A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

Answer: Option B

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

As per question, we get
\begin{aligned}
\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\
=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
=> (2x - 5)(x - 9) = x(x - 5)\\
=> x^2 - 18x + 45 = 0
\end{aligned}

After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15

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:

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 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 C

Explanation:

When we have question like one is filling the tank and other is empting it, then we subtraction as,

Filled in 1 hour = 1/4
Empties in 1 hour = 1/9

Net filled in 1 hour = 1/4 - 1/9
= 5/36

So cistern will be filled in 36/5 hours i.e. 7.2 hours