Four identical taps are fitted on a tank at equal intervals of height: one at the bottom and the others at one-quarter, one-half and three-quarters of the height of the tank. One tap alone can empty the full tank in 12 hours. If all four taps are opened when the tank is full, how long will it take to empty the whole tank?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6 h 15 min
A tap can drain only the water above its own level. So all four taps work on the top quarter, three on the next quarter, two on the next and only the bottom tap on the last quarter.
One tap empties \(\frac{1}{12}\) of the tank per hour
Top quarter, 4 taps: \(\displaystyle \frac{1}{4}\div\frac{4}{12}=\frac{3}{4}\) hour
Second quarter, 3 taps: \(\displaystyle \frac{1}{4}\div\frac{3}{12}=1\) hour
Third quarter, 2 taps: \(\displaystyle \frac{1}{4}\div\frac{2}{12}=\frac{3}{2}\) hours
Bottom quarter, 1 tap: \(\displaystyle \frac{1}{4}\div\frac{1}{12}=3\) hours
Total \(\displaystyle =\frac{3}{4}+1+\frac{3}{2}+3=6\frac{1}{4}\) hours \(=6\) h 15 min