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
Rates of filling:
A : $\frac{1}{12}$ of tank per hour
B : $\frac{1}{15}$ of tank per hour
C : $\frac{1}{20}$ of tank per hour
A is always open.
B and C open alternately for 1 hour each.
Work done in a 2-hour cycle:
Hour 1: A + B = $\frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20}$
Hour 2: A + C = $\frac{1}{12} + \frac{1}{20} = \frac{5+3}{60} = \frac{8}{60} = \frac{2}{15}$
Total in 2 hours:
$\frac{3}{20} + \frac{2}{15} = \frac{9 + 8}{60} = \frac{17}{60}$
Work done in 2 hours = $\frac{17}{60}$
After 3 such cycles (6 hours):
$3 \times \frac{17}{60} = \frac{51}{60} = \frac{17}{20}$
Remaining part:
$1 - \frac{17}{20} = \frac{3}{20}$
Next hour (A + B):
Work = $\frac{3}{20}$ which exactly equals remaining.
So tank becomes full after:
$6 + 1 = 7$ hours
The tank will be full in 7 hours.