Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 3)
3.
A pump can fill a tank with water in 2 hours. Because of a leak, it took 2
hours to fill the tank. The leak can drain all the water of the tank in:
hours to fill the tank. The leak can drain all the water of the tank in:Answer: Option
Explanation:
| Work done by the leak in 1 hour = |  |
1 | - | 3 |  |
= | 1 | . |
| 2 | 7 | 14 |
Leak will empty the tank in 14 hrs.
Discussion:
90 comments Page 2 of 9.
Hari said:
7 years ago
Assume they are 2 different pipes.
So total units of water to be filled is lcm of 2 and 7/3 =42units.
The leak takes 1/3time extra than 1st pipe.
So work done by the leak is 1/3 * 42 = 14.
So total units of water to be filled is lcm of 2 and 7/3 =42units.
The leak takes 1/3time extra than 1st pipe.
So work done by the leak is 1/3 * 42 = 14.
(1)
Anonymous said:
5 years ago
Work done by pipe A in 1 hour = 1/2
Work done by pipe B in 1 hour = 1/x
Total work done is 7/3.
Therefore,
=> 1/2 + 1/x = 1/7/3.
=> 1/2 + 1/x = 1*3/7.
=> 1/2 - 3/7 = 1/x.
=> 1/x = 1/14.
Hence the pipe will empty the tank in 14hrs.
Work done by pipe B in 1 hour = 1/x
Total work done is 7/3.
Therefore,
=> 1/2 + 1/x = 1/7/3.
=> 1/2 + 1/x = 1*3/7.
=> 1/2 - 3/7 = 1/x.
=> 1/x = 1/14.
Hence the pipe will empty the tank in 14hrs.
(1)
Mr Pai said:
5 years ago
Time taken by the pump to fill the tank = 2 hours
Work done by the pump in 1 hour = 1/2
Time taken by the pump to fill with leakage = 7/3
Work done by the pump with leakage in 1 hour = 3/7
Time taken by the leakage to empty the tank = x hours
Work done by the leakage in 1 hour = 1/x.
Therefore, (1/2)-(1/x) = 3/7.
-1/x = (3/7)-(1/2).
-1/x = (6-7)/14.
x = 14 hours.
Work done by the pump in 1 hour = 1/2
Time taken by the pump to fill with leakage = 7/3
Work done by the pump with leakage in 1 hour = 3/7
Time taken by the leakage to empty the tank = x hours
Work done by the leakage in 1 hour = 1/x.
Therefore, (1/2)-(1/x) = 3/7.
-1/x = (3/7)-(1/2).
-1/x = (6-7)/14.
x = 14 hours.
(1)
R.Jana said:
4 years ago
How to do by using the ratio process? Please explain.
(1)
KALAISELVAN E said:
4 years ago
@Jad.
Thanks for explaining.
Thanks for explaining.
(1)
Pranav said:
8 years ago
Ans : D
Tank fill without leak = 2 hours
Tank fill with leak = 2 1/3 hours
= 2/1 * 1/3
= 7/3 hours
(imp)
The time taken to drain a tank is
= Part filled in 1 hour (without leak) - partly filled in 1 hour (with leak).
= 1/2 - 1/ 7/3
=> 1/ 7/3 = 3/7
= 1/2 - 3/7
= 1/14.
So, time taken to drain the tank is 14 hours.
Tank fill without leak = 2 hours
Tank fill with leak = 2 1/3 hours
= 2/1 * 1/3
= 7/3 hours
(imp)
The time taken to drain a tank is
= Part filled in 1 hour (without leak) - partly filled in 1 hour (with leak).
= 1/2 - 1/ 7/3
=> 1/ 7/3 = 3/7
= 1/2 - 3/7
= 1/14.
So, time taken to drain the tank is 14 hours.
Mrunal said:
8 years ago
Here is a different solution.
Let tank capacity is 42 liters.
Pump A filling rate is = 42 liters/ 2 hours= 21 lph.
Pump A filling rate due to the leak is = 42 liters/ 7/3 hours = 18 lph.
Now the leaking rate is 21-18= 3 lph.
Leak can empty tank in = 42 liters/ 3 lph= 14 hrs.
Hope this is simple and easy to understand.
Let tank capacity is 42 liters.
Pump A filling rate is = 42 liters/ 2 hours= 21 lph.
Pump A filling rate due to the leak is = 42 liters/ 7/3 hours = 18 lph.
Now the leaking rate is 21-18= 3 lph.
Leak can empty tank in = 42 liters/ 3 lph= 14 hrs.
Hope this is simple and easy to understand.
Swati said:
8 years ago
Why 42/3? @Anuj.
Manuj said:
7 years ago
From, where 2 1/3 came?
Jamsheed said:
7 years ago
Please tell me if a small pump drained a tank in x hours. In 3 hours how much it will drained?
Anyone solve this.
Anyone solve this.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers