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1. 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.

A. 8 min

B. 3 min

C. 5.6 min

D. 4.5 min

Solution:

Let the obstruction remain for x min.
Hence,
Part of cistern filled in X min + part of cistern filled in 3 min = full cistern

[\frac{7 x}{8 \times 12} + \frac{5 x}{6 \times 16}]

+

[\frac{3}{12} + \frac{3}{16}]

= 1

\frac{12 x}{96} + \frac{7}{16} = 1

Thus,
X = 4.5 min.

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