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 3 of 9.
Vinay arora said:
1 decade ago
In 120 mints it complete = 1 tank
but it take 140 mints now so,
in 1 mint it fill =1/120
extra time it take be= 140-120= 20 mints
in 20 mint it fill= 20 * (1/120) =1/6
1/6 tank fill in 2 Hr 20 mint that is 7/3 so
1/6 of tank fill= 7/3
1 tank fill in= (7/3)*6= 14 hr
ok buddy......
but it take 140 mints now so,
in 1 mint it fill =1/120
extra time it take be= 140-120= 20 mints
in 20 mint it fill= 20 * (1/120) =1/6
1/6 tank fill in 2 Hr 20 mint that is 7/3 so
1/6 of tank fill= 7/3
1 tank fill in= (7/3)*6= 14 hr
ok buddy......
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.
SAI TRIVENI said:
1 decade ago
As per question a pump can fill and due to leakage it took 1/3 of time extra.
As we can say {PUMP-(PUMP+LEAKAGE)== IF WE SIMPLEFY THIS == PUMP-PUMP-LEAKAGE THIS GIVE RESULT AS = -LEAKAGE.
I CAN'T UNDERSTAND WHY U REVERSED THE VALUE IT MUST BE 7/3 BUT U TOOK 3/7 HOW?
As we can say {PUMP-(PUMP+LEAKAGE)== IF WE SIMPLEFY THIS == PUMP-PUMP-LEAKAGE THIS GIVE RESULT AS = -LEAKAGE.
I CAN'T UNDERSTAND WHY U REVERSED THE VALUE IT MUST BE 7/3 BUT U TOOK 3/7 HOW?
Raj said:
1 decade ago
It is very simple.
Pump takes 2hr to fill
& due to leakage it takes 7/3 hrs
calculate it in terms of 1 hr.
pump will fill 1/2 part of tank
& due to leakage 3/7 part of tank
so leakage in 1 hr will be
1/2 - 3/7 = 1/14 part,
so full tank will be empty in
14 hr.
Pump takes 2hr to fill
& due to leakage it takes 7/3 hrs
calculate it in terms of 1 hr.
pump will fill 1/2 part of tank
& due to leakage 3/7 part of tank
so leakage in 1 hr will be
1/2 - 3/7 = 1/14 part,
so full tank will be empty in
14 hr.
PGCR ( Sri Lanka) said:
1 decade ago
Just because of the leak it took 2 1/3 hours.
Normally it takes only 2 hours.
Hence,
1 1 1
- - - = -
2 X (2 1/3)
OR
1 1 1
- - - = -
2 (2 1/3) X
So, X = 14 hours.
Normally it takes only 2 hours.
Hence,
1 1 1
- - - = -
2 X (2 1/3)
OR
1 1 1
- - - = -
2 (2 1/3) X
So, X = 14 hours.
MAARA said:
11 months ago
If A=2 I mean 2 hours.
And, it takes 2 hours and 20 minutes to fill the tank so per 1 hour there is a drain of 10 minutes for two hours there is a drain of 20 minutes.
So, a * b/a-b = 2 * 7/3/2 - 7/3.
So 14/3/-1/3.
And by inverse 14 is the correct answer.
And, it takes 2 hours and 20 minutes to fill the tank so per 1 hour there is a drain of 10 minutes for two hours there is a drain of 20 minutes.
So, a * b/a-b = 2 * 7/3/2 - 7/3.
So 14/3/-1/3.
And by inverse 14 is the correct answer.
(2)
Juned shah said:
1 decade ago
/*Other method to solve this problem */.
A, A-l.
Time 120, 140.
Time/min 7, 6.
Capacity 840, 840 of tank.
So due to leakage difference in time/min (7-6) = 1.
So 840/1 = 840 min.
Convert it in hr.
= 840/60 = 14 hr.
& This is our require answer.
A, A-l.
Time 120, 140.
Time/min 7, 6.
Capacity 840, 840 of tank.
So due to leakage difference in time/min (7-6) = 1.
So 840/1 = 840 min.
Convert it in hr.
= 840/60 = 14 hr.
& This is our require answer.
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)
JISHNU C V said:
9 years ago
@Mouni.
Work done for 1 hr without leak = 1/2,
Time taken to fill the tank with leak = 2 1/3hrs = 7/3 hrs,
Work done with leak in 1 hr = 3/7,
Work done by leak in 1 hr = 1/2 - 3/7 = 1/14,
So, tank will be empty by the leak in 14 hrs.
Work done for 1 hr without leak = 1/2,
Time taken to fill the tank with leak = 2 1/3hrs = 7/3 hrs,
Work done with leak in 1 hr = 3/7,
Work done by leak in 1 hr = 1/2 - 3/7 = 1/14,
So, tank will be empty by the leak in 14 hrs.
Rachel Vaz said:
7 years ago
Through LCM method:
Pump A fills in = 2hrs.
Leaking pump = 7/3 hrs.
Take LCM of 2 and 3 = 6 units,
Therefore,
Pump A = 3 u/h,
Leaking pump = 14 u/h,
Hence, the leaking pump can drain all the water in = 14 hours.
Pump A fills in = 2hrs.
Leaking pump = 7/3 hrs.
Take LCM of 2 and 3 = 6 units,
Therefore,
Pump A = 3 u/h,
Leaking pump = 14 u/h,
Hence, the leaking pump can drain all the water in = 14 hours.
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