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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.
Answer & Solution
Correct Answer:
Option
D
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.
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LoginThen how to come X=4.5 min
Above eqn slove X value in negative