Question

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 the full flow in pipe A and \(\frac{5}{6}\) of the full in pipe B. The obstruction, is removed after some time and the tank is now filled in 3 min from that moment. For how many minutes was the obstruction there? 

(a) 8 min 

(b) 3 min 

(c) 5.6 min 

(d) 4.5 min

Answer 1

(d) 4.5 min

Let the obstruction remain for x minutes only

\(\therefore\) Part of cistern filled in x minutes + Part of cistern filled in 3 minutes = cistern filled 

\(\Big[\big(\frac{7}{8}\times\frac{X}{12}\big)+\big(\frac{5}{6}\times\frac{X}{16}\big)\Big]\) + \(\Big[\frac{3}{12}+\frac{3}{16}\Big]\) = 1

\(\Rightarrow\) \(\frac{12X}{96}\) + \(\frac{7}{16}\) = 1 \(\Rightarrow\) \(\frac{12X}{96}\) = \(\frac{9}{16}\)

\(\Rightarrow\) x = \(\frac{9\times96}{16\times12}\) = 4.5 min