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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. 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.
2. 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
3. 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}
4. 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?
 15 bukets
 17 bukets
 18 bukets
 19 bukets
Answer: Option C
Explanation:
Capacity of the tank = (12*13.5) litres
= 162 litres
Capacity of each bucket = 9 litres.
So we can get answer by dividing total capacity of tank by total capacity of bucket.
Number of buckets needed = (162/9) = 18 buckets
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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