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 $\frac{5}{6}$ of the full in pipe $B$ the obstruction, is removed after some time and the tank is how 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

The correct option is 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
$[(\frac{7}{8}\times \frac{x}{12})+(\frac{5}{6}\times \frac{x}{16})]+[\frac{3}{12}+\frac{3}{16}]=1$
$\Rightarrow \frac{12x}{96}+\frac{7}{16}=1$

$\Rightarrow \frac{12x}{96}=\frac{9}{16}$
$\Rightarrow x=\frac{9\times 96}{16\times 12}=4.5$ min