Question Detail

Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.

\begin{aligned} 3\frac{9}{5} \end{aligned}
\begin{aligned} 3\frac{9}{17} \end{aligned}
\begin{aligned} 3\frac{7}{5} \end{aligned}
\begin{aligned} 3\frac{7}{17} \end{aligned}

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}

1. 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.

2 hours 30 mins
2 hours 45 mins
3 hours 30 mins
3 hours 45 mins

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

2. A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other half.

15 mins
20 mins
25 mins
30 mins

Answer: Option D

Explanation:

Let the total time be x mins.
Part filled in first half means in x/2 = 1/40

Part filled in second half means in x/2 = \begin{aligned}
\frac{1}{60}+\frac{1}{40} \\
= \frac{1}{24} \\
\text{ Total = } \\
\frac{x}{2}*\frac{1}{40} + \frac{x}{2}*\frac{1}{24} = 1 \\

=> \frac{x}{2} \left(\frac{1}{40}+\frac{1}{24} \right) = 1 \\
=> \frac{x}{2}*\frac{1}{15} = 1 \\
=> x = 30 mins
\end{aligned}

3. Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.

\begin{aligned} 3\frac{9}{5} \end{aligned}
\begin{aligned} 3\frac{9}{17} \end{aligned}
\begin{aligned} 3\frac{7}{5} \end{aligned}
\begin{aligned} 3\frac{7}{17} \end{aligned}

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}

4. There are two pipes which are functioning simultaneouly to fill a tank in 12 hours, if one pipe fills the tank 10 hours faster than other then how many hours second pipe will take to fill the tank ?

30 hours
35 hours
40 hours
42 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}

5. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

3 hours
5 hours
7 hours
10 hours

Answer: Option B

Explanation:

(A+B)'s 2 hour's work when opened =
\begin{aligned}
\frac{1}{6}+\frac{1}{4} = \frac{5}{12} \\

(A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \\
= \frac{5}{6}

\text{Remaining work = } 1-\frac{5}{6} \\
= \frac{1}{6} \\

\text{Now, its A turn in 5 th hour} \\
\frac{1}{6} \text{ work will be done by A in 1 hour}\\
\text{Total time = }4+1 = 5 hours
\end{aligned}