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.
Chakri said:
1 decade ago
Please tell me why it's 7/3 changed into 3/7?
Ashu said:
1 decade ago
@chakri
If you take 7/3 hrs to complete 1 job
Then, what length of job will you take to complete in 1 hour
( what you do here is...use reciprocal value when u shift the value)
that is
7/3=>1
1=>3/7
If you take 7/3 hrs to complete 1 job
Then, what length of job will you take to complete in 1 hour
( what you do here is...use reciprocal value when u shift the value)
that is
7/3=>1
1=>3/7
Ashu said:
1 decade ago
PGCR great logic. What type of logic you used here?
Please explain.
Please explain.
Deepi said:
1 decade ago
A takes 2 hrs lets take as x
Leakage time lets y
2 1/3 hrs=7/3hrs
=> (x*y)/(y-x)=7/3
=> (2y)/(y-2)=7/3
=> 2y=(y-2)(7/3)
=> 2y=(7y/3)-14/3
=> -y/3=-14/3
=> y=14
Leakage time lets y
2 1/3 hrs=7/3hrs
=> (x*y)/(y-x)=7/3
=> (2y)/(y-2)=7/3
=> 2y=(y-2)(7/3)
=> 2y=(7y/3)-14/3
=> -y/3=-14/3
=> y=14
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.
Allhad said:
1 decade ago
To all who don't understand how 7/3.
Actually time given is 2hrs and 20min i.e. 2hrs+20min (1/3 hrs).
Actually time given is 2hrs and 20min i.e. 2hrs+20min (1/3 hrs).
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......
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?
Raja said:
1 decade ago
@Sai.
Its actually 1/(7/3).
Its actually 1/(7/3).
Abhilash said:
1 decade ago
Actually tank could have filled in 2 hr.
Part will be 1/2.
Let the time taken by the leakage is y.
Therefore, part is 1/y.
3/7 = (1/2-1/y) here 3/7 is the total time taken to fill the tank (in parts).
1/y = ((7-6))/14.
y = 14.
Part will be 1/2.
Let the time taken by the leakage is y.
Therefore, part is 1/y.
3/7 = (1/2-1/y) here 3/7 is the total time taken to fill the tank (in parts).
1/y = ((7-6))/14.
y = 14.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers