Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ?

Answer: Option C

Explanation:

Work done by the inlet in 1 hour =
\begin{aligned}
\frac{1}{6}-\frac{1}{8} = \frac{1}{24} \\
\text{Work done by inlet in 1 min} \\
= \frac{1}{24}*\frac{1}{60}\\
= \frac{1}{1440} \\
=> \text{Volume of 1/1440 part = 4 liters} \\

\end{aligned}

Volume of whole = (1440 * 4) litres = 5760 litres.

Answer: Option C

Explanation:

Part filled without leak in 1 hour = 1/9
Part filled with leak in 1 hour = 1/10

Work done by leak in 1 hour \begin{aligned}
= \frac{1}{9} - \frac{1}{10} \\
= \frac{1}{90}
\end{aligned}

We used subtraction as it is getting empty.

So total time to empty the cistern is 90 hours

Answer: Option A

Explanation:

Lets suppose tank got filled by first pipe in X hours,
So, second pipe will fill the tank in X + 10 hours.

\begin{aligned}
=> \frac{1}{X} + \frac{1}{X} + 10 = \frac{1}{12} \\
=> X = 20
\end{aligned}

Answer: Option B

Explanation:

In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as

Part filled by A in 1 min = 1/20
Part filled by B in 1 min = 1/30

Part filled by (A+B) in 1 min = 1/20 + 1/30
= 1/12

So both pipes can fill the tank in 12 mins.

Answer: Option D

Explanation:

We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as

Work done for 1 hour without leak = 1/3
Work done with leak =
\begin{aligned}
3\frac{1}{2} = \frac{7}{2} \\
\text{Work done with leak in 1 hr= }\frac{2}{7} \\
\text{Work done by leak in 1 hr }\\
= \frac{1}{3} - \frac{2}{7} = \frac{1}{21}
\end{aligned}

So tank will be empty by the leak in 21 hours.