Squares and Square Roots

$$\begin{aligned}(-8)^2\\[1em] = (-8) \times (-8)\\[1em] = 64\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}-\sqrt{18 - \sqrt{11-7}}\\[1em] = -\sqrt{18 - \sqrt{4}}\\[1em] = -\sqrt{18 - 2}\\[1em] = -\sqrt{16}\\[1em] = -4\end{aligned}$$

Hence, option 2 is the correct option.

$$\begin{aligned}\sqrt{2.50 \times 10^3}\\[1em] = \sqrt{2.50 \times 1000}\\[1em] = \sqrt{2500}\\[1em] = 50\end{aligned}$$

Hence, option 3 is the correct option.

Finding prime factors of 48

48 = (2 x 2) x (2 x 2) x 3

Since the prime factor 3 is not in pair.

The given number should be multiplied by 3.

Hence, option 2 is the correct option.

Finding prime factors of 175

175 = (5 x 5) x 7

Since the prime factor 7 is not in pair.

The given number should be divided by 7.

Hence, option 2 is the correct option.

$$\begin{aligned}59^2\\[1em] = 59 \times 59\\[1em] = 3481\end{aligned}$$

$59^2 = 3481$

$$\begin{aligned}6.3^2\\[1em] = 6.3 \times 6.3\\[1em] = 39.69\end{aligned}$$

$6.3^2 = 39.69$

$$\begin{aligned}\Big[15\frac{2}{3}\Big]^2\\[1em] = \Big[\frac{47}{3}\Big]^2\\[1em] = \frac{47 \times 47}{3 \times 3}\\[1em] = \frac{2209}{9}\\[1em] = 245\frac{4}{9}\end{aligned}$$

$15\frac{2}{3} = 245\frac{4}{9}$

$$\begin{aligned}11025 = (3 \times 3) \times (5 \times 5) \times (7 \times 7)\\[1em] ⇒ \sqrt{11025} = 3 \times 5 \times 7\\[1em] ⇒ \sqrt{11025} = 105\end{aligned}$$

Square root of 11025 is 105.

$$\begin{aligned}396900 = (2 \times 2) \times (3 \times 3) \times (3 \times 3) \times (5 \times 5) \times (7 \times 7)\\[1em] ⇒ \sqrt{396900} = 2 \times 3 \times 3 \times 5 \times 7\\[1em] ⇒ \sqrt{396900} = 630\end{aligned}$$

Square root of 396900 is 630.

$$\begin{aligned}194481 = (3 \times 3) \times (3 \times 3) \times (7 \times 7) \times (7 \times 7)\\[1em] ⇒ \sqrt{194481} = 3 \times 3 \times 7 \times 7\\[1em] ⇒ \sqrt{194481} = 441\end{aligned}$$

Square root of 194481 is 441.

Finding prime factors of 2592

$2592 = (2 \times 2) \times (2 \times 2) \times 2 \times (3 \times 3)$

Since the prime factor 2 is not in pair.

2 should be multiplied with 2592 so that the product is a perfect square.

Finding prime factors of 12748

$12748 = (2 \times 2) \times 3187$

Since the prime factor 3187 is not in pair.

3187 should be multiplied with 12748 so that the product is a perfect square.

Finding prime factors of 10368

$10368 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 2 \times (3 \times 3) \times (3 \times 3)$

Since the prime factor 2 is not in pair.

$$\begin{aligned}10368 ÷ 2 = 5184\\[1em] \sqrt{5184} = 2 \times 2 \times 2 \times 3 \times 3\\[1em] \sqrt{5184} = 72\end{aligned}$$

2 should be divided by 10368 so that the 5184 is a perfect square and square root of 5184 is 72.

$$\begin{aligned}0.1764 = \frac{1764}{10000}\\[1em] \sqrt{\frac{1764}{10000}} = \frac{\sqrt{1764}}{\sqrt{10000}}\\[1em] = \frac{\sqrt{(2\times2)\times(3\times3)\times(7\times7)}}{\sqrt{(2\times2)\times(2\times2)\times(5\times5)\times(5\times5)}}\\[1em] = \frac{2\times3\times7}{2\times2\times5\times5}\\[1em] = \frac{42}{100}\\[1em] = 0.42\end{aligned}$$

$\sqrt{0.1764} = 0.42$

$$\begin{aligned}\sqrt{96\frac{1}{25}} = \sqrt{\frac{2401}{25}}\\[1em] = \frac{\sqrt{2401}}{\sqrt{25}}\\[1em] = \frac{\sqrt{(7\times7)\times(7\times7)}}{\sqrt{(5\times5)}}\\[1em] = \frac{7\times7}{5}\\[1em] = \frac{49}{5}\\[1em] = 9\frac{4}{5}\end{aligned}$$

$\sqrt{96\frac{1}{25}} = 9\frac{4}{5}$

$$\begin{aligned}\sqrt{0.0169} = \sqrt{\frac{169}{10000}}\\[1em] = \frac{\sqrt{169}}{\sqrt{10000}}\\[1em] = \frac{\sqrt{13\times13}}{\sqrt{10\times10}}\\[1em] = \frac{13}{100}\\[1em] = 0.13\end{aligned}$$

$\sqrt{0.0169} = 0.13$

$$\begin{aligned}\sqrt{\frac{14.4}{22.5}}\\[1em] = \frac{\sqrt{144 \times 10}}{\sqrt{225 \times 10}}\\[1em] = \frac{\sqrt{144}}{\sqrt{225}}\\[1em] = \frac{\sqrt{(2\times2)\times(2\times2)\times(3\times3)}}{\sqrt{(3\times3)\times(5\times5)}}\\[1em] = \frac{2\times2\times3}{3\times5}\\[1em] = \frac{12}{15}\\[1em] = 0.8\end{aligned}$$

$\sqrt{\frac{14.4}{22.5}} = 0.8$

$$\begin{aligned}\sqrt{\frac{0.225}{28.9}}\\[1em] = \sqrt{\frac{0.225 \times 10000}{28.9 \times 10000}}\\[1em] = \frac{\sqrt{225 \times 10}}{\sqrt{289 \times 1000}}\\[1em] = \frac{\sqrt{225}}{\sqrt{28900}}\\[1em] = \frac{\sqrt{(3\times3)\times(5\times5)}}{\sqrt{(2\times2)\times(5\times5)\times(17\times17)}}\\[1em] = \frac{3\times5}{2\times5\times17}\\[1em] = \frac{15}{170}\\[1em] = \frac{3}{34}\end{aligned}$$

$\sqrt{\frac{0.225}{28.9}} = \frac{3}{34}$

$$\begin{aligned}\sqrt{\frac{25}{32} \times 2\frac{13}{18} \times 0.25}\\[1em] = \sqrt{\frac{25}{32} \times \frac{49}{18} \times \frac{25}{100}}\\[1em] = \sqrt{\frac{(5\times5)\times(7\times7)\times(5\times5)}{(2\times2)\times(2\times2)\times(2\times2)\times(3\times3)\times(2\times2)\times(5\times5)}}\\[1em] = \frac{5\times7\times5}{2\times2\times2\times2\times3\times5}\\[1em] = \frac{175}{240}\\[1em] = \frac{35}{48}\end{aligned}$$

$\sqrt{\frac{25}{32} \times 2\frac{13}{18} \times 0.25} = \frac{35}{48}$

$$\begin{aligned}\sqrt{1\frac{4}{5} \times 14\frac{21}{44} \times 2\frac{7}{55}}\\[1em] = \sqrt{\frac{9}{5} \times \frac{637}{44} \times \frac{117}{55}}\\[1em] = \sqrt{\frac{(3\times3)\times(7\times7)\times(13\times13)\times(3\times3)}{(5\times5)\times(2\times2)\times(11\times11)}}\\[1em] = \frac{3\times7\times13\times3}{5\times2\times11}\\[1em] = \frac{819}{110}\\[1em] = 7\frac{49}{110}\end{aligned}$$

$\sqrt{1\frac{4}{5} \times 14\frac{21}{44} \times 2\frac{7}{55}} = 7\frac{49}{110}$

$$\begin{aligned}\sqrt{3^2 \times 6^3 \times 24}\\[1em] = \sqrt{3\times 3 \times 6\times 6\times 6 \times 24}\\[1em] = \sqrt{(3\times 3) \times (2 \times 2)\times (2\times 2) \times (3 \times 3) \times (2 \times 2) \times (3\times 3)}\\[1em] = 3 \times 2\times 2\times 3 \times 2 \times 3\\[1em] = 216\end{aligned}$$

$\sqrt{3^2 \times 6^3 \times 24} = 216$

$$\begin{aligned}\sqrt{(0.5)^3 \times 6 \times 3^5}\\[1em] = \sqrt{\Big(\frac{5}{10}\Big)^3 \times 6 \times 3^5}\\[1em] = \sqrt{\Big(\frac{125}{1000}\Big) \times 6 \times 243}\\[1em] = \sqrt{\Big(\frac{125}{500}\Big) \times 3 \times 243}\\[1em] = \sqrt{\Big(\frac{25}{100}\Big) \times 729}\\[1em] = \Big(\frac{5}{10}\Big) \times 27\\[1em] = \Big(\frac{135}{10}\Big)\\[1em] = 13.5\end{aligned}$$

$\sqrt{(0.5)^3 \times 6 \times 3^5} = 13.5$

$$\begin{aligned}\sqrt{\Big[5 + 2\frac{21}{25}\Big] \times \frac{0.169}{1.6}}\\[1em] = \sqrt{\Big[5 + \frac{71}{25}\Big] \times \frac{169 \times 10}{16 \times 1000}}\\[1em] = \sqrt{\Big[\frac{5}{1} + \frac{71}{25}\Big] \times \frac{169 \times 10}{16 \times 1000}}\\[1em] = \sqrt{\Big[\frac{5 \times 25}{1 \times 25} + \frac{71}{25}\Big] \times \frac{169}{16 \times 100}}\\[1em] = \sqrt{\Big[\frac{125}{25} + \frac{71}{25}\Big] \times \frac{169}{16 \times 100}}\\[1em] = \sqrt{\Big[\frac{125 + 71}{25}\Big] \times \frac{169}{1600}}\\[1em] = \sqrt{\Big[\frac{196}{25}\Big] \times \frac{169}{1600}}\\[1em] = \sqrt{\Big[\frac{14\times 14}{5 \times 5}\Big] \times \frac{13 \times 13}{40 \times 40}}\\[1em] = \Big[\frac{14}{5}\Big] \times \frac{13}{40}\\[1em] = \Big[\frac{14 \times 13}{5 \times 40}\Big]\\[1em] = \Big[\frac{182}{200}\Big]\\[1em] = \Big[\frac{91}{100}\Big]\\[1em] = 0.91\end{aligned}$$

$\sqrt{\Big[5 + 2\frac{21}{25}\Big] \times \frac{0.169}{1.6}} = 0.91$

$$\begin{aligned}\sqrt{5\Big[2\frac{3}{4} - \frac{3}{10}\Big]}\\[1em] = \sqrt{5\Big[\frac{11}{4} - \frac{3}{10}\Big]}\end{aligned}$$

LCM of 4 and 10 is 2 x 2 x 5 = 20.

$$\begin{aligned}= \sqrt{5\Big[\frac{11\times 5}{4 \times 5} - \frac{3 \times 2}{10 \times 2}\Big]}\\[1em] = \sqrt{5\Big[\frac{55}{20} - \frac{6}{20}\Big]}\\[1em] = \sqrt{5\Big[\frac{55 - 6}{20}\Big]}\\[1em] = \sqrt{5\Big[\frac{49}{20}\Big]}\\[1em] = \sqrt{\Big[\frac{5 \times 49}{20}\Big]}\\[1em] = \sqrt{\Big[\frac{49}{4}\Big]}\\[1em] = \sqrt{\Big[\frac{7 \times 7}{2 \times 2}\Big]}\\[1em] = \frac{7}{2}\\[1em] = 3\frac{1}{2}\end{aligned}$$

$\sqrt{5\Big[2\frac{3}{4} - \frac{3}{10}\Big]} = 3\frac{1}{2}$

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

$\sqrt{248 + \sqrt{52 +\sqrt{144}}} = 16$

Let number of days he had been on tour be $x$.

Rupees spent every day = number of days he had been on tour = $x$

Total rupees spent = ₹1,296

Hence,

$$\begin{aligned}x \times x = ₹1,296\\[1em] ⇒ x^2 = 1,296\\[1em] ⇒ x = \sqrt{1,296}\\[1em] ⇒ x = \sqrt{36 \times 36}\\[1em] ⇒ x = 36\end{aligned}$$

So, the man was on tour for 36 days.

Let the number of columns be $x$.

The number of rows = the number of columns = $x$

Students arranged in P.T. display = 745 - 16 = 729

Hence,

$$\begin{aligned}x \times x = 729\\[1em] ⇒ x^2 = 729\\[1em] ⇒ x = \sqrt{729}\\[1em] ⇒ x = \sqrt{27 \times 27}\\[1em] ⇒ x = 27\end{aligned}$$

Hence, the number of rows = 27.

$12^2 = 144$

$21^2 = 441$

12 and 21 are mirror images of each other.

And, 144 and 441 are mirror images of each other.

$102^2 = 10404$

$201^2 = 40401$

102 and 201 are mirror images of each other.

And, 10404 and 40401 are mirror images of each other.

12 and 21; 102 and 201 are 2 such pairs.

LCM of 3, 4, 5 and 6 is 2 x 2 x 3 x 5 = 60

As 3 and 5 are not in pair. To make it a perfect square, it must be multiplied with 3 x 5 = 15.

So, required perfect square = $60 \times 15 = 900$.

$$\begin{aligned}\sqrt{7.84} + \sqrt{78400}\\[1em] = \sqrt{\frac{784}{100}} + \sqrt{784 \times 100}\\[1em] = {\frac{\sqrt{784}}{\sqrt{100}}} + ({\sqrt{784} \times \sqrt{100}})\\[1em] = {\frac{28}{10}} + ({28 \times 10})\\[1em] = 2.8 + 280\\[1em] = 282.8\end{aligned}$$

$\sqrt{7.84} + \sqrt{78400} = 282.8$.

$$\begin{aligned}\sqrt{0.0784} + \sqrt{0.000784}\\[1em] = \sqrt{\frac{784}{10000}} + \sqrt{\frac{784}{1000000}}\\[1em] = {\frac{\sqrt{784}}{\sqrt{10000}}} + {\frac{\sqrt{784}}{\sqrt{1000000}}}\\[1em] = {\frac{28}{100}} + {\frac{28}{1000}}\\[1em] = 0.28 + 0.028\\[1em] = 0.308\end{aligned}$$

$\sqrt{0.0784} + \sqrt{0.000784} = 0.308$

If $\sqrt{5} = 2.24$, then

$$\begin{aligned}\sqrt{20}\\[1em] = \sqrt{4 \times 5}\\[1em] = 2\sqrt{5}\\[1em] = 2 \times 2.24\\[1em] = 4.48\end{aligned}$$

Hence, option 2 is the correct option.

If $\sqrt{27.8} = 5.27$, then

$$\begin{aligned}\sqrt{2780}\\[1em] = \sqrt{27.8 \times 100}\\[1em] = 10\sqrt{27.8}\\[1em] = 10 \times 5.27\\[1em] = 52.7\end{aligned}$$

Hence, option 2 is the correct option.

Since 4² < 23 < 5²,

The least number that must be added to 23 to make the resulting number a perfect square = 5² - 23 = 25 - 23 = 2.

Hence, option 2 is the correct option.

Since 4² < 23 < 5²,

The least number that must be subtracted from 23 to make the resulting number a perfect square = 23 - 4² = 23 - 16 = 7.

Hence, option 1 is the correct option.

Square root of 4761 by division method:

$\sqrt{4761} = 69$

Square root of 7744 by division method:

$\sqrt{7744} = 88$

Square root of 15129 by division method:

$\sqrt{15129} = 123$

Square root of 0.2916 by division method:

$\sqrt{0.2916} = 0.54$

Square root of 0.001225 by division method:

$\sqrt{0.001225} = 0.035$

Square root of 0.023104 by division method:

$\sqrt{0.023104} = 0.152$

Square root of 27.3529 by division method:

$\sqrt{27.3529} = 5.23$

Square root of 4.2025 by division method:

$\sqrt{4.2025} = 2.05$

Square root of 531.7636 by division method:

$\sqrt{531.7636} = 23.06$

Square root of 0.007225 by division method:

$\sqrt{0.007225} = 0.085$

Square root of 245 correct to two places of decimal by division method:

$\sqrt{245} = 15.65$

Square root of 496 correct to three places of decimal by division method:

$\sqrt{496} = 22.271$

Square root of 82.6 correct to two places of decimal by division method:

$\sqrt{82.6} = 9.08$

Square root of 0.065 correct to three places of decimal by division method:

$\sqrt{0.065} = 0.254$

Square root of 5.2005 correct to two places of decimal by division method:

$\sqrt{5.2005} = 2.28$

Square root of 0.602 correct to two places of decimal by division method:

$\sqrt{0.602} = 0.77$

$$\begin{aligned}\sqrt{3\frac{4}{5}}\\[1em] = \sqrt{\frac{19}{5}}\\[1em] = \sqrt{3.8}\end{aligned}$$

Square root of 3.8 correct to two places of decimal by division method:

$\therefore \sqrt{3.8} = 1.94$

$\sqrt{3\frac{4}{5}} = 1.94$

$$\begin{aligned}\sqrt{6\frac{7}{8}}\\[1em] = \sqrt{\frac{55}{8}}\\[1em] = \sqrt{6.875}\end{aligned}$$

Square root of 6.875 correct to two places of decimal by division method:

$\therefore \sqrt{6.875} = 2.62$

$\sqrt{6\frac{7}{8}} = 2.62$

Using division method:

Clearly, if 12 is subtracted from 796, the remainder will be a perfect square.

Since, 796 - 12 = 784

and, $\sqrt{784} = 28$

Hence, 12 should be subtracted from 796.

Using division method:

Clearly, if 37 is subtracted from 1886, the remainder will be a perfect square.

Since, 1886 - 37 = 1849

and, $\sqrt{1849} = 43$

Hence, 37 should be subtracted from 1886.

Using division method:

Clearly, if 88 is subtracted from 23497, the remainder will be a perfect square.

Since, 23497 - 88 = 23409

and, $\sqrt{23409} = 153$

Hence, 88 should be subtracted from 23497.

Using division method:

Clearly, 511 is greater than 22².

∴ On adding the required number to 511, we shall be getting 23² i.e, 529.

So, the required number = 529 - 511 = 18.

Hence, 18 should be added to 511.

Using division method:

Clearly, 7172 is greater than 84².

∴ On adding the required number to 7172, we shall be getting 85² i.e, 7225.

So, the required number = 7225 - 7172 = 53.

Hence, 53 should be added to 7172.

Using division method:

Clearly, 55078 is greater than 235².

∴ On adding the required number to 55078, we shall be getting 235² i.e, 55225.

So, the required number = 55225 - 55078 = 147.

Hence, 147 should be added to 55078.

Square root of 7 correct to two decimal places using division method:

$\sqrt{7} = 2.645 \approx 2.65$

$$\begin{aligned}\sqrt{\frac{4 + \sqrt{7}}{4 - \sqrt{7}}}\\[1em] = \sqrt{\frac{(4 + \sqrt{7})(4 + \sqrt{7})}{(4 - \sqrt{7})(4 + \sqrt{7})}}\\[1em] = \sqrt{\frac{(4 + \sqrt{7})^2}{4^2 - (\sqrt{7})^2}}\\[1em] = \sqrt{\frac{(4 + \sqrt{7})^2}{16 - 7}}\\[1em] = \sqrt{\frac{(4 + \sqrt{7})^2}{9}}\\[1em] = {\frac{(4 + \sqrt{7})}{3}}\\[1em]\end{aligned}$$

Putting the value of $\sqrt{7}$, we get

$$\begin{aligned}= {\frac{(4 + 2.65)}{3}}\\[1em] = {\frac{(6.65)}{3}}\\[1em] = 2.22\end{aligned}$$

Hence, $\sqrt{7} = 2.65$ and $\sqrt{\frac{4 + \sqrt{7}}{4 - \sqrt{7}}} = 2.22$.

Square root of 5 correct to two decimal places using division method:

$\sqrt{5} = 2.236 \approx 2.24$

$$\begin{aligned}{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}\\[1em] = \sqrt{\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}}\\[1em] = \sqrt{\frac{(3 - \sqrt{5})^2}{3^2 - (\sqrt{5})^2}}\\[1em] = \sqrt{\frac{(3 - \sqrt{5})^2}{9 - 5}}\\[1em] = \sqrt{\frac{(3 - \sqrt{5})^2}{4}}\\[1em] = {\frac{(3 - \sqrt{5})}{2}}\\[1em]\end{aligned}$$

Putting the value of $\sqrt{5}$,we get

$$\begin{aligned}= {\frac{(3 - 2.24)}{2}}\\[1em] = {\frac{(0.76)}{2}}\\[1em] = 0.38\end{aligned}$$

Hence, $\sqrt{5} = 2.24$ and $\sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}} = 0.38$.

$$\begin{aligned}\sqrt{\frac{1764}{2809}}\\[1em] = {\frac{\sqrt{1764}}{\sqrt{2809}}}\end{aligned}$$

Square root of 1764 using division method:

$\sqrt{1764} = 42$

Square root of 2809 using division method:

$\sqrt{2809} = 53$

$\therefore {\frac{\sqrt{1764}}{\sqrt{2809}}} = {\frac{42}{53}}$

$\sqrt{\frac{1764}{2809}} = {\frac{42}{53}}$

$$\begin{aligned}\sqrt{\frac{507}{4107}}\\[1em] = \sqrt{\frac{169}{1369}}\\[1em] = {\frac{\sqrt{169}}{\sqrt{1369}}}\\[1em] = {\frac{13}{37}}\end{aligned}$$

Square root of 169 using division method:

$\sqrt{169} = 13$

Square root of 1369 using division method:

$\sqrt{1369} = 37$

$\therefore {\frac{\sqrt{169}}{\sqrt{1369}}} = {\frac{13}{37}}$

$\sqrt{\frac{507}{4107}} = {\frac{13}{37}}$

$$\begin{aligned}\sqrt{108 \times 2028}\\[1em] = \sqrt{108} \times \sqrt{2028}\\[1em] = \sqrt{3 \times 36} \times \sqrt{3 \times 676}\\[1em] = \sqrt{3} \times \sqrt{36} \times \sqrt{3} \times \sqrt{676}\\[1em] = \sqrt{3} \times 6 \times \sqrt{3} \times 26\\[1em] = (\sqrt{3} \times \sqrt{3}) \times 6 \times 26\\[1em] = 3 \times 6 \times 26\\[1em] = 468\end{aligned}$$

Hence, $\sqrt{108 \times 2028} = 468$.

$$\begin{aligned}0.01 + \sqrt{0.0064}\\[1em] = 0.01 + \sqrt{\frac{64}{10000}}\\[1em] = 0.01 + {\frac{\sqrt{64}}{\sqrt{10000}}}\\[1em] = 0.01 + {\frac{8}{100}}\\[1em] = 0.01 + 0.08\\[1em] = 0.09\end{aligned}$$

Hence, 0.01 + $\sqrt{0.0064} = 0.09$

Square root of 7.832 correct to two places of decimal by division method:

$\sqrt{7.832} = 2.798 \approx 2.80$

Square root of 7.832 correct to two significant digits by division method:

$\sqrt{7.832} = 2.79 \approx 2.8$

∴ If 49 is subtracted from 1205, the resulting number will be a perfect square.

Hence, 49 should be subtracted from 1205.

Clearly, 1205 is greater than 34².

∴ On adding the required number to 1205, we shall be getting 35² i.e, 1225

So, the required number = 1225 - 1205 = 20.

Hence, 20 should be added to 1205.

∴ If 12 is subtracted from 2037, the resulting number will be a perfect square.

Hence, 12 should be subtracted from 2025.

Clearly, 5483 is greater than 74².

∴ On adding the required number to 5483, we shall be getting 75² i.e, 5625

So, the required number = 5625 - 5483 = 142.

Hence, 142 should be added to 5483.

According to the property of square numbers, for any natural number n,

$(n+1)^2 - n^2 = (n+1) + n$

$$\begin{aligned}18^2 - 17^2\\[1em] = 18 + 17\\[1em] = 35\end{aligned}$$

Hence, option 3 is the correct option.

As we know that, the sum of first n odd natural numbers = n².

$$\begin{aligned}1 + 3 + 5 + 7 \\[1em] = 4^2\\[1em] = (2^2)^2 \\[1em] = 2^4\end{aligned}$$

Hence, option 2 is the correct option.

According to the property of square numbers, if the digit at the unit's place of a number is 4 or 6, then its square will always have 6 at its unit's place.

Hence, option 4 is the correct option.

According to the property of square numbers, if a number ends with n zeroes, its square ends with 2n zeroes.

So, if a number ends with 5 zeroes, its square ends with 2 x 5 = 10 zeroes.

Hence, option 3 is the correct option.

According to the property of square numbers, the ending digit of the square of a number is 0, 1, 4, 5, 6 or 9.

Hence, (i) 3051, (iii) 5684 and (v) 50699 can be square of a number.

According to the property of square number, if a number has 1 or 9 in its unit's place, then square of this number always has 1 at its unit place.

Hence, (ii) 81 and (iii) 139 will have 1 at their unit's place.

According to the property of square number, if a number has 1 or 9 in its unit's place, then square of this number always has 1 at its unit place.

Hence, (i) 32², (ii) 57² and (v) 265² will not have 1 at their unit's place.

According to the property of square number, if a number has 4 or 6 in its unit's place, then square of this number always has 6 at its unit place.

Hence, (i) 35, (ii) 23 and (v) 98 will not have 6 at their unit's place.

According to the property of square number, if a number has 4 or 6 in its unit's place, then square of this number always has 6 at its unit place.

Hence, (i) 26², (iii) 34² and (v) 244² will have 6 at their unit's place.

According to the property of square numbers, if a number ends with n zeroes, its square ends with 2n zeroes.

So, if a number ends with 3 zeroes, its square ends with 2 x 3 = 6 zeroes.

According to the property of square numbers, if a number ends with n zeroes, its square ends with 2n zeroes.

So, if the square of a number ends with 10 zeroes, the number will have $\frac{10}{2} = 5$ zeroes.

According to the property of square numbers, if a number ends with n zeroes, its square ends with 2n zeroes.

2n will always be an even number.

Hence, it is not possible for a square of a number to end with 5 zeroes.

According to the property of square numbers, a number having 2, 3, 7 or 8 in its unit's place is never a perfect square.

∴ (i) 2162, (ii) 6843, (iii) 9637 and (iv) 6598 are not perfect square.

(i) 23

Digit at ones place = 3

Square of 3 is 9.

So, digit at ones place of square number = 9

Hence, square of 23 is odd.

(ii) 54

Digit at ones place = 4

Square of 4 is 16.

So, digit at ones place of square number = 6

Hence, square of 54 is even.

(iii) 76

Digit at ones place = 6

Square of 6 is 36.

So, digit at ones place of square number = 6

Hence, square of 76 is even.

(iv) 75

Digit at ones place = 5

Square of 5 is 25.

So, digit at ones place of square number = 5

Hence, square of 54 is odd.

According to the property of square numbers, if a number ends with n zeroes, its square ends with 2n zeroes.

2n will always be an even number. The numbers 640, 81000 and 3600000 have 1, 3 and 5 zeroes respectively.

Hence, the numbers 640, 81000 and 3600000 are not a perfect square.

According to the property of square numbers, for any natural number n,

$(n+1)^2 - n^2 = (n+1) + n$

$$\begin{aligned}37^2 - 36^2\\[1em] = 37 + 36\\[1em] = 73\end{aligned}$$

Hence, $37^2 - 36^2 = 73$

According to the property of square numbers, for any natural number n,

$(n+1)^2 - n^2 = (n+1) + n$

$$\begin{aligned}85^2 - 84^2\\[1em] = 85 + 84\\[1em] = 169\end{aligned}$$

Hence, $85^2 - 84^2 = 169$

According to the property of square numbers, for any natural number n,

$(n+1)^2 - n^2 = (n+1) + n$

$$\begin{aligned}101^2 - 100^2\\[1em] = 101 + 100\\[1em] = 201\end{aligned}$$

Hence, $101^2 - 100^2 = 201$

As we know that, the sum of first n odd natural numbers = n².

$$\begin{aligned}1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23\\[1em] = 12^2\\[1em] = 144\end{aligned}$$

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

As we know that, the sum of first n odd natural numbers = n².

$$\begin{aligned}1 + 3 + 5 + 9 + ............... + 39 + 41\\[1em] = 21^2\\[1em] = 441\end{aligned}$$

Hence, 1 + 3 + 5 + 7 + 9 + ............... + 39 + 41 = 441.

As we know that, the sum of first n odd natural numbers = n².

$$\begin{aligned}1 + 3 + 5 + 9 + ............... + 51 + 53\\[1em] = 27^2\\[1em] = 729\end{aligned}$$

Hence, 1 + 3 + 5 + 7 + 9 + ............... + 51 + 53 = 729.

3² + 4² = 5²

5² + 12² = 13²

10² + 24² = 26²

Hence, (3,4,5), (5,12,13) and (10,24,26) are three sets of Pythagorean triplets.

$$\begin{aligned}6^2 + 8^2\\[1em] = 100\\[1em] = 10^2\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}\sqrt{1210000}\\[1em] = \sqrt{11 \times 11 \times 10^4}\\[1em] = 11 \times 10^2 \\[1em] = 1100\end{aligned}$$ Number of digits in $\sqrt{1210000}$ = 4.

Hence, option 2 is the correct option.

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

Hence, option 2 is the correct option.

Area of square plot = l²

$$\begin{aligned}l^2 = 441m^2\\[1em] ⇒ l = \sqrt{441m^2}\\[1em] ⇒ l = 21m\end{aligned}$$

Perimeter of square plot = 4 x l

$$\begin{aligned}= 4 \times 21m\\[1em] = 84m\end{aligned}$$

Hence, option 2 is the correct option.

Finding prime factors of 3675, we get :

⇒ 3675 = 3 x (5 x 5) x (7 x 7)

Since the prime factor 3 is not in pair.

∴ 3675 is not a perfect square.

So, statement 1 is true.

⇒ 3675 = 3 x (5 x 5) x (7 x 7)

Multiplying by 3 on both sides, we get :

⇒ 3675 × 3 = (3 x 3) x (5 x 5) x (7 x 7)

⇒ 11025 = (3 x 3) x (5 x 5) x (7 x 7)

Since all prime factor are in pair. Therefore, 3675 x 3 is a perfect square.

⇒ 3675 = 3 x (5 x 5) x (7 x 7)

Dividing by 3 on both sides, we get :

⇒ $\frac{3675}{3}$ = (5 x 5) x (7 x 7)

⇒ 1225 = (5 x 5) x (7 x 7)

Since all prime factor are in pair. Therefore, 3675 ÷ 3 is a perfect square.

So, statement 2 is true.

Hence, option 1 is the correct option.

We know that,

The square root of any number n is that number which when multiplied by itself gives n as the product.

⇒ $\sqrt{n^2} = \sqrt{n \times n} = n$

So, reason (R) is true.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{\frac{121}{361}}\\[1em] \Rightarrow \sqrt{\frac{11}{19} \times \frac{11}{19}}\\[1em] \Rightarrow \sqrt{\frac{11^2}{19^2}}\\[1em] \Rightarrow \frac{\sqrt{11^2}}{\sqrt{19^2}}\\[1em] \Rightarrow \frac{11}{19}\end{aligned}$$

So, assertion (A) is true and reason (R) clearly explains assertion.

Hence, option 1 is the correct option.

Finding prime factors of 49, we get :

⇒ 49 = (7 x 7)

Since all prime factors are in pair.

∴ 49 is a perfect square.

49 divided by 3 leaves quotient 16 and remainder 1.

⇒ 49 = 3 x 16 + 1

So, assertion (A) is true.

For each perfect squares,

1 divided by 3 leaves quotient 0 and remainder 1.

⇒ 1 = 3 x 0 + 1

4 divided by 3 leaves quotient 1 and remainder 1.

⇒ 4 = 3 x 1 + 1

16 divided by 3 leaves quotient 5 and remainder 1.

⇒ 16 = 3 x 5 + 1

25 divided by 3 leaves quotient 8 and remainder 1.

⇒ 25 = 3 x 8 + 1

36 divided by 3 leaves quotient 12 and remainder 0.

⇒ 36 = 3 x 12 + 0

Therefore when each of the perfect square numbers 1, 4, 9, ............. is divided by 3, the remainder is not always 1.

So, reason (R) is false.

Hence, option 3 is the correct option.

According to assertion,

⇒ 12² + 5² = 13²

Solving L.H.S. of the above equation, we get :

⇒ 12² + 5²

⇒ 144 + 25

⇒ 169

⇒ 13 x 13

⇒ 13²

Since, L.H.S. = R.H.S.

So, assertion (A) is true.

According to reason,

⇒ (n + 1)² - n² = (n + 1) + n.

Solving L.H.S. of the above equation, we get :

⇒ (n + 1)² - n²

Using formula; a² - b² = (a - b)(a + b), we get :

= [(n + 1) - n][(n + 1) + n]

= [n + 1 - n][n + 1 + n]

= 1 x [(n + 1) + n]

= (n + 1) + n

Since, L.H.S. = R.H.S.

So, reason (R) is true but it does not explain assertion (A).

Hence, option 2 is the correct option.

We know that,

The sum of first n odd natural numbers = n².

So, reason (R) is true.

According to assertion :

1 + 3 + 5 + 7 + ....... + 21 = 10²

Solving the L.H.S. of above equation, we get :

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

The above is a sum of the first 11 odd natural numbers. So, by formula

⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 11² = 121.

So, assertion (A) is false.

Hence, option 4 is the correct option.

$$\begin{aligned}(21)^2\\[1em] = 21\times21\\[1em] = 441\\[1em] = 220 + 221\end{aligned}$$

Hence, the sum of 220 and 221 is equivalent to 21².

$$\begin{aligned}\sqrt{10\frac{86}{121}}\\[1em] = \sqrt{\frac{1296}{121}}\\[1em] = {\frac{\sqrt{1296}}{\sqrt{121}}}\\[1em] = {\frac{\sqrt{(2\times2)\times(2\times2)\times(3\times3)\times(3\times3)}}{\sqrt{(11\times11)}}}\\[1em] = {\frac{2\times2\times3\times3}{11}}\\[1em] = {\frac{36}{11}}\\[1em] = 3{\frac{3}{11}}\end{aligned}$$

Hence, $\sqrt{10\frac{86}{121}}= 3{\frac{3}{11}}$

Finding prime factors of 336

$336 = (2 \times 2) \times (2 \times 2) \times 3 \times 7$

Since the prime factor 3 and 7 are not in pair. So we need one more 3 and one more 7 : 3 × 7 = 21.

Hence, 336 is not a perfect square and 21 is the smallest number that should be multiplied with 336 so that the product is a perfect square.

∴ If 19 is subtracted from 980, the resulting number will be a perfect square.

Hence, 19 should be subtracted from 980 and $\sqrt{961} = 31$.

The smallest 4-digit number = 1000

As we can see 1000 is not a perfect square.

Clearly, 1000 is greater than 31².

∴ On adding the required number to 1000, we shall be getting 32² i.e, 1024

So, the required number = 1024 -1000 = 24.

1000 + 24 = 1024 is a perfect square.

The greatest 4-digit number = 9999

Clearly, 9999 is greater than 99².

∴ On subtracting the required number to 9999, we shall be getting 99² i.e, 9801.

So, the required number = 9999 - 198 = 9801.

Hence, 1024 and 9801 are the smallest and greatest 4-digit number.

(i)

$$\begin{aligned}\sqrt{0.1849} + \sqrt{18.49}\\[1em] = \sqrt{\frac{1849}{10000}} + \sqrt{\frac{1849}{100}}\\[1em] = \frac{\sqrt{1849}}{\sqrt{10000}} + \frac{\sqrt{1849}}{\sqrt{100}}\\[1em] = \frac{43}{100} + \frac{43}{10}\\[1em] = 0.43 + 4.3\\[1em] = 4.73\end{aligned}$$

Hence, $\sqrt{0.1849} + \sqrt{18.49} = 4.73$

(ii)

$$\begin{aligned}\sqrt{184900} - \sqrt{4 \times 18.49}\\[1em] = \sqrt{1849 \times 100} - \sqrt{4 \times \frac{1849}{100}}\\[1em] = {\sqrt{1849} \times \sqrt{100}} - \sqrt{4} \times \sqrt{\frac{1849}{100}}\\[1em] = {43 \times 10} - 2 \times {\frac{43}{10}}\\[1em] = 430 - 2 \times 4.3\\[1em] = 430 - 8.6\\[1em] = 421.4\end{aligned}$$

Hence, $\sqrt{184900} - \sqrt{4 \times 18.49} = 421.4$

Let the first number be $x$

The second number is four times the other = $4x$

Hence,

$$\begin{aligned}x \times 4x = 256\\[1em] ⇒ 4x^2 = 256\\[1em] ⇒ x^2 = \frac{256}{4}\\[1em] ⇒ x^2 = 64\\[1em] ⇒ x = \sqrt{64}\\[1em] ⇒ x = 8\\[1em] ⇒ 4x = 4 \times 8 = 32\end{aligned}$$

The numbers are 8 and 32.

Area of square plot = l²

$$\begin{aligned}l^2 = 2116m^2\\[1em] ⇒ l = \sqrt{2116m^2}\\[1em] ⇒ l = 46m\end{aligned}$$

Perimeter of square plot = 4 x l

$$\begin{aligned}= 4 \times 46m\\[1em] = 184m\end{aligned}$$

The man take 5 rounds = 5 x 184m = 920m.

Hence, the total distance travelled by the man = 920m.

LCM of 6, 9 and 15 is 2 x 3 x 3 x 5 = 90

As 2 and 5 are not in pair. To make it a perfect square, it must be multiplied with 2 x 5 = 10.

∴ Required perfect square = 90 x 10 = 900.

The smallest square number which is divisible by 6, 9 and 15 is 900.