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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 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
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. 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}
4. 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}
5. 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
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