The pump fills the tank in 8 minutes, so its filling rate is 18\frac{1}{8}81 of the tank per minute. The tap empties the tank in 7 minutes, so its emptying rate is 17\frac{1}{7}71 of the tank per minute. When both the pump and the tap are opened together, the net rate of change in the tank's water level is the filling rate minus the emptying rate:
Net rate=18−17=7−856=−156 of the tank per minute\text{Net rate}=\frac{1}{8}-\frac{1}{7}=\frac{7-8}{56}=-\frac{1}{56}\text{ of the tank per minute}Net rate=81−71=567−8=−561 of the tank per minute
The negative sign means the tank is being emptied overall at a rate of 156\frac{1}{56}561 of the tank per minute. Since the tank is initially one- fourth full (14\frac{1}{4}41), the amount of water to be emptied before the tank is empty is:
14 of the tank\frac{1}{4}\text{ of the tank}41 of the tank
The time ttt it will take to empty the tank from one-fourth full at the net emptying rate is:
t=14156=14×56=14 minutest=\frac{\frac{1}{4}}{\frac{1}{56}}=\frac{1}{4}\times 56=14\text{ minutes}t=56141=41×56=14 minutes
So, the tank will be emptied in 14 minutes when both the pump and the tap are opened together starting from one-fourth full. This matches the calculated logic and confirms the answer.