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Three pipes A,B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?

Correct Answer: 18 min
Let pipe C alone can empty the cistern in x min.A fills cistern in 1 min = $$\frac{1}{{10}}$$B fills cistern in 1 min = $$\frac{1}{{15}}$$A and B together fill in 1 min $$ = \frac{{10 \times 15}}{{10 + 15}} = \frac{{150}}{{25}} = 6\,{\text{min}}$$Since, waste pipe was left open for 6 min then,6 min, $$\frac{6}{x}$$ part of cistern will be emptied by CNow,$$\frac{6}{x}$$ part of the cistern would be filled by A and B in 2 min.Hence,cistern will be filled in $$\frac{3}{x}$$ min. And $$\frac{x}{3}$$ = 6x = 18 min.

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Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to $$\frac{7}{8}$$ of full flow in pipe A and $$\frac{5}{6}$$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.

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