Three pipes A, B and C can fill a tank in 12 hours, 15 hours and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, in how many hours will the tank be full?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 7 hours
Part of the tank filled in one hour by each pipe:
\(A: \frac{1}{12},\quad B: \frac{1}{15},\quad C: \frac{1}{20}\)
B opens first (in the order named), then C, and so on. Work in each 2-hour cycle:
\( \displaystyle \text{1st hour (A + B)} = \frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{9}{60} \)
\( \displaystyle \text{2nd hour (A + C)} = \frac{1}{12} + \frac{1}{20} = \frac{5 + 3}{60} = \frac{8}{60} \)
\( \displaystyle \text{One cycle} = \frac{17}{60} \)
After 3 cycles (6 hours):
\( \displaystyle 3 \times \frac{17}{60} = \frac{51}{60} \), so \( \displaystyle \frac{9}{60} \) is still empty.
In the 7th hour A and B are open again and fill exactly \(\frac{9}{60}\). So the tank is full after 7 hours.
Tip: Never just divide 1 by the cycle rate (\(\frac{60}{17} \times 2 \approx 7.06\) hours). Check the last part hour by hour, because the pipe working in that hour decides the time.