Two pipes can fill a cistern, individually, in 70 min and 30 min, respectively. There is a pipe at the bottom of the cistern to empty it. If all three pipes are open simultaneously, the cistern gets filled in 28 min. How long will the pipe at the bottom take to empty the full cistern if no other pipe is then open?

Correct option is B

Given:

Pipe A fills the cistern in 70 min

rate =$$\frac{1}{70}$$​ cistern/min

Pipe B fills the cistern in 30 min

rate =$$\frac{1}{30}$$​ cistern/min

Bottom pipe empties in x min

rate =$$\frac{1}{x}$$​ cistern/min (emptying)

All three together fill the cistern in 28 min

net rate =$$\frac{1}{28}$$​ cistern/min

Solution:

Combined rate = sum of filling rates - emptying rate

​$$\frac{1}{70}+\frac{1}{30}-\frac{1}{x}=\frac{1}{28} \\ \ \\\frac{1}{21}-\frac{1}{x}=\frac{1}{28} \\ \ \\\frac{1}{x}=\frac{1}{21}-\frac{1}{28} \\ \ \\=\frac{4-3}{84}=\frac{1}{84}$$​​

x = 84 minutes

Alternate method:
Assume cistern capacity = LCM(70,30,28) = 420 units.

A fills 6 units/min, B fills 14 units/min, let bottom pipe empty = e units/min.
Together for 28 min: ($$6+14-e)\times 28=420 \implies (20-e)\times 28=420 \implies 20-e=15 \implies e=5$$​ units/min.
Time to empty =$$\dfrac{420}{5}=$$​84 min.