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.
Suresh said:
4 years ago
Convert Hours into minutes.
A can fill in 2hrs i.e 120mins.
Leakage i.e a-b 2 1/3 is equal to 140mins.
Then capacity (LCM) is 840.
Efficiency : A is 7 and a-b is 6 and the difference is 1.
It can empty one part per minute out of 840 minutes. Then 840 parts can be emptied in 840 minutes.
Now convert 840 minutes to hours.
840/60 is 14 hours.
A can fill in 2hrs i.e 120mins.
Leakage i.e a-b 2 1/3 is equal to 140mins.
Then capacity (LCM) is 840.
Efficiency : A is 7 and a-b is 6 and the difference is 1.
It can empty one part per minute out of 840 minutes. Then 840 parts can be emptied in 840 minutes.
Now convert 840 minutes to hours.
840/60 is 14 hours.
(19)
R.Jana said:
4 years ago
How to do by using the ratio process? Please explain.
(1)
Dhivyanisan said:
4 years ago
Time is taken by pump without leak = 2hrs.
Its discharge capacity says a= 1/2.
Time is taken by a pump with leak = 2 (1/3)hrs or 7/3 hrs.
It's discharge capacity say b= 1/(7/3) = 3/7.
So discharge capacity of the leak is a-b = 1/14.
Time taken by the leak to empty tank = 1/(1/14) = 14 hrs.
Its discharge capacity says a= 1/2.
Time is taken by a pump with leak = 2 (1/3)hrs or 7/3 hrs.
It's discharge capacity say b= 1/(7/3) = 3/7.
So discharge capacity of the leak is a-b = 1/14.
Time taken by the leak to empty tank = 1/(1/14) = 14 hrs.
SUDHARSHAN NAIDU said:
5 years ago
The efficiency of a pipe in 1 hour is 1/2 = 0.5lit/h.
The efficiency of a pipe with leakage 1/2.33 = 0.429,
The difference between 0.5 - 0.429 is 0.07.
The efficiency of a leak is 0.07lits/ hour.
So, check with an option in how many hours the 1-litre tank empty we get the answer.
The efficiency of a pipe with leakage 1/2.33 = 0.429,
The difference between 0.5 - 0.429 is 0.07.
The efficiency of a leak is 0.07lits/ hour.
So, check with an option in how many hours the 1-litre tank empty we get the answer.
Reddy said:
5 years ago
How 3/7 with leakage? Explain, please.
Jothigonzi said:
5 years ago
@Deepi.
Nice explanation and easier to understand. Thanks.
Nice explanation and easier to understand. Thanks.
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)
Hika said:
5 years ago
Can anyone explain the answer in LCM method?
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)
Karthik said:
5 years ago
There is a simple formula.
a(1+(a/x)).
a= 2 hours.
x= 2 1/3- 2= 1/3.
Solution:
2(1+(2/ (1/3) )= 14.
a(1+(a/x)).
a= 2 hours.
x= 2 1/3- 2= 1/3.
Solution:
2(1+(2/ (1/3) )= 14.
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