Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?

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:

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

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 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 B

Explanation:

In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as

Part filled by A in 1 min = 1/20
Part filled by B in 1 min = 1/30

Part filled by (A+B) in 1 min = 1/20 + 1/30
= 1/12

So both pipes can fill the tank in 12 mins.