A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.

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}

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 C

Explanation:

Capacity of the tank = (12*13.5) litres
= 162 litres
Capacity of each bucket = 9 litres.
So we can get answer by dividing total capacity of tank by total capacity of bucket.
Number of buckets needed = (162/9) = 18 buckets

Answer: Option A

Explanation:

Let the slower pipe alone fill the tank in x minutes
then faster will fill in x/3 minutes.

Part filled by slower pipe in 1 minute = 1/x
Part filled by faster pipe in 1 minute = 3/x

Part filled by both in 1 minute = \begin{aligned}
\frac{1}{x} + \frac{3}{x}= \frac{1}{36} \\
=> \frac{4}{x} = \frac{1}{36} \\
x = 36*4 = 144 mins
\end{aligned}

Answer: Option D

Explanation:

Let the total time be x mins.
Part filled in first half means in x/2 = 1/40

Part filled in second half means in x/2 = \begin{aligned}
\frac{1}{60}+\frac{1}{40} \\
= \frac{1}{24} \\
\text{ Total = } \\
\frac{x}{2}*\frac{1}{40} + \frac{x}{2}*\frac{1}{24} = 1 \\

=> \frac{x}{2} \left(\frac{1}{40}+\frac{1}{24} \right) = 1 \\
=> \frac{x}{2}*\frac{1}{15} = 1 \\
=> x = 30 mins
\end{aligned}