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:

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 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.

Answer: Option C

Explanation:

How we can solve this question ?
First we will calculate the work done for 10 mins, then we will get the remaining work, then we will find answer with one tap work, As

Part filled by Tap A in 1 min = 1/20
Part filled by Tap B in 1 min = 1/60

(A+B)'s 10 mins work =
\begin{aligned}
10*\left(\frac{1}{20}+\frac{1}{60}\right) \\
= 10*\frac{4}{60} = \frac{2}{3} \\
\text{Remaining work = } 1-\frac{2}{3} \\
= \frac{1}{3} \\

\text{METHOD 1} \\
=> \frac{1}{60}:\frac{1}{3}=1:X \\
=> X = 20 \\

\text{METHOD 2} \\

1/60 \text{ part filled by B in} = 1 min \\
1/3 \text{ part will be filled in} \\
= \frac{\frac{1}{3}}{\frac{1}{60}} \\
= \frac{60}{3} = 20
\end{aligned}

Answer: Option A

Explanation:

Part filled in 3 minutes =
\begin{aligned}
3*\left(\frac{1}{12} + \frac{1}{15}\right) \\
= 3*\frac{9}{60} = \frac{9}{20}\\
\text{Remaining part }= 1-\frac{9}{20} \\
= \frac{11}{20} \\
=> \frac{1}{15}:\frac{11}{20}=1:X \\
=> X = \frac{11}{20}*\frac{15}{1} \\
=> X = 8.25 mins
\end{aligned}

So it will take further 8 mins 15 seconds to fill the bucket.

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.