, there are 45 chairs in each row.

The least number divisible by each one of 8, 9, 10 and 15 is their L.C.M.

Finding the L.C.M. of 8, 9, 10 and 15 :

$$\begin{array}{l|rrrr} 2 & 8 & 9 & 10 & 15 \\ \hline 2 & 4 & 9 & 5 & 15 \\ \hline 2 & 2 & 9 & 5 & 15 \\ \hline 3 & 1 & 9 & 5 & 15 \\ \hline 3 & 1 & 3 & 5 & 5 \\ \hline 5 & 1 & 1 & 5 & 5 \\ \hline & 1 & 1 & 1 & 1 \\ \end{array}$$

L.C.M. of 8, 9, 10 and 15 = 2 × 2 × 2 × 3 × 3 × 5 = 360

By prime factorisation we get :

360 = (2 × 2) × (3 × 3) × 2 × 5

To make it a perfect square, it must be multiplied by 2 × 5, i.e., by 10.

$$\begin{aligned}\Rightarrow 360 \times 10 = 3600\\[1em] \Rightarrow 3600 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{3600} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)} \\[1em] \Rightarrow \sqrt{3600} = 2 \times 2 \times 3 \times 5\\[1em] \Rightarrow \sqrt{3600} = 60\end{aligned}$$

Hence, the least square number which is exactly divisible by each of the numbers 8, 9, 10 and 15 is 3600.

According to the properties of perfect squares, a number ending in 2, 3, 7 or 8 is never a perfect square and a number ending in an odd number of zeros is never a perfect square.

(i) The number 2162 ends in 2.

Hence, 2162 is not a perfect square as its unit digit is 2.

(ii) The number 6843 ends in 3.

Hence, 6843 is not a perfect square as its unit digit is 3.

(iii) The number 9637 ends in 7.

Hence, 9637 is not a perfect square as its unit digit is 7.

(iv) The number 6598 ends in 8.

Hence, 6598 is not a perfect square as its unit digit is 8.

(v) The number 640 ends in one zero.

Hence, 640 is not a perfect square as it has an odd number of zeros at the end.

(vi) The number 49000 ends in three zeros.

Hence, 49000 is not a perfect square as it has an odd number of zeros at the end.

(vii) The number 3600000 ends in five zeros.

Hence, 3600000 is not a perfect square as it has an odd number of zeros at the end.

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 676 into prime factors, we get :

$$\begin{array}{l|r} 2 & 676 \\ \hline 2 & 338 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

∴ 676 = (2 × 2) × (13 × 13)

$$\begin{aligned}\Rightarrow \sqrt{676} = \sqrt{(2 \times 2) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{676} = 2 \times 13\\[1em] \Rightarrow \sqrt{676} = 26\end{aligned}$$

Since 26 is an even number.

Hence, 676 is the square of an even number.

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 729 into prime factors, we get :

$$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

∴ 729 = (3 × 3) × (3 × 3) × (3 × 3)

$$\begin{aligned}\Rightarrow \sqrt{729} = \sqrt{(3 \times 3) \times (3 \times 3) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{729} = 3 \times 3 \times 3\\[1em] \Rightarrow \sqrt{729} = 27\end{aligned}$$

Since 27 is an odd number.

Hence, 729 is not the square of an even number.

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 1089 into prime factors, we get :

$$\begin{array}{l|r} 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

∴ 1089 = (3 × 3) × (11 × 11)

$$\begin{aligned}\Rightarrow \sqrt{1089} = \sqrt{(3 \times 3) \times (11 \times 11)} \\[1em] \Rightarrow \sqrt{1089} = 3 \times 11\\[1em] \Rightarrow \sqrt{1089} = 33\end{aligned}$$

Since 33 is an odd number.

Hence, 1089 is not the square of an even number.

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 2304 into prime factors, we get :

$$\begin{array}{l|r} 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

∴ 2304 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)

$$\begin{aligned}\Rightarrow \sqrt{2304} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{2304} = 48\end{aligned}$$

Since 48 is an even number.

Hence, 2304 is the square of an even number.

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 5625 into prime factors, we get :

$$\begin{array}{l|r} 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

∴ 5625 = (3 × 3) × (5 × 5) × (5 × 5)

$$\begin{aligned}\Rightarrow \sqrt{5625} = \sqrt{(3 \times 3) \times (5 \times 5) \times (5 \times 5)} \\[1em] \Rightarrow \sqrt{5625} = 3 \times 5 \times 5\\[1em] \Rightarrow \sqrt{5625} = 75\end{aligned}$$

Since 75 is an odd number.

Hence, 5625 is not the square of an even number.

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 9216 into prime factors, we get :

$$\begin{array}{l|r} 2 & 9216 \\ \hline 2 & 4608 \\ \hline 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

∴ 9216 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)

$$\begin{aligned}\Rightarrow \sqrt{9216} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{9216} = 2 \times 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{9216} = 96\end{aligned}$$

Since 96 is an even number.

Hence, 9216 is the square of an even number.

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 484 into prime factors, we get :

$$\begin{array}{l|r} 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

∴ 484 = (2 × 2) × (11 × 11)

$$\begin{aligned}\Rightarrow \sqrt{484} = \sqrt{(2 × 2) × (11 × 11)} \\[1em] \Rightarrow \sqrt{484} = 2 \times 11\\[1em] \Rightarrow \sqrt{484} = 22\end{aligned}$$

Since 22 is an even number.

Hence, 484 is not the square of an odd number.

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 961 into prime factors, we get :

$$\begin{array}{l|r} 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$$

∴ 961 = (31 × 31)

$$\begin{aligned}\Rightarrow \sqrt{961} = \sqrt{31 \times 31} \\[1em] \Rightarrow \sqrt{961} = 31\end{aligned}$$

Since 31 is an odd number.

Hence, 961 is the square of an odd number.

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 4225 into prime factors, we get :

$$\begin{array}{l|r} 5 & 4225 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

∴ 4225 = (5 × 5) × (13 × 13)

$$\begin{aligned}\Rightarrow \sqrt{4225} = \sqrt{(5 \times 5) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{4225} = 5 \times 13\\[1em] \Rightarrow \sqrt{4225} = 65\end{aligned}$$

Since 65 is an odd number.

Hence, 4225 is the square of an odd number.

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 7396 into prime factors, we get :

$$\begin{array}{l|r} 2 & 7396 \\ \hline 2 & 3698 \\ \hline 43 & 1849 \\ \hline 43 & 43 \\ \hline & 1 \\ \end{array}$$

∴ 7396 = (2 × 2) × (43 × 43)

$$\begin{aligned}\Rightarrow \sqrt{7396} = \sqrt{(2 \times 2) \times (43 \times 43)} \\[1em] \Rightarrow \sqrt{7396} = 2 \times 43\\[1em] \Rightarrow \sqrt{7396} = 86\end{aligned}$$

Since 86 is an even number.

Hence, 7396 is not the square of an odd number.

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 6241 into prime factors, we get :

$$\begin{array}{l|r} 79 & 6241 \\ \hline 79 & 79 \\ \hline & 1 \\ \end{array}$$

∴ 6241 = (79 × 79)

$$\begin{aligned}\Rightarrow \sqrt{6241} = \sqrt{79 \times 79} \\[1em] \Rightarrow \sqrt{6241} = 79\end{aligned}$$

Since 79 is an odd number.

Hence, 6241 is the square of an odd number.

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 8649 into prime factors, we get :

$$\begin{array}{l|r} 3 & 8649 \\ \hline 3 & 2883 \\ \hline 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$$

∴ 8649 = (3 × 3) × (31 × 31)

$$\begin{aligned}\Rightarrow \sqrt{8649} = \sqrt{(3 \times 3) \times (31 \times 31)} \\[1em] \Rightarrow \sqrt{8649} = 3 \times 31\\[1em] \Rightarrow \sqrt{8649} = 93\end{aligned}$$

Since 93 is an odd number.

Hence, 8649 is the square of an odd number.

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, the given numbers are the first 11 odd natural numbers.

⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

⇒ 11²

⇒ 121

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121.

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, the last odd number is 49.

$$\begin{aligned}\Rightarrow 2n - 1 = 49 \\[1em] \Rightarrow 2n = 50 \\[1em] \Rightarrow n = 25\end{aligned}$$

So, the given numbers are the first 25 odd natural numbers.

⇒ 1 + 3 + 5 + 7 + ..... + 47 + 49

⇒ 25²

⇒ 625

Hence, 1 + 3 + 5 + 7 + ..... + 47 + 49 = 625.

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, the last odd number is 99.

$$\begin{aligned}\Rightarrow 2n - 1 = 99 \\[1em] \Rightarrow 2n = 100 \\[1em] \Rightarrow n = 50\end{aligned}$$

So, the given numbers are the first 50 odd natural numbers.

⇒ 1 + 3 + 5 + 7 + ..... + 97 + 99

⇒ 50²

⇒ 2500

Hence, 1 + 3 + 5 + 7 + ..... + 97 + 99 = 2500.

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, 144 = 12², so 144 is the sum of the first 12 odd natural numbers.

⇒ 144

⇒ 12²

⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Hence, 144 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23.

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, 225 = 15², so 225 is the sum of the first 15 odd natural numbers.

⇒ 225

⇒ 15²

⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

Hence, 225 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29.

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$$\begin{aligned}\Rightarrow 2m = 6\\[1em] \Rightarrow m = 3\end{aligned}$$

∴ $m^2 - 1 = 3^2 - 1 = 9 - 1 = 8$ and $m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$.

Hence, (6, 8, 10) is a Pythagorean triplet whose smallest member is 6.

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$$\begin{aligned}\Rightarrow 2m = 14\\[1em] \Rightarrow m = 7\end{aligned}$$

∴ $m^2 - 1 = 7^2 - 1 = 49 - 1 = 48$ and $m^2 + 1 = 7^2 + 1 = 49 + 1 = 50$.

Hence, (14, 48, 50) is a Pythagorean triplet whose smallest member is 14.

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$$\begin{aligned}\Rightarrow 2m = 16\\[1em] \Rightarrow m = 8\end{aligned}$$

∴ $m^2 - 1 = 8^2 - 1 = 64 - 1 = 63$ and $m^2 + 1 = 8^2 + 1 = 64 + 1 = 65$.

Hence, (16, 63, 65) is a Pythagorean triplet whose smallest member is 16.

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$$\begin{aligned}\Rightarrow 2m = 20\\[1em] \Rightarrow m = 10\end{aligned}$$

∴ $m^2 - 1 = 10^2 - 1 = 100 - 1 = 99$ and $m^2 + 1 = 10^2 + 1 = 100 + 1 = 101$.

Hence, (20, 99, 101) is a Pythagorean triplet whose smallest member is 20.

Solving,

⇒ (29)² - (28)²

⇒ (29 + 28) × (29 - 28)

⇒ 57 × 1

⇒ 57

Hence, (29)² - (28)² = 57.

Solving,

⇒ (56)² - (55)²

⇒ (56 + 55) × (56 - 55)

⇒ 111 × 1

⇒ 111

Hence, (56)² - (55)² = 111.

Solving,

⇒ (82)² - (81)²

⇒ (82 + 81) × (82 - 81)

⇒ 163 × 1

⇒ 163

Hence, (82)² - (81)² = 163.

Solving,

⇒ (108)² - (107)²

⇒ (108 + 107) × (108 - 107)

⇒ 215 × 1

⇒ 215

Hence, (108)² - (107)² = 215.

Solving,

⇒ (221)² - (220)²

⇒ (221 + 220) × (221 - 220)

⇒ 441 × 1

⇒ 441

Hence, (221)² - (220)² = 441.

Solving,

⇒ (204)²

⇒ (200 + 4)²

⇒ (200)² + 2 × 200 × 4 + (4)²

⇒ 40000 + 1600 + 16

⇒ 41616

Hence, (204)² = 41616.

Solving,

⇒ (307)²

⇒ (300 + 7)²

⇒ (300)² + 2 × 300 × 7 + (7)²

⇒ 90000 + 4200 + 49

⇒ 94249

Hence, (307)² = 94249.

Solving,

⇒ (430)²

⇒ (400 + 30)²

⇒ (400)² + 2 × 400 × 30 + (30)²

⇒ 160000 + 24000 + 900

⇒ 184900

Hence, (430)² = 184900.

Solving,

⇒ (94)²

⇒ (100 - 6)²

⇒ (100)² - 2 × 100 × 6 + (6)²

⇒ 10000 - 1200 + 36

⇒ 8836

Hence, (94)² = 8836.

Solving,

⇒ (192)²

⇒ (200 - 8)²

⇒ (200)² - 2 × 200 × 8 + (8)²

⇒ 40000 - 3200 + 64

⇒ 36864

Hence, (192)² = 36864.

Solving,

⇒ (389)²

⇒ (400 - 11)²

⇒ (400)² - 2 × 400 × 11 + (11)²

⇒ 160000 - 8800 + 121

⇒ 151321

Hence, (389)² = 151321.

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 59 × 61

⇒ (60 - 1) × (60 + 1)

⇒ (60)² - (1)²

⇒ 3600 - 1

⇒ 3599

Hence, 59 × 61 = 3599.

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 88 × 92

⇒ (90 - 2) × (90 + 2)

⇒ (90)² - (2)²

⇒ 8100 - 4

⇒ 8096

Hence, 88 × 92 = 8096.

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 94 × 106

⇒ (100 - 6) × (100 + 6)

⇒ (100)² - (6)²

⇒ 10000 - 36

⇒ 9964

Hence, 94 × 106 = 9964.

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 48 × 52

⇒ (50 - 2) × (50 + 2)

⇒ (50)² - (2)²

⇒ 2500 - 4

⇒ 2496

Hence, 48 × 52 = 2496.

(i) The perfect squares 4, 9, 121 and 625 have one, one, three and three digits respectively.

Hence, the given statement is false.

(ii) A perfect square can only have an even number of zeros at the end but the converse is not true.

For example, 300, 1200 and 50000 end with an even number of zeros but these are not perfect squares.

Hence, the given statement is false.

(iii) The square of a prime number $p$ is $p \times p$, which has $p$ as a factor other than 1 and itself.

For example, 2 is a prime number but 2² = 4, which is not a prime number.

Hence, the given statement is false.

(iv) The perfect squares from 1 to 100 are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100, which are 10 in number.

Hence, the given statement is true.

(v) The numbers 9 and 16 are perfect squares.

9 + 16 = 25 = 5², which is a perfect square.

But, the numbers 4 and 9 are also perfect squares and 4 + 9 = 13, which is not a perfect square.

Hence, the given statement is false.

(vi) The numbers 25 and 9 are perfect squares.

25 - 9 = 16 = 4², which is a perfect square.

But, the numbers 9 and 4 are also perfect squares and 9 - 4 = 5, which is not a perfect square.

Hence, the given statement is false.

(vii) Let the two perfect squares be $a^2$ and $b^2$.

$a^2 \times b^2 = (ab)^2$, which is a perfect square.

For example, 4 × 9 = 36 = 6², which is a perfect square.

Hence, the given statement is true.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{576} = 24}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{961} = 31}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{1444} = 38}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{4489} = 67}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{6241} = 79}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{5476} = 74}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{9025} = 95}$.

Marking periods and using the division method, we have :

Hence, $\mathbf{\sqrt{7569} = 87}$.

Let the side of the square field be $x$ metres.

Area of the square field = $x^2$

Area of the square field = 7744 sq. metres (given)

$$\begin{aligned}\Rightarrow x^2 = 7744\\[1em] \Rightarrow x = \sqrt{7744}\end{aligned}$$

Marking periods and using the division method, we have :

⇒ $x = 88$

Side of the square field = 88 metres

Perimeter of the square field = 4 × side = 4 × 88 = 352 metres

Hence, the perimeter of the square field is 352 metres.

Let us try to find the square root of 1104.

Clearly, (33)² is less than 1104 by 15.

So, the least number to be subtracted from 1104 to get a perfect square = 15.

Resulting perfect square number = 1104 - 15 = 1089 and $\sqrt{1089} = 33$

Hence, 15 should be subtracted from 1104, the perfect square so obtained is 1089 and $\mathbf{\sqrt{1089} = 33}$.

Let us try to find the square root of 7956.

Clearly, (89)² is less than 7956 by 35.

So, the least number to be subtracted from 7956 to get a perfect square = 35.

Resulting perfect square number = 7956 - 35 = 7921 and $\sqrt{7921} = 89$

Hence, 35 should be subtracted from 7956, the perfect square so obtained is 7921 and $\mathbf{\sqrt{7921} = 89}$.

Let us try to find the square root of 6203.

Clearly, (78)² < 6203 < (79)².

∴ Required number to be added = (79)² - 6203 = 6241 - 6203 = 38.

Resulting number = 6203 + 38 = 6241 = (79)² and $\sqrt{6241} = 79$

Hence, 38 should be added to 6203, the perfect square so obtained is 6241 and $\mathbf{\sqrt{6241} = 79}$.

Let us try to find the square root of 7348.

Clearly, (85)² < 7348 < (86)².

∴ Required number to be added = (86)² - 7348 = 7396 - 7348 = 48.

Resulting number = 7348 + 48 = 7396 = (86)².

and, $\sqrt{7396} = 86$

Hence, 48 should be added to 7348, the perfect square so obtained is 7396 and $\mathbf{\sqrt{7396} = 86}$.

Greatest number of six digits = 999999

Let us try to find the square root of 999999

Clearly, (999)² is less than 999999 by 1998.

So, the least number to be subtracted is 1998.

Required number = 999999 - 1998 = 998001 and $\sqrt{998001} = 999$

Hence, the greatest number of six digits which is a perfect square is 998001 and $\mathbf{\sqrt{998001} = 999}$.

Least number of four digits = 1000.

Let us find its square root.

Clearly, (31)² < 1000 < (32)².

∴ Required number to be added = (32)² - 1000 = 1024 - 1000 = 24.

Required number = 1000 + 24 = 1024 = (32)².

Hence, the least number of four digits which is a perfect square is 1024.

Let us try to find the square root of 69192.

Clearly, (263)² is less than 69192 by 23.

So, the least number to be subtracted from 69192 to get a perfect square = 23.

Resulting perfect square number = 69192 - 23 = 69169 = (263)².

Hence, 69192 must be decreased by 23 to make it a perfect square.

Let us try to find the square root of 69192.

Clearly, (263)² < 69192 < (264)².

∴ Required number to be added = (264)² - 69192 = 69696 - 69192 = 504.

Resulting number = 69192 + 504 = 69696 = (264)².

Hence, 69192 must be increased by 504 to make it a perfect square.

Resolving 69192 into prime factors, we get :

$$\begin{array}{l|r} 2 & 69192 \\ \hline 2 & 34596 \\ \hline 2 & 17298 \\ \hline 3 & 8649 \\ \hline 3 & 2883 \\ \hline 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$$

69192 = (2 × 2) × (3 × 3) × (31 × 31) × 2

Thus, 2, 2 and 3, 3 and 31, 31 exist in pairs while one 2 is alone.

So, we should multiply the given number by 2 to get a perfect square number.

$$\begin{aligned}\Rightarrow 69192 \times 2 = 138384\\[1em] \Rightarrow 138384 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (31 \times 31)\\[1em] \Rightarrow \sqrt{138384} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (31 \times 31)} \\[1em] \Rightarrow \sqrt{138384} = 2 \times 2 \times 3 \times 31\\[1em] \Rightarrow \sqrt{138384} = 372\end{aligned}$$

Hence, 69192 must be multiplied by 2 to make it a perfect square.

Resolving 69192 into prime factors, we get :

$$\begin{array}{l|r} 2 & 69192 \\ \hline 2 & 34596 \\ \hline 2 & 17298 \\ \hline 3 & 8649 \\ \hline 3 & 2883 \\ \hline 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$$

69192 = (2 × 2) × (3 × 3) × (31 × 31) × 2

Clearly, in the product of prime factors, the factor 2 does not exist in pair.

So, we must divide the given number by 2.

$$\begin{aligned}\Rightarrow 69192 \div 2 = 34596\\[1em] \Rightarrow 34596 = (2 \times 2) \times (3 \times 3) \times (31 \times 31)\\[1em] \Rightarrow \sqrt{34596} = \sqrt{(2 \times 2) \times (3 \times 3) \times (31 \times 31)} \\[1em] \Rightarrow \sqrt{34596} = 2 \times 3 \times 31\\[1em] \Rightarrow \sqrt{34596} = 186\end{aligned}$$

Hence, 69192 must be divided by 2 to make it a perfect square.

Here, the number of decimal places is already even. So, mark the periods and proceed as follows :

Hence, $\mathbf{\sqrt{2.89} = 1.7}$.

Here, the number of decimal places is already even. So, mark the periods and proceed as follows :

Hence, $\mathbf{\sqrt{33.64} = 5.8}$.

Here, the number of decimal places is already even. So, mark the periods and proceed as follows :

Hence, $\mathbf{\sqrt{156.25} = 12.5}$.

Here, the number of decimal places is already even. So, mark the periods and proceed as follows :

Hence, $\mathbf{\sqrt{75.69} = 8.7}$.

Here, the number of decimal places is already even. So, mark the periods and proceed as follows :

Hence, $\mathbf{\sqrt{5.4289} = 2.33}$.

Here, the number of decimal places is already even. So, mark the periods and proceed as follows :

Hence, $\mathbf{\sqrt{18.1476} = 4.26}$.

Since the square root is required to be correct to 2 places of decimal, we shall find it up to 3 places of decimal and round it off to 2 places of decimal.

For finding the square root up to 3 decimal places, we need 3 pairs of decimal places.

So, we write 2 = 2.000000.

Now, mark off periods and extract the square root as shown below :

$\sqrt{2} = 1.414$ (up to 3 places of decimal)

$\sqrt{2} = 1.41$ (correct up to 2 places of decimal)

Hence, $\mathbf{\sqrt{2} = 1.41}$.

Since the square root is required to be correct to 2 places of decimal, we shall find it up to 3 places of decimal and round it off to 2 places of decimal.

So, we write 0.4 = 0.400000.

Now, mark off periods and extract the square root as shown below :

$\sqrt{0.4} = 0.632$ (up to 3 places of decimal)

$\sqrt{0.4} = 0.63$ (correct up to 2 places of decimal)

Hence, $\mathbf{\sqrt{0.4} = 0.63}$.

Since the square root is required to be correct to 2 places of decimal, we shall find it up to 3 places of decimal and round it off to 2 places of decimal.

So, we write 2.8 = 2.800000.

Now, mark off periods and extract the square root as shown below :

$\sqrt{2.8} = 1.673$ (up to 3 places of decimal)

$\sqrt{2.8} = 1.67$ (correct up to 2 places of decimal)

Hence, $\mathbf{\sqrt{2.8} = 1.67}$.

By formula,

Area of the rectangle = length × breadth

Area of the rectangle = 13.6 x 3.4

Area of the rectangle = 46.24 m²

Let the length of each side of the square be $x$ metres.

Area of the square = Area of the rectangle

$$\begin{aligned}\Rightarrow x^2 = 46.24\\[1em] \Rightarrow x = \sqrt{46.24}\end{aligned}$$

Marking periods and using the division method, we have :

⇒ $x = 6.8$

Hence, the length of each side of the square is 6.8 metres.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{1 + \frac{27}{169}} = 1 + \frac{x}{13}\\[1em] \Rightarrow \sqrt{\frac{169 + 27}{169}} = 1 + \frac{x}{13}\\[1em] \Rightarrow \sqrt{\frac{196}{169}} = 1 + \frac{x}{13}\\[1em] \Rightarrow \frac{\sqrt{196}}{\sqrt{169}} = 1 + \frac{x}{13}\\[1em] \Rightarrow \frac{14}{13} = 1 + \frac{x}{13}\\[1em] \Rightarrow 1 + \frac{1}{13} = 1 + \frac{x}{13}\\[1em] \Rightarrow \frac{1}{13} = \frac{x}{13}\\[1em] \Rightarrow x = 1\end{aligned}$$

Hence, $\bm{x = 1}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{3\frac{13}{36}}\\[1em] \Rightarrow \sqrt{\frac{121}{36}}\\[1em] \Rightarrow \frac{\sqrt{121}}{\sqrt{36}}\\[1em] \Rightarrow \frac{\sqrt{11 \times 11}}{\sqrt{(2 \times 2) \times (3 \times 3)}}\\[1em] \Rightarrow \frac{11}{6}\\[1em] \Rightarrow 1\frac{5}{6}\end{aligned}$$

Hence, $\mathbf{\sqrt{3\frac{13}{36}} = 1\frac{5}{6}}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{4\frac{73}{324}}\\[1em] \Rightarrow \sqrt{\frac{1369}{324}}\\[1em] \Rightarrow \frac{\sqrt{1369}}{\sqrt{324}}\\[1em] \Rightarrow \frac{\sqrt{37 \times 37}}{\sqrt{(2 \times 2) \times (3 \times 3) \times (3 \times 3)}}\\[1em] \Rightarrow \frac{37}{18}\\[1em] \Rightarrow 2\frac{1}{18}\end{aligned}$$

Hence, $\mathbf{\sqrt{4\frac{73}{324}} = 2\frac{1}{18}}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{3\frac{942}{2209}}\\[1em] \Rightarrow \sqrt{\frac{7569}{2209}}\\[1em] \Rightarrow \frac{\sqrt{7569}}{\sqrt{2209}}\\[1em] \Rightarrow \frac{\sqrt{(3 \times 3) \times (29 \times 29)}}{\sqrt{47 \times 47}}\\[1em] \Rightarrow \frac{3 \times 29}{47}\\[1em] \Rightarrow \frac{87}{47}\\[1em] \Rightarrow 1\frac{40}{47}\end{aligned}$$

Hence, $\mathbf{\sqrt{3\frac{942}{2209}} = 1\frac{40}{47}}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{\frac{1089}{4624}}\\[1em] \Rightarrow \frac{\sqrt{1089}}{\sqrt{4624}}\\[1em] \Rightarrow \frac{\sqrt{(3 \times 3) \times (11 \times 11)}}{\sqrt{(2 \times 2) \times (2 \times 2) \times (17 \times 17)}}\\[1em] \Rightarrow \frac{3 \times 11}{2 \times 2 \times 17}\\[1em] \Rightarrow \frac{33}{68}\end{aligned}$$

Hence, $\mathbf{\sqrt{\frac{1089}{4624}} = \frac{33}{68}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\sqrt{243}}{\sqrt{867}}\\[1em] \Rightarrow \sqrt{\frac{243}{867}}\\[1em] \Rightarrow \sqrt{\frac{\cancel{3} \times 81}{\cancel{3} \times 289}}\\[1em] \Rightarrow \frac{\sqrt{81}}{\sqrt{289}}\\[1em] \Rightarrow \frac{\sqrt{9 \times 9}}{\sqrt{17 \times 17}}\\[1em] \Rightarrow \frac{9}{17}\end{aligned}$$

Hence, $\mathbf{\frac{\sqrt{243}}{\sqrt{867}} = \frac{9}{17}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\sqrt{1183}}{\sqrt{2023}}\\[1em] \Rightarrow \sqrt{\frac{1183}{2023}}\\[1em] \Rightarrow \sqrt{\frac{\cancel{7} \times 169}{\cancel{7} \times 289}}\\[1em] \Rightarrow \sqrt{\frac{169}{289}}\\[1em] \Rightarrow \frac{\sqrt{169}}{\sqrt{289}}\\[1em] \Rightarrow \frac{\sqrt{13 \times 13}}{\sqrt{17 \times 17}}\\[1em] \Rightarrow \frac{13}{17}\end{aligned}$$

Hence, $\mathbf{\frac{\sqrt{1183}}{\sqrt{2023}} = \frac{13}{17}}$.

Marking periods and using the division method, we have :

∴ $\sqrt{15625} = 125$

Now, evaluating $\sqrt{156.25} + \sqrt{1.5625}$,

$$\begin{aligned}\Rightarrow \sqrt{156.25} + \sqrt{1.5625}\\[1em] \Rightarrow \sqrt{\frac{15625}{100}} + \sqrt{\frac{15625}{10000}}\\[1em] \Rightarrow \frac{\sqrt{15625}}{\sqrt{100}} + \frac{\sqrt{15625}}{\sqrt{10000}}\\[1em] \Rightarrow \frac{125}{10} + \frac{125}{100}\\[1em] \Rightarrow 12.5 + 1.25\\[1em] \Rightarrow 13.75\end{aligned}$$

Hence, $\mathbf{\sqrt{15625} = 125}$ and $\mathbf{\sqrt{156.25} + \sqrt{1.5625} = 13.75}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{99} \times \sqrt{396}\\[1em] \Rightarrow \sqrt{99 \times 396}\\[1em] \Rightarrow \sqrt{(3 \times 3) \times 11 \times (2 \times 2) \times (3 \times 3) \times 11}\\[1em] \Rightarrow \sqrt{(2 \times 2) \times (3 \times 3) \times (3 \times 3) \times (11 \times 11)}\\[1em] \Rightarrow 2 \times 3 \times 3 \times 11\\[1em] \Rightarrow 198\end{aligned}$$

Hence, $\mathbf{\sqrt{99} \times \sqrt{396} = 198}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{147} \times \sqrt{243}\\[1em] \Rightarrow \sqrt{147 \times 243}\\[1em] \Rightarrow \sqrt{3 \times (7 \times 7) \times (3 \times 3) \times (3 \times 3) \times 3}\\[1em] \Rightarrow \sqrt{(3 \times 3) \times (3 \times 3) \times (3 \times 3) \times (7 \times 7)}\\[1em] \Rightarrow 3 \times 3 \times 3 \times 7\\[1em] \Rightarrow 189\end{aligned}$$

Hence, $\mathbf{\sqrt{147} \times \sqrt{243} = 189}$.

Converting the decimals into a fraction of whole numbers:

$$\begin{aligned}\Rightarrow \sqrt{\frac{0.289}{0.00121}}\\[1em] \Rightarrow \sqrt{\frac{28900}{121}}\\[1em] \Rightarrow \frac{\sqrt{28900}}{\sqrt{121}}\\[1em] \Rightarrow \frac{\sqrt{(2 \times 2) \times (5 \times 5) \times (17 \times 17)}}{\sqrt{11 \times 11}}\\[1em] \Rightarrow \frac{2 \times 5 \times 17}{11}\\[1em] \Rightarrow \frac{170}{11}\\[1em] \Rightarrow 15\frac{5}{11}\end{aligned}$$

Hence, $\mathbf{\sqrt{\frac{0.289}{0.00121}} = \frac{170}{11} = 15\frac{5}{11}}$.

Converting the decimals into a fraction of whole numbers:

$$\begin{aligned}\Rightarrow \sqrt{\frac{48.4}{0.289}}\\[1em] \Rightarrow \sqrt{\frac{48400}{289}}\\[1em] \Rightarrow \frac{\sqrt{48400}}{\sqrt{289}}\\[1em] \Rightarrow \frac{\sqrt{(2 \times 2) \times (5 \times 5) \times (11 \times 11) \times (2 \times 2)}}{\sqrt{17 \times 17}}\\[1em] \Rightarrow \frac{2 \times 5 \times 11 \times 2}{17}\\[1em] \Rightarrow \frac{220}{17}\\[1em] \Rightarrow 12\frac{16}{17}\end{aligned}$$

Hence, $\mathbf{\sqrt{\frac{48.4}{0.289}} = \frac{220}{17} = 12\frac{16}{17}}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{0.01 + \sqrt{0.0064}}\\[1em] [\because \sqrt{0.0064} = 0.08]\\[1em] \Rightarrow \sqrt{0.01 + 0.08}\\[1em] \Rightarrow \sqrt{0.09}\\[1em] \Rightarrow 0.3\end{aligned}$$

Hence, $\mathbf{\sqrt{0.01 + \sqrt{0.0064}} = 0.3}$.

Evaluating,

⇒ $\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.21} + \sqrt{0.0009}$

⇒ 0.1 + 0.9 + 1.1 + 0.03

⇒ 2.13

Hence, $\mathbf{\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.21} + \sqrt{0.0009} = 2.13}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}}\\[1em] [\because \sqrt{9} = 3]\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{19 - 3}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{16}}}\\[1em] [\because \sqrt{16} = 4]\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + 4}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{25}}\\[1em] [\because \sqrt{25} = 5]\\[1em] \Rightarrow \sqrt{41 - 5}\\[1em] \Rightarrow \sqrt{36}\\[1em] \Rightarrow 6\end{aligned}$$

Hence, $\mathbf{\sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}} = 6}$.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{10 + \sqrt{25 + \sqrt{108 + \sqrt{154 + \sqrt{225}}}}}\\[1em] [\because \sqrt{225} = 15]\\[1em] \Rightarrow \sqrt{10 + \sqrt{25 + \sqrt{108 + \sqrt{154 + 15}}}}\\[1em] \Rightarrow \sqrt{10 + \sqrt{25 + \sqrt{108 + \sqrt{169}}}}\\[1em] [\because \sqrt{169} = 13]\\[1em] \Rightarrow \sqrt{10 + \sqrt{25 + \sqrt{108 + 13}}}\\[1em] \Rightarrow \sqrt{10 + \sqrt{25 + \sqrt{121}}}\\[1em] [\because \sqrt{121} = 11]\\[1em] \Rightarrow \sqrt{10 + \sqrt{25 + 11}}\\[1em] \Rightarrow \sqrt{10 + \sqrt{36}}\\[1em] [\because \sqrt{36} = 6]\\[1em] \Rightarrow \sqrt{10 + 6}\\[1em] \Rightarrow \sqrt{16}\\[1em] \Rightarrow 4\end{aligned}$$

Hence, $\mathbf{\sqrt{10 + \sqrt{25 + \sqrt{108 + \sqrt{154 + \sqrt{225}}}}} = 4}$.

Let the number be $x$.

Three-fifths of the square of the number = 126.15

$$\begin{aligned}\Rightarrow \frac{3}{5}x^2 = 126.15\\[1em] \Rightarrow x^2 = \frac{126.15 \times 5}{3}\\[1em] \Rightarrow x^2 = \frac{630.75}{3}\\[1em] \Rightarrow x^2 = 210.25\\[1em] \Rightarrow x = \sqrt{210.25}\end{aligned}$$

Marking periods and using the division method, we have :

⇒ $x = 14.5$

Hence, the required number is 14.5.

The cube of a number is that number raised to the power 3.

⇒ (16)³

⇒ 16 × 16 × 16

⇒ 4096

Hence, (16)³ = 4096.

The cube of a number is that number raised to the power 3.

⇒ (30)³

⇒ 30 × 30 × 30

⇒ 27000

Hence, (30)³ = 27000.

The cube of a number is that number raised to the power 3.

⇒ (1.2)³

⇒ 1.2 × 1.2 × 1.2

⇒ 1.728

Hence, (1.2)³ = 1.728.

The cube of a number is that number raised to the power 3.

⇒ (0.7)³

⇒ 0.7 × 0.7 × 0.7

⇒ 0.343

Hence, (0.7)³ = 0.343.

The cube of a number is that number raised to the power 3.

⇒ (0.06)³

⇒ 0.06 × 0.06 × 0.06

⇒ 0.000216

Hence, (0.06)³ = 0.000216.

The cube of a number is that number raised to the power 3.

$$\begin{aligned}\Rightarrow \left(\frac{2}{5}\right)^3\\[1em] \Rightarrow \frac{2}{5} \times \frac{2}{5} \times \frac{2}{5}\\[1em] \Rightarrow \frac{8}{125}\end{aligned}$$

Hence, $\mathbf{\left(\frac{2}{5}\right)^3 = \frac{8}{125}}$.

The cube of a number is that number raised to the power 3.

$$\begin{aligned}\Rightarrow \left(\frac{1}{9}\right)^3\\[1em] \Rightarrow \frac{1}{9} \times \frac{1}{9} \times \frac{1}{9}\\[1em] \Rightarrow \frac{1}{729}\end{aligned}$$

Hence, $\mathbf{\left(\frac{1}{9}\right)^3 = \frac{1}{729}}$.

Converting mixed fraction into improper fraction:

$1\frac{3}{5} = \frac{8}{5}$

The cube of a number is that number raised to the power 3.

$$\begin{aligned}\Rightarrow \left(1\frac{3}{5}\right)^3\\[1em] \Rightarrow \left(\frac{8}{5}\right)^3\\[1em] \Rightarrow \frac{8}{5} \times \frac{8}{5} \times \frac{8}{5}\\[1em] \Rightarrow \frac{512}{125}\\[1em] \Rightarrow 4\frac{12}{125}\end{aligned}$$

Hence, $\mathbf{\left(1\frac{3}{5}\right)^3 = 4\frac{12}{125}}$.

Resolving 729 into prime factors, we get :

$$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

729 = (3 × 3 × 3) × (3 × 3 × 3)

Since 729 can be expressed as the product of triplets of equal factors, so 729 is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{729} = \sqrt[3]{(3 × 3 × 3) × (3 × 3 × 3)} \\[1em] \Rightarrow \sqrt[3]{729} = 3 \times 3\\[1em] \Rightarrow \sqrt[3]{729} = 9\end{aligned}$$

Hence, 729 is a perfect cube and $\mathbf{\sqrt[3]{729} = 9}$.

Resolving 1331 into prime factors, we get :

$$\begin{array}{l|r} 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

1331 = (11 × 11 × 11)

Since 1331 can be expressed as the product of triplets of equal factors, so 1331 is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{1331} = \sqrt[3]{11 \times 11 \times 11} \\[1em] \Rightarrow \sqrt[3]{1331} = 11\end{aligned}$$

Hence, 1331 is a perfect cube and $\mathbf{\sqrt[3]{1331} = 11}$.

Resolving 5324 into prime factors, we get :

$$\begin{array}{l|r} 2 & 5324 \\ \hline 2 & 2662 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

5324 = (11 × 11 × 11) × 2 × 2

Since 2 × 2 is left after grouping the prime factors in triplets, so 5324 cannot be expressed as the product of triplets of equal factors.

Hence, 5324 is not a perfect cube.

Resolving 3375 into prime factors, we get :

$$\begin{array}{l|r} 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

3375 = (3 × 3 × 3) × (5 × 5 × 5)

Since 3375 can be expressed as the product of triplets of equal factors, so 3375 is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{3375} = \sqrt[3]{(3 \times 3 \times 3) \times (5 \times 5 \times 5)} \\[1em] \Rightarrow \sqrt[3]{3375} = 3 \times 5\\[1em] \Rightarrow \sqrt[3]{3375} = 15\end{aligned}$$

Hence, 3375 is a perfect cube and $\mathbf{\sqrt[3]{3375} = 15}$.

Resolving 9261 into prime factors, we get :

$$\begin{array}{l|r} 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

9261 = (3 × 3 × 3) × (7 × 7 × 7)

Since 9261 can be expressed as the product of triplets of equal factors, so 9261 is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{9261} = \sqrt[3]{(3 \times 3 \times 3) \times (7 \times 7 \times 7)} \\[1em] \Rightarrow \sqrt[3]{9261} = 3 \times 7\\[1em] \Rightarrow \sqrt[3]{9261} = 21\end{aligned}$$

Hence, 9261 is a perfect cube and $\mathbf{\sqrt[3]{9261} = 21}$.

Resolving 5832 into prime factors, we get :

$$\begin{array}{l|r} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)

Since 5832 can be expressed as the product of triplets of equal factors, so 5832 is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{5832} = \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)} \\[1em] \Rightarrow \sqrt[3]{5832} = 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt[3]{5832} = 18\end{aligned}$$

Hence, 5832 is a perfect cube and $\mathbf{\sqrt[3]{5832} = 18}$.

Resolving 1728 into prime factors, we get :

$$\begin{array}{l|r} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

Since 1728 can be expressed as the product of triplets of equal factors, so 1728 is a perfect cube.

$$\begin{aligned}\Rightarrow \sqrt[3]{1728} = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)} \\[1em] \Rightarrow \sqrt[3]{1728} = 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt[3]{1728} = 12\end{aligned}$$

Hence, 1728 is a perfect cube and $\mathbf{\sqrt[3]{1728} = 12}$.

Resolving 10584 into prime factors, we get :

$$\begin{array}{l|r} 2 & 10584 \\ \hline 2 & 5292 \\ \hline 2 & 2646 \\ \hline 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

10584 = (2 × 2 × 2) × (3 × 3 × 3) × 7 × 7

Since 7 × 7 is left after grouping the prime factors in triplets, so 10584 cannot be expressed as the product of triplets of equal factors.

Hence, 10584 is not a perfect cube.

Resolving 1323 into prime factors, we get :

$$\begin{array}{l|r} 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

1323 = (3 × 3 × 3) × 7 × 7

Clearly, to make it a perfect cube it must be multiplied by 7.

⇒ 1323 × 7 = 9261

⇒ 9261 = (3 × 3 × 3) × (7 × 7 × 7), which is a perfect cube.

Hence, the smallest number by which 1323 must be multiplied so that the product is a perfect cube is 7.

Resolving 1600 into prime factors, we get :

$$\begin{array}{l|r} 2 & 1600 \\ \hline 2 & 800 \\ \hline 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

1600 = (2 × 2 × 2) × (2 × 2 × 2) × 5 × 5

Clearly, to make it a perfect cube, it must be divided by (5 × 5), i.e., 25.

⇒ 1600 ÷ 25 = 64

⇒ 64 = (2 × 2 × 2) × (2 × 2 × 2), which is a perfect cube.

Hence, the smallest number by which 1600 must be divided so that the quotient is a perfect cube is 25.

Resolving 2560 into prime factors, we get :

$$\begin{array}{l|r} 2 & 2560 \\ \hline 2 & 1280 \\ \hline 2 & 640 \\ \hline 2 & 320 \\ \hline 2 & 160 \\ \hline 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

2560 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 5

Clearly, to make it a perfect cube it must be multiplied by (5 × 5), i.e., 25.

$$\begin{aligned}\Rightarrow 2560 \times 25 = 64000\\[1em] \Rightarrow 64000 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5), \text{ which is a perfect cube.}\end{aligned}$$

Hence, the smallest number by which 2560 must be multiplied so that the product is a perfect cube is 25.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{\frac{216}{2197}}\\[1em] \Rightarrow \frac{\sqrt[3]{216}}{\sqrt[3]{2197}}\\[1em] \Rightarrow \frac{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}}{\sqrt[3]{13 \times 13 \times 13}}\\[1em] \Rightarrow \frac{2 \times 3}{13}\\[1em] \Rightarrow \frac{6}{13}\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{\frac{216}{2197}} = \frac{6}{13}}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{4\frac{508}{1331}}\\[1em] \Rightarrow \sqrt[3]{\frac{5832}{1331}}\\[1em] \Rightarrow \frac{\sqrt[3]{5832}}{\sqrt[3]{1331}}\\[1em] \Rightarrow \frac{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}}{\sqrt[3]{11 \times 11 \times 11}}\\[1em] \Rightarrow \frac{2 \times 3 \times 3}{11}\\[1em] \Rightarrow \frac{18}{11}\\[1em] \Rightarrow 1\frac{7}{11}\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{4\frac{508}{1331}} = 1\frac{7}{11}}$.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{42.875}\\[1em] \Rightarrow \sqrt[3]{\frac{42875}{1000}}\\[1em] \Rightarrow \frac{\sqrt[3]{42875}}{\sqrt[3]{1000}}\\[1em] \Rightarrow \frac{\sqrt[3]{(5 \times 5 \times 5) \times (7 \times 7 \times 7)}}{\sqrt[3]{10 \times 10 \times 10}}\\[1em] \Rightarrow \frac{5 \times 7}{10}\\[1em] \Rightarrow \frac{35}{10}\\[1em] \Rightarrow 3.5\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{42.875} = 3.5}$.

We know that 11² = 121 and 17² = 289, which lie between 120 and 300.

Also, 10² = 100 < 120 and 18² = 324 > 300.

So, the perfect squares lying between 120 and 300 are :

121, 144, 169, 196, 225, 256 and 289

These are 7 in number.

Hence, option 3 is the correct option.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{19 - 3}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + \sqrt{16}}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{21 + 4}}\\[1em] \Rightarrow \sqrt{41 - \sqrt{25}}\\[1em] \Rightarrow \sqrt{41 - 5}\\[1em] \Rightarrow \sqrt{36}\\[1em] \Rightarrow 6\end{aligned}$$

Hence, option 3 is the correct option.

Evaluating,

⇒ $\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.44} + \sqrt{0.0009}$

⇒ 0.1 + 0.9 + 1.2 + 0.03

⇒ 2.23

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{52}{x} = \sqrt{\frac{169}{289}}\\[1em] \Rightarrow \frac{52}{x} = \frac{\sqrt{169}}{\sqrt{289}}\\[1em] \Rightarrow \frac{52}{x} = \frac{13}{17}\\[1em] \Rightarrow 13x = 52 \times 17\\[1em] \Rightarrow x = \frac{52 \times 17}{13}\\[1em] \Rightarrow x = 4 \times 17\\[1em] \Rightarrow x = 68\end{aligned}$$

Hence, option 4 is the correct option.

Let the required number be $x$.

$$\begin{aligned}\Rightarrow x \div \sqrt{0.25} = 25\\[1em] \Rightarrow x \div 0.5 = 25\\[1em] \Rightarrow x = 25 \times 0.5\\[1em] \Rightarrow x = 12.5\end{aligned}$$

Hence, option 1 is the correct option.

Let the value of $\star$ be $x$.

$$\begin{aligned}\Rightarrow \left(\frac{x}{15}\right)\left(\frac{x}{135}\right) = 1\\[1em] \Rightarrow \frac{x^2}{2025} = 1\\[1em] \Rightarrow x^2 = 2025\\[1em] \Rightarrow x = \sqrt{2025}\\[1em] \Rightarrow x = \sqrt{(3 \times 3) \times (3 \times 3) \times (5 \times 5)}\\[1em] \Rightarrow x = 3 \times 3 \times 5\\[1em] \Rightarrow x = 45\end{aligned}$$

Hence, option 4 is the correct option.

We know that 1² = 1 and 9² = 81, so the square of a number ends in 1 only when the number ends in 1 or 9.

The numbers from 1 to 50 ending in 1 are 1, 11, 21, 31 and 41.

The numbers from 1 to 50 ending in 9 are 9, 19, 29, 39 and 49.

∴ Number of such numbers = 5 + 5 = 10

Required percentage = $\frac{10}{50}$ x 100%

Required percentage = 20%

Hence, option 4 is the correct option.

According to the property of perfect squares, a number ending in 2, 3, 7 or 8 is never a perfect square.

So, 8 cannot be the unit digit of a perfect square number.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\sqrt{288}}{\sqrt{128}}\\[1em] \Rightarrow \sqrt{\frac{288}{128}}\\[1em] \Rightarrow \sqrt{\frac{\cancel{32} \times 9}{\cancel{32} \times 4}}\\[1em] \Rightarrow \sqrt{\frac{9}{4}}\\[1em] \Rightarrow \frac{\sqrt{9}}{\sqrt{4}}\\[1em] \Rightarrow \frac{3}{2}\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{72} \times \sqrt{98}\\[1em] \Rightarrow \sqrt{72 \times 98}\\[1em] \Rightarrow \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 7 \times 7}\\[1em] \Rightarrow \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)}\\[1em] \Rightarrow 2 \times 2 \times 3 \times 7\\[1em] \Rightarrow 84\end{aligned}$$

Hence, option 4 is the correct option.

Resolving 294 into prime factors, we get :

$$\begin{array}{l|r} 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

294 = (7 × 7) × 2 × 3

Since the prime factors 2 and 3 are left unpaired, so 294 should be multiplied by 2 × 3, i.e., 6.

Hence, option 3 is the correct option.

Resolving 1470 into prime factors, we get :

$$\begin{array}{l|r} 2 & 1470 \\ \hline 3 & 735 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

1470 = (7 × 7) × 2 × 3 × 5

Since the prime factors 2, 3 and 5 are left unpaired, so 1470 should be divided by 2 × 3 × 5, i.e., 30.

Hence, option 4 is the correct option.

Let us try to find the square root of 7700.

Clearly, (87)² < 7700 < (88)².

∴ Required number to be added = (88)² - 7700 = 7744 - 7700 = 44.

Hence, option 1 is the correct option.

Solving,

⇒ $\sqrt{110.25} \times \sqrt{0.01} \div \sqrt{0.0025} - \sqrt{420.25}$

⇒ 10.5 × 0.1 ÷ 0.05 - 20.5

⇒ 1.05 ÷ 0.05 - 20.5

⇒ 21 - 20.5

⇒ 0.5

Hence, option 1 is the correct option.

Greatest number of four digits = 9999.

Let us try to find the square root of 9999.

Clearly, (99)² is less than 9999 by 198.

So, the least number to be subtracted is 198.

Required number = 9999 - 198 = 9801

Hence, option 2 is the correct option.

Let the number of rows be $x$.

Then, the number of trees in each row = number of rows = $x$.

∴ Total number of trees = $x \times x = x^2$.

But, total number of trees = 17956 (given).

$$\begin{aligned}\Rightarrow x^2 = 17956\\[1em] \Rightarrow x = \sqrt{17956}\end{aligned}$$

Marking periods and using the division method, we have :

⇒ $x = 134$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{4\frac{12}{125}}\\[1em] \Rightarrow \sqrt[3]{\frac{512}{125}}\\[1em] \Rightarrow \frac{\sqrt[3]{512}}{\sqrt[3]{125}}\\[1em] \Rightarrow \frac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{5 \times 5 \times 5}}\\[1em] \Rightarrow \frac{2 \times 2 \times 2}{5}\\[1em] \Rightarrow \frac{8}{5}\\[1em] \Rightarrow 1\frac{3}{5}\end{aligned}$$

Hence, option 2 is the correct option.

Resolving 21600 into prime factors, we get :

$$\begin{array}{l|r} 2 & 21600 \\ \hline 2 & 10800 \\ \hline 2 & 5400 \\ \hline 2 & 2700 \\ \hline 2 & 1350 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

21600 = (2 × 2 × 2) × (3 × 3 × 3) × 2 × 2 × 5 × 5

Since 2 × 2 and 5 × 5 are left after grouping the prime factors in triplets, so 21600 must be multiplied by 2 × 5, i.e., 10.

⇒ 21600 x 10 = 216000

⇒ 216000 = (2 x 2 x 2) x (2 x 2 x 2) x (3 x 3 x 3) x (5 x 5 x 5), which is a perfect cube.

Hence, option 2 is the correct option.

Resolving 3600 into prime factors, we get :

$$\begin{array}{l|r} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

3600 = (2 × 2 × 2) × 2 × 3 × 3 × 5 × 5

Since 2, 3 × 3 and 5 × 5 are left after grouping the prime factors in triplets, so 3600 must be divided by 2 × 3 × 3 × 5 × 5, i.e., 450.

⇒ 3600 ÷ 450 = 8

⇒ 8 = (2 × 2 × 2), which is a perfect cube.

Hence, option 4 is the correct option.

(i) The perfect squares lying between 1 and 100 are 4, 9, 16, 25, 36, 49, 64 and 81, which are 8 in number.

Hence, there are 8 perfect squares between 1 and 100.

(ii) The perfect cubes lying between 1 and 100 are 8, 27 and 64, which are 3 in number.

Hence, there are 3 perfect cubes between 1 and 100.

(iii) According to the property of perfect squares, a number ending in 2, 3, 7 or 8 is never a perfect square.

Hence, a perfect square number cannot have 2, 3, 7 and 8 as the unit digit.

(iv) We know that 7³ = 343, which ends in the digit 3.

For example, 17³ = 4913 and 27³ = 19683, which end in the digit 3.

Hence, the cube of a number ending in 7 will end in the digit 3.

(v) 1 m = 100 cm

$$\begin{aligned}\Rightarrow 1 \text{ m}^2 = 100 \times 100 = 10000 \text{ cm}^2\\[1em] \Rightarrow 1 \text{ m}^3 = 100 \times 100 \times 100 = 1000000 \text{ cm}^3\end{aligned}$$

Hence, 1 m² = 10000 cm² and 1 m³ = 1000000 cm³.

(vi) (100)³ = 100 × 100 × 100 = 1000000

The number 1000000 has 6 zeroes.

Hence, (100)³ has 6 zeroes.

(i) False

Reason — The numbers 4 and 9 are perfect squares but 4 + 9 = 13, which is not a perfect square.

(ii) True

Reason — Let the two perfect squares be $a^2$ and $b^2$. Then, $a^2 \times b^2 = (ab)^2$, which is a perfect square. For example, 4 × 9 = 36 = 6².

(iii) False

Reason — For the natural number 1, 1² = 1, which is equal to the number itself and not greater than it.

(iv) False

Reason — For the natural number 1, 1² = 1 and 1³ = 1. So, the square of 1 is equal to its cube.

(v) False

Reason — (200)³ = 200 × 200 × 200 = 8000000 and not 80000. Also, 80000 = (2 × 2 × 2) × (2 × 2 × 2) × 2 × (5 × 5 × 5) × 5 is not a perfect cube, so its cube root is not a whole number.

We know that there are $2n$ non-perfect square numbers between the squares of two consecutive numbers $n$ and $(n + 1)$.

So, Reason (R) is true.

Here, $n = 21$

Natural numbers between 21² and 22² = 2 × 21 = 42

So, Assertion (A) is true and Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Let $a = -2$.

$$\begin{aligned}\Rightarrow a^2 = (-2)^2 = 4\\[1em] \Rightarrow a^3 = (-2)^3 = -8\end{aligned}$$

Here, $a^3$ is less than $a^2$.

So, Assertion (A) is false.

The square of a negative integer is positive as the product of two negative integers is positive and the cube of a negative integer is negative as the product of three negative integers is negative.

So, Reason (R) is true.

Hence, option 4 is the correct option.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}\\[1em] \Rightarrow \sqrt[3]{3 \times 3 \times 3} + \sqrt[3]{0.2 \times 0.2 \times 0.2} + \sqrt[3]{0.4 \times 0.4 \times 0.4}\\[1em] \Rightarrow 3 + 0.2 + 0.4\\[1em] \Rightarrow 3.6\end{aligned}$$

So, Assertion (A) is true.

The cube of a negative number is negative, so every negative number has a cube root which is negative.

For example, $\sqrt[3]{-27} = -3$.

So, Reason (R) is false.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{\frac{3}{7}}\\[1em] \Rightarrow \sqrt{\frac{3 \times 7}{7 \times 7}}\\[1em] \Rightarrow \sqrt{\frac{21}{49}}\\[1em] \Rightarrow \frac{\sqrt{21}}{\sqrt{49}}\\[1em] \Rightarrow \frac{4.578}{7}\\[1em] \Rightarrow 0.654\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{256} \div \sqrt{x} = 2\\[1em] \Rightarrow \frac{16}{\sqrt{x}} = 2\\[1em] \Rightarrow \sqrt{x} = \frac{16}{2}\\[1em] \Rightarrow \sqrt{x} = 8\\[1em] \Rightarrow x = 8^2\\[1em] \Rightarrow x = 64\end{aligned}$$

Hence, option 2 is the correct option.

Evaluating,

$$\begin{aligned}\Rightarrow \sqrt{248 + \sqrt{52 + \sqrt{144}}}\\[1em] [\because \sqrt{144} = 12]\\[1em] \Rightarrow \sqrt{248 + \sqrt{52 + 12}}\\[1em] \Rightarrow \sqrt{248 + \sqrt{64}}\\[1em] [\because \sqrt{64} = 8]\\[1em] \Rightarrow \sqrt{248 + 8}\\[1em] \Rightarrow \sqrt{256}\\[1em] \Rightarrow 16\end{aligned}$$

Hence, option 3 is the correct option.

Let $x = -5$.

$$\begin{aligned}\Rightarrow y = x^2 = (-5)^2 = 25\\[1em] \Rightarrow \sqrt{y} = \sqrt{25} = 5\end{aligned}$$

Here, $x = -5$ but $\sqrt{y}$ = 5.

So, $x$ is not equal to $\sqrt{y}$ when $x$ is a negative integer.

Hence, option 2 is the correct option.

We know that there are $2n$ non-perfect square numbers between the squares of two consecutive numbers $n$ and $(n + 1)$.

⇒ $2n = 24$

⇒ $n = 12$

So, the two consecutive perfect squares are 12² and 13².

⇒ 12² + 13²

⇒ 144 + 169

⇒ 313

Hence, option 4 is the correct option.

Let the number be 0.5.

⇒ (0.5)³ = 0.5 × 0.5 × 0.5 = 0.125

Here, 0.125 < 0.5, so the cube is smaller than the number itself.

Again, let the number be -2.

⇒ (-2)³ = -8

Here, -8 < -2, so the cube is smaller than the number itself.

Hence, option 1 is the correct option.

Let the three distinct numbers be 4, 6 and 9.

None of the numbers 4, 6 and 9 is a perfect cube.

⇒ 4 x 6 x 9 = 216

⇒ 216 = (2 x 2 x 2) x (3 x 3 x 3) = 6³, which is a perfect cube.

So, the product of three distinct non-cube numbers can be a perfect cube.

Hence, option 2 is the correct option.

Given, $x$ is a positive rational number such that $x \lt \frac{1}{x}$.

$$\begin{aligned}\Rightarrow x^2 \lt 1\\[1em] \Rightarrow x \lt 1\end{aligned}$$

So, $x^3 \lt 1$.

Given, $y$ is a positive rational number such that $y \gt \frac{1}{y}$.

$$\begin{aligned}\Rightarrow y^2 \gt 1\\[1em] \Rightarrow y \gt 1\end{aligned}$$

So, $y^3$ > 1.

Since $x^3 \lt 1$ and $y^3 \gt 1$, so $x^3 \lt y^3$.

∴ $x^3 - y^3$ is negative.

Hence, option 2 is the correct option.

Solving,

Resolving each number into prime factors,

4 = 2², 27 = 3³, 108 = 2² × 3³ and 256 = 2⁸

$$\begin{aligned}\Rightarrow x^2 = 4 \times 27 \times 108 \times 256\\[1em] \Rightarrow x^2 = 2^2 \times 3^3 \times (2^2 \times 3^3) \times 2^8\\[1em] \Rightarrow x^2 = 2^{(2 + 2 + 8)} \times 3^{(3 + 3)}\\[1em] \Rightarrow x^2 = 2^{12} \times 3^6\\[1em] \Rightarrow x = \sqrt{2^{12} \times 3^6}\\[1em] \Rightarrow x = 2^6 \times 3^3\\[1em] \Rightarrow x = 64 \times 27\\[1em] \Rightarrow x = 1728\end{aligned}$$

Now, finding the cube root of 1728,

$$\begin{aligned}\Rightarrow \sqrt[3]{x}\\[1em] \Rightarrow \sqrt[3]{1728}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 3\\[1em] \Rightarrow 12\end{aligned}$$

Hence, option 3 is the correct option.

Since $x$ is a perfect cube, let $x = a^3$.

Since $y$ is a perfect square, let $y = b^2$.

Taking $x = 8$ and $y = 9$,

(i) $x \times y = 8 \times 9 = 72$, which is not a perfect cube.

(ii) $x \times y^2 = 8 \times 81 = 648$, which is not a perfect cube.

(iii) $x^2 \times y^2 = 64 \times 81 = 5184$, which is not a perfect cube.

(iv) $x^2 \times y^3 = (a^3)^2 \times (b^2)^3 = a^6 \times b^6 = (a^2b^2)^3$, which is a perfect cube.

For example, $x^2 \times y^3 = 64 \times 729 = 46656 = (36)^3$, which is a perfect cube.

Hence, option 3 is the correct option.