A pipe can fill a water tank in 36 minutes, and another can fill it in 48 minutes, but a third pipe can empty it in 18 minutes. The first two pipes are kept open for 16 minutes at the beginning, and then the third pipe is also opened. How long does it take to empty the tank?

Correct option is B

Given:
Pipe A fills in 36 minutes
Pipe B fills in 48 minutes
Pipe C empties in 18 minutes
First 16 minutes: A and B open
Then A, B, C open together
Formula Used:
Work = Rate × Time
Net Rate = Sum of individual rates
Solution:
Rate of A = $$\frac{1}{36}$$​ tank/min

Rate of B = $$\frac{1}{48}$$​ tank/min
Work in first 16 minutes = ($$\frac{1}{36} +\frac{1}{48}$$​) × 16 = $$\frac 79$$​ of tank

Net rate of A, B, C = $$\frac{1}{36}+\frac{1}{48}-\frac{1}{18}=-\frac{1}{144}$$​ tank/min
Time to empty = $$\frac{\frac{7}{9}}{\frac{1}{144}}$$​= 112 minutes 
Alternate Solution: 
LCM of 36, 48, 18 = 144
So, assume tank capacity = 144 units.
Rate of A = 144 ÷ 36 = 4 units/min
Rate of B = 144 ÷ 48 = 3 units/min
Rate of C = −(144 ÷ 18) = −8 units/min
Work done in one minutes (only A and B open) = A + B = 4 + 3 = 7 units/min
Work done in 16 minutes = 7 × 16 = 112 units
So after 16 minutes, 112 units of water are in the tank.
A + B + C = 4 + 3 − 8 = −1 unit/min
(negative means tank is getting emptied)
Time = Work ÷ Rate = 112 ÷ 1 = 112 minutes