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Question Detail
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
1. 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}{x-5} = \frac{1}{x-9} \\
=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
=> (2x - 5)(x - 9) = x(x - 5)\\
=> x^2 - 18x + 45 = 0
\end{aligned}
After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15
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. Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ?
- 2.5 hours
- 2 hours
- 3.5 hours
- 3 hours
Answer: Option D
Explanation:
Part filled by A in 1 hour = 1/5
Part filled by B in 1 hour = 1/10
Part filled by C in 1 hour = 1/30
Part filled by (A+B+C) in 1 hour =
\begin{aligned}
\frac{1}{5}+\frac{1}{10}+\frac{1}{30} \\
= \frac{1}{3} \\
\end{aligned}
So all pipes will fill the tank in 3 hours.
4. 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.
5. 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}
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