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Question Detail
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 = 11/2 = 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
1. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in
 144 mins
 140 mins
 136 mins
 132 minw
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}
2. A cistern can be filled in 9 hours but due to a leak at its bottom it takes 10 hours. If the cistern is full, then the time that the leak will take to make it empty will be ?
 20 hours
 19 hours
 90 hours
 80 hours
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
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. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is
 5780 litres
 5770 litres
 5760 litres
 5750 litres
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.
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