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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 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. Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?
 8 min 15 sec
 7 min 15 sec
 6 min 15 sec
 5 min 15 sec
Answer: Option A
Explanation:
Part filled in 3 minutes =
\begin{aligned}
3*\left(\frac{1}{12} + \frac{1}{15}\right) \\
= 3*\frac{9}{60} = \frac{9}{20}\\
\text{Remaining part }= 1\frac{9}{20} \\
= \frac{11}{20} \\
=> \frac{1}{15}:\frac{11}{20}=1:X \\
=> X = \frac{11}{20}*\frac{15}{1} \\
=> X = 8.25 mins
\end{aligned}
So it will take further 8 mins 15 seconds to fill the bucket.
4. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?
 3 hours
 5 hours
 7 hours
 10 hours
Answer: Option B
Explanation:
(A+B)'s 2 hour's work when opened =
\begin{aligned}
\frac{1}{6}+\frac{1}{4} = \frac{5}{12} \\
(A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \\
= \frac{5}{6}
\text{Remaining work = } 1\frac{5}{6} \\
= \frac{1}{6} \\
\text{Now, its A turn in 5 th hour} \\
\frac{1}{6} \text{ work will be done by A in 1 hour}\\
\text{Total time = }4+1 = 5 hours
\end{aligned}
5. 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}
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