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 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:

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:

There are two important points to learn in this type of question,
First, if both will open then tank will be empty first.

Second most important thing is,
If we are calculating filling of tank then we will subtract as (filling-empting)

If we are calculating empting of thank then we will subtract as (empting-filling)

So lets come on the question now,

Part to emptied 2/5

Part emptied in 1 minute =
\begin{aligned}
\frac{1}{6} - \frac{1}{10} \\
= \frac{1}{15} \\
=> \frac{1}{15}:\frac{2}{5}::1:x \\
=> \frac{2}{5}*15 = 6 mins

\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

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}