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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. 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}
2. 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.
3. 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}
4. 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}
5. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time ?
 10 mins
 15 mins
 20 mins
 25 mins
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}
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