Answer: Option B. -> 14
$$\eqalign{
& \Leftrightarrow {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}} \cr
& \Leftrightarrow {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}} \cr
& \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{150}} \cr
& \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}} \cr
& \Leftrightarrow n = {5^{64}} \times {2^{64}} \times {2^{11}} \cr
& \Leftrightarrow n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}} \cr
& \Leftrightarrow n = {10^{64}} \times 2048 \cr} $$
∴ Sum of digits of n
= 2 + 0 + 4 + 8
= 14

Answer: Option A. -> 0.1764
$$\eqalign{
& \Leftrightarrow \frac{{\sqrt x }}{{\sqrt {441} }} = 0.02 \cr
& \Leftrightarrow \frac{{\sqrt x }}{{21}} = 0.02 \cr
& \Leftrightarrow \sqrt x = 0.02 \times 21 \cr
& \Leftrightarrow \sqrt x = 0.42 \cr
& \Leftrightarrow x = {\left( {0.42} \right)^2} \cr
& \Leftrightarrow x = 0.1764{\text{ }} \cr} $$

Answer: Option D. -> 12
$$\eqalign{
& \Leftrightarrow \sqrt {{3^n}} = 729 \cr
& \Leftrightarrow \sqrt {{3^n}} = {3^6} \cr
& \Leftrightarrow {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2} \cr
& \Leftrightarrow {3^n} = {3^{12}} \cr
& \Leftrightarrow n = 12 \cr} $$

Answer: Option B. -> 0.1
$$\eqalign{
& \Leftrightarrow \frac{{0.13}}{{{p^2}}} = 13 \cr
& \Leftrightarrow {p^2} = \frac{{0.13}}{{13}} \cr
& \Leftrightarrow {p^2} = \frac{1}{{100}} \cr
& \Leftrightarrow p = \sqrt {\frac{1}{{100}}} \cr
& \Leftrightarrow p = \frac{1}{{10}} \cr
& \Leftrightarrow p = 0.1 \cr} $$

Answer: Option C. -> 6
$$\eqalign{
& \,\,\,\,\,\,1|\overline 1 \,\,\overline {58} \,\,\overline {76} \,(126 \cr
& \,\,\,\,\,\,\,\,\,|1 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,22|\,\,\,\,\,\,58 \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,44 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& 246\,|\,\,\,\,\,\,\,14\,76 \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,14\,76 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,x \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& \therefore \sqrt {15876} = 126 \cr} $$

Answer: Option A. -> 13
$$\eqalign{
& {\text{Let the given number be }}x \cr
& {\text{Then,}} \cr
& \Leftrightarrow {x^2} - 25 = {\left( {x - 25} \right)^2}{\text{ }} \cr
& \Leftrightarrow {x^2} - 25 = {x^2} + 625 - 50x \cr
& \Leftrightarrow 50x = 650 \cr
& \Leftrightarrow x = 13 \cr} $$

Answer: Option C. -> 7
$$\eqalign{
& {\left( {11} \right)^2} = 121{\text{ }} \cr
& {\text{And }} \cr
& {\left( {17} \right)^2} = 289 \cr} $$
So, the perfect squares between 120 and 300 are the squares of numbers from 11 to 17.
Clearly, these are 7 in number.

Answer: Option E. -> 20%
The squares of numbers having 1 and 9 as the unit's digit end in the digit 1.
$$\eqalign{
& {\text{Such numbers are,}} \cr
& 1,9,11,19,21,29,31,39,41,49{\text{ i}}{\text{.e}}{\text{.,}} \cr
& {\text{There are 10 such numbers}}{\text{.}} \cr
& \therefore {\text{Required percentage}} \cr
& = \left( {\frac{{10}}{{50}} \times 100} \right)\% \cr
& = 20\% \cr} $$

Answer: Option B. -> $$\frac{{173}}{{100}}$$
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,1|\overline 3 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.732 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,\,\,\,27|\,\,2\,\,\,00 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,1\,\,\,89 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,343\,|\,\,\,\,\,\,\,\,\,11\,\,00 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,10\,\,29 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& 3492\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,71\,\,00 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,69\,\,84 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,16 \cr} $$
$$\eqalign{
& \therefore \sqrt 3 \cr
& = 1.73 \cr
& = \frac{{173}}{{100}} \cr} $$

Answer: Option D. -> 24
The first perfect square number after 50 is 64 $$\left( {64 = {8^2}} \right)$$   and the last perfect square number before 1000 is 961 $$\left[ {961 = {{\left( {31} \right)}^2}} \right]$$
So, the perfect squares between 50 and 1000 are the squares of numbers from 8 to 31.
(31 - 8) + 1 = 24
Clearly, these are 24 in number.