Answer & Solution
Answer: Option C
Solution:
$$\eqalign{
& {\text{Suppose,}}\,{\text{first}}\,{\text{pipe}}\,{\text{alone}}\,{\text{takes}}\,x\,{\text{hours}}\,{\text{to}}\,{\text{fll}}\,{\text{the}}\,\,{\text{tank}}{\text{.}} \cr
& {\text{Then,}} \cr
& {\text{Second}}\,{\text{and}}\,{\text{third}}\,{\text{pipes}}\,{\text{will}}\,{\text{take}}\,\left( {x - 5} \right){\text{and}}\left( {x - 9} \right)\,{\text{hours}}\,{\text{respectively}}\,{\text{to}}\,{\text{fill}}\,{\text{the}}\,{\text{tank}}. \cr
& \therefore \frac{1}{x} + \frac{1}{{\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr
& \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr
& \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr
& \Rightarrow {x^2} - 18x + 45 = 0 \cr
& \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr
& \Rightarrow x = 15\,\,\,\,\,\,\left[ {{\text{neglecting}}\,x = 3} \right] \cr} $$
