A cistern has two inlet pipes, A and B, and one outlet pipe, C.

Pipe A fills the cistern in 15 hours
Pipe B fills the cistern in 18 hours
Pipe C empties the cistern in 27 hours

Their hourly work rates:

A = $\frac{1}{15}$
B = $\frac{1}{18}$
C = $-\frac{1}{27}$

1. First 5 hours (A and B together)

Work done per hour:

$\frac{1}{15} + \frac{1}{18}$

LCM of 15 and 18 = 90

$= \frac{6}{90} + \frac{5}{90}$
$= \frac{11}{90}$

Work done in 5 hours:

$5 \times \frac{11}{90} = \frac{55}{90} = \frac{11}{18}$

Cistern filled = $\frac{11}{18}$

2. Next 12 hours (B and C together)

Work rate:

$\frac{1}{18} - \frac{1}{27}$

LCM of 18 and 27 = 54

$= \frac{3}{54} - \frac{2}{54}$
$= \frac{1}{54}$

Work done in 12 hours:

$12 \times \frac{1}{54} = \frac{12}{54} = \frac{2}{9}$

Total filled now:

$\frac{11}{18} + \frac{2}{9}$

$= \frac{11}{18} + \frac{4}{18}$

$= \frac{15}{18} = \frac{5}{6}$

3. Remaining part

Remaining work:

$1 - \frac{5}{6} = \frac{1}{6}$

Pipe B fills $\frac{1}{18}$ of the cistern in 1 hour.

Time required:

$\frac{\frac{1}{6}}{\frac{1}{18}} = 3$

✅ Pipe B will take 3 hours to fill the remaining part.

Correct option: 3 hours / 3 घंटे