Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ?

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 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 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:

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 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}