Answer: Option C. -> 14
$$\eqalign{
& x = \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 - 1} }} \times \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 + 1} }} \cr
& \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{ {3 - 1} }} \cr
& \,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2} \cr
& \,\,\,\,\,\, = 2 + \sqrt 3 \cr
& y = \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 + 1} }} \times \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 - 1} }} \cr
& \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{ {3 - 1} }} \cr
& \,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr
& \,\,\,\,\,\, = 2 - \sqrt 3 \cr
& \therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$

Answer: Option C. -> 77
Money collected = (59.29 x 100) paise = 5929 paise.
∴ Number of members = $$\sqrt {5929} $$   = 77

Answer: Option B. -> 2
$$\eqalign{
& \sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} \cr
& = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} \cr
& = \sqrt {49 - 45} \cr
& = \sqrt 4 \cr
& = 2 \cr} $$

Answer: Option B. -> 7.826
$$\eqalign{
& \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr
& = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr
& = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr
& = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr
& = \frac{{35\sqrt 5 }}{{10}} \cr
& = \frac{{7 \times 2.236}}{2} \cr
& = 7 \times 1.118 \cr
& = 7.826 \cr} $$

Answer: Option A. -> 5
$$\eqalign{
& {\text{Give}}\,{\text{Expression}} \cr
& = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr
& = 5 \cr} $$

Answer: Option C. -> $$\frac{4}{3}$$
$$\eqalign{
& {\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right)^2} \cr
& = {\left( {\sqrt 3 } \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - 2 \times \sqrt 3 \times \frac{1}{{\sqrt 3 }} \cr
& = 3 + \frac{1}{3} - 2 \cr
& = 1 + \frac{1}{3} \cr
& = \frac{4}{3} \cr} $$

Answer: Option D. -> None of these
A number ending in 8 can never be a perfect square.

Answer: Option B. -> 100
$$\eqalign{
& {\text{Let}}\sqrt {0.0169 \times x} = 1.3 \cr
& {\text{Then}},\,0.0169x = {\left( {1.3} \right)^2} = 1.69 \cr
& \Rightarrow x = \frac{{1.69}}{{0.0169}} = 100 \cr} $$

Answer: Option A. -> 253
$$\eqalign{
& \,\,\,\,\,\,\,\,2|\overline 6 \overline {40} \overline {09} \,(\,\,253 \cr
& \,\,\,\,\,\,\,\,\,\,\,|4 \cr
& \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr
& \,\,\,\,45\,|240 \cr
& \,\,\,\,\,\,\,\,\,\,\,|225 \cr
& \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr
& 503\,\,|\,\,\,\,\,\,1509 \cr
& \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,1509 \cr
& \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr
& \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,x \cr
& \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr
& \therefore \sqrt {64009} = 253 \cr} $$

Answer: Option B. -> 203
$$\eqalign{
& \,\,\,\,\,2|4\,\,\overline {12} \,\,\overline {09} \,(203 \cr
& \,\,\,\,\,\,\,\,\,|4 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,40|\,\,\,\,\,\,\,\,12 \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,0 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,12\,09 \cr
& 403|\,\,\,\,\,\,\,\,\,12\,09 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,x \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - - \, \cr
& \therefore \sqrt {41209} = 203 \cr} $$