Two pipes A and B can fill a tank in 12 minutes and 28 minutes respectively. If both the pipes are opened together. After how much time B should be closed so that the tank is full in 10 minutes ?

Assume the tank capacity be 84 units (i.e L.C.M of 12&28)

A can fill a tank$=12\min \Rightarrow for1\min =\frac{1}{12}part=\frac{84}{12}=7units$

B can fill a tank $=28\min \Rightarrow for1\min =\frac{84}{28}=3units$

Time for  which ‘B’ is opened ‘x’ minutes

$$\begin{gathered} \Rightarrow \frac{5\times 14+x\times 3}{84}=1 \\ \Rightarrow 70+3x=84 \\ \Rightarrow x=\frac{14}{3} \end{gathered}$$$7\times 10+3\times x=84$

$70+3x=84$

$3x=84-70$

$3x=14$

$x=\frac{14}{3}\text{\hspace{0.17em}\hspace{0.17em}minutes}$ 

Or

 Pipe B open in x minutes

Pipe A can fill tank in 1 min = $\frac{1}{12}$, in 10 min = $\frac{10}{12}=\frac{5}{6}$

A+B$\Rightarrow \frac{5}{6}+\frac{x}{28}=1$

$\Rightarrow \frac{5\times 14+3\times x}{84}=1$

$\Rightarrow x=\frac{14}{3}$