An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?

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 D

Explanation:

Part filled by A in 1 hour = 1/5
Part filled by B in 1 hour = 1/10
Part filled by C in 1 hour = 1/30

Part filled by (A+B+C) in 1 hour =
$$\begin{aligned} \frac{1}{5}+\frac{1}{10}+\frac{1}{30} \\ = \frac{1}{3} \\ \end{aligned}$$

So all pipes will fill the tank in 3 hours.

Answer: Option A

Explanation:

Part filled in 3 minutes =
$$\begin{aligned} 3*\left(\frac{1}{12} + \frac{1}{15}\right) \\ = 3*\frac{9}{60} = \frac{9}{20}\\ \text{Remaining part }= 1-\frac{9}{20} \\ = \frac{11}{20} \\ => \frac{1}{15}:\frac{11}{20}=1:X \\ => X = \frac{11}{20}*\frac{15}{1} \\ => X = 8.25 mins \end{aligned}$$

So it will take further 8 mins 15 seconds to fill the bucket.

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 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