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

E. None of these

Answer: Option D

Solution(By Examveda Team)

Let the obstruction remain for x min.
Hence,
Part of cistern filled in X min + part of cistern filled in 3 min = full cistern
$$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$$    $$ + $$ $$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$$   = 1
$$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$$
Thus,
X = 4.5 min.