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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 by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time cistern will get filled ?
 7 hours
 7.1 hours
 7.2 hours
 7.3 hours
Answer: Option C
Explanation:
When we have question like one is filling the tank and other is empting it, then we subtraction as,
Filled in 1 hour = 1/4
Empties in 1 hour = 1/9
Net filled in 1 hour = 1/4  1/9
= 5/36
So cistern will be filled in 36/5 hours i.e. 7.2 hours
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. 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
4. 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}
5. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank ?
 10 mins
 12 mins
 15 mins
 20 mins
Answer: Option B
Explanation:
In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as
Part filled by A in 1 min = 1/20
Part filled by B in 1 min = 1/30
Part filled by (A+B) in 1 min = 1/20 + 1/30
= 1/12
So both pipes can fill the tank in 12 mins.
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