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?

7 hours

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.