Discussion of Question with ID = 076 under Pipes-and-Cisterns

This is the discussion forum for this question. If you find any mistakes in the solution, or if you have a better solution, then this is the right place to discuss. A healthy discussion helps all of us, so you are requested to be polite and soft, even if you disagree with the views of others. The question and its current solution has also been given on this page.



A tank is filled in 15 minutes by three taps running together. Tap A is twice as fast as tap B, and tap B is twice as fast as tap C. How much time will tap A take to fill the tank?


105 mins.


106 mins.


104 mins.


107 mins.

Ans: a

Let the time taken by tap A be x mins. Then 15 minutes work of all the taps should add to 1. So we have, $15 × 1/x + 15 × 2/x + 15 × 4/x$ = 1, which is same as $15 × 7/x$ = 1. Solving, we get x = 105 mins.

Comments and Discussion