Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.

Answer: Option D

Explanation:

Half tank will be filled in 3 hours
Lets calculate remaining half,

Part filled by the four taps in 1 hour = 4*(1/6) = 2/3

Remaining part after 1/2 filled = 1-1/2 = 1/2

$$\begin{aligned} \frac{2}{3}:\frac{1}{2}::1:X \\ => X = \left( \frac{1}{2}*1*{3}{2} \right) \\ => X = \frac{3}{4} hrs = 45 \text{ mins} \\ \end{aligned}$$
Total time = 3 hours + 45 mins = 3 hours 45 mins

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 B

Explanation:

(A+B)'s 2 hour's work when opened =

$$\begin{aligned} \frac{1}{6}+\frac{1}{4} = \frac{5}{12} \\ (A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \\ = \frac{5}{6} \text{Remaining work = } 1-\frac{5}{6} \\ = \frac{1}{6} \\ \text{Now, its A turn in 5 th hour} \\ \frac{1}{6} \text{ work will be done by A in 1 hour}\\ \text{Total time = }4+1 = 5 hours \end{aligned}$$

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.

Answer: Option B

Explanation:

Net part filled in 1 hour =

$$\begin{aligned} \left(\frac{1}{5}+\frac{1}{6}-\frac{1}{12}\right) \\ = \frac{17}{60} hrs \\ = 3\frac{9}{17} \end{aligned}$$