Answer: Option B. -> $$\frac{4}{3}$$
$$\eqalign{
& {\text{ }}a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr
& \,\,\,\,\,\, {\text{ = }}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \times \frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{5 + 1 + 2\sqrt 5 }}{4} \cr
& \,\,\,\,\,\,\, = \left( {\frac{{3 + \sqrt 5 }}{2}} \right) \cr
& {\text{ }}b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& \,\,\,\,\,\,\, = \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \times \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{5 + 1 - 2\sqrt 5 }}{4} \cr
& \,\,\,\,\,\,\, = \left( {\frac{{3 - \sqrt 5 }}{2}} \right) \cr
& \,\,\,\,\,\,\, \therefore {a^2} + {b^2} \cr
& \,\,\,\,\,\,\, = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} + {\left( {\frac{{3 - \sqrt 5 }}{2}} \right)^2} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2}}}{4} + \frac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{9 + 2.3.\sqrt 5 + 5 + 9 - 2.3.\sqrt 5 + 5}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{2\left( {9 + 5} \right)}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{28}}{4} \cr
& \,\,\,\,\,\,\, = 7 \cr
& {\text{Also, }} \cr
& ab = \frac{{\left( {3 + \sqrt 5 } \right)}}{2} \times \frac{{\left( {3 - \sqrt 5 } \right)}}{2} \cr
& \,\,\,\,\,\,\, = \frac{{\left( {9 - 5} \right)}}{4} \cr
& \,\,\,\,\,\,\, = 1 \cr
& \,\,\,\,\,\,\, \therefore \left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right) \cr
& \,\,\,\,\,\,\, = \frac{{\left( {{a^2} + {b^2}} \right) + ab}}{{\left( {{a^2} + {b^2}} \right) - ab}} \cr
& \,\,\,\,\,\,\, = \frac{{7 + 1}}{{7 - 1}} \cr
& \,\,\,\,\,\,\, = \frac{8}{6} \cr
& \,\,\,\,\,\,\, = \frac{4}{3} \cr} $$

Answer: Option D. -> 36
Let the total number of camels be x
Then,
$$\eqalign{
& \Leftrightarrow x - \left( {\frac{x}{4} + 2\sqrt x } \right) = 15 \cr
& \Leftrightarrow \frac{{3x}}{4} - 2\sqrt x = 15 \cr
& \Leftrightarrow 3x - 8\sqrt x = 60 \cr
& \Leftrightarrow 8\sqrt x = 3x - 60 \cr
& \Leftrightarrow 64x = {\left( {3x - 60} \right)^2} \cr
& \Leftrightarrow 64x = 9{x^2} + 3600 - 360x \cr
& \Leftrightarrow 9{x^2} - 424x + 3600 = 0 \cr
& \Leftrightarrow 9{x^2} - 324x - 100x + 3600 = 0 \cr
& \Leftrightarrow 9x\left( {x - 36} \right) - 100\left( {x - 36} \right) = 0 \cr
& \Leftrightarrow \left( {x - 36} \right)\left( {9x - 100} \right) = 0 \cr
& \Leftrightarrow x = 36\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because x \ne \frac{{100}}{9}} \right] \cr} $$

Answer: Option B. -> 19
$$\eqalign{
& {\text{Required number to be added}} \cr
& = {9^3} - 710 \cr
& = 729 - 710 \cr
& = 19 \cr} $$

Answer: Option C. -> 125
$$\eqalign{
& {\text{Given,}} \cr
& = \root 4 \of {{{\left( {625} \right)}^3}} \cr
& = {\left( {625} \right)^{\frac{3}{4}}} \cr
& = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr
& = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr
& = {5^3} \cr
& = 125 \cr} $$

Answer: Option D. -> $$3x - 5$$
$$\eqalign{
& {\text{We have to find}} \cr
& = \sqrt {9{x^2} + 25 - 30x} \cr} $$
 $$ = \sqrt {{{\left( {3x} \right)}^2} - 2.3x.5 + {{\left( { - 5} \right)}^2}} $$
  $$\,\,\,\,\,\,\,\left\{ {\because {a^2} - 2ab + {b^2} = {{\left( {a + b} \right)}^2}} \right\}$$
$$\eqalign{
& = \sqrt {{{\left( {3x - 5} \right)}^2}} \cr
& = 3x - 5 \cr} $$

Answer: Option C. -> 0.5223
$$\eqalign{
& = \sqrt {\frac{3}{{11}}} \cr
& = \sqrt {\frac{{3 \times 11}}{{11 \times 11}}} \cr
& = \frac{{\sqrt {33} }}{{11}} \cr
& = \frac{{5.745}}{{11}} \cr
& = 0.5223 \cr} $$

Answer: Option B. -> $$\frac{1}{{16}}$$
$$\eqalign{
& \Leftrightarrow \sqrt y = 4x \cr
& \Leftrightarrow y = {\left( {4x} \right)^2} \cr
& \Leftrightarrow y = 16{x^2} \cr
& \Leftrightarrow \frac{{{x^2}}}{y} = \frac{1}{{16}} \cr} $$