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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. A water tank is twofifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?
 6 min to empty
 7 min to full
 6 min to full
 7 min to empty
Answer: Option A
Explanation:
There are two important points to learn in this type of question,
First, if both will open then tank will be empty first.
Second most important thing is,
If we are calculating filling of tank then we will subtract as (fillingempting)
If we are calculating empting of thank then we will subtract as (emptingfilling)
So lets come on the question now,
Part to emptied 2/5
Part emptied in 1 minute =
\begin{aligned}
\frac{1}{6}  \frac{1}{10} \\
= \frac{1}{15} \\
=> \frac{1}{15}:\frac{2}{5}::1:x \\
=> \frac{2}{5}*15 = 6 mins
\end{aligned}
2. 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}
3. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?
 10 hours
 15 hours
 17 hours
 18 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}{x5} = \frac{1}{x9} \\
=> \frac{x5+x}{x(x5)} = \frac{1}{x9}\\
=> (2x  5)(x  9) = x(x  5)\\
=> x^2  18x + 45 = 0
\end{aligned}
After solving this euation, we get
(x15)(x+3) = 0,
As value can not be negative, so x = 15
4. 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
5. 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}
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