Indian Institute of Foreign Trade 2017 Solved Paper

Choice (B) Let the time taken by first bore well be x hours, then the time taken by second bore well will be x – 10 and the time taken by the third bore well will be x – 18. Now since first two bore wells operating together take same time as taken by the third bore well we can say
$\frac{1}{\mathit{x}}+\frac{1}{\mathit{x}-10}=\frac{1}{\mathit{x}-18}$
$\Rightarrow {\mathit{x}}^2-36\mathit{x}+180=0$
⇒ x = 30 or x = 6. Now x cannot be 6 otherwise the time taken by the second and third bore well will be negative.Thus, x = 30. Therefore the time required by the third bore well to fill the tank alone will be x – 18 = 30 – 18 = 12 hours.