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# A pool can be filled in 4 hours and drained in 5 hours. The

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A pool can be filled in 4 hours and drained in 5 hours. The [#permalink]

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04 Jan 2008, 06:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1pm and some time alter the drain that empties the pool was also opened. If the pool was filled by 11pm and not earlier, when was the drain opened?

2 pm
2:50 pm
2:30 pm
3:30 pm
5pm
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Manager
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04 Jan 2008, 07:10
guys help to solve that problem, very interesting

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Director
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04 Jan 2008, 07:15
The pool fills up in 10 hours.
The number of hours it's only being filled = x
The number of hours it's being filled AND drained = (10-x)

Each hour it's only being filled the pool fills up by 1/4
Each hour it's being filled AND drained it fills up by 1/20 (1/4-1/5)

1/4(x) + 1/20(10-x) = 1 (1 represents the pool being full)
1/4x + 1/2 - 1/20x = 1
5x + 10 - x = 20
4x + 10 = 20
4x = 10
x = 2.5

If the valve is opened at 1:00 and it's being filled only for 2.5 hours: 1+2.5=3.5 or 3:30 pm

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CEO
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04 Jan 2008, 10:54
very nice. OA is D

RT=Cap

r1*t1 + r2*t2 = 1
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05 Jan 2008, 05:45
1
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3.30 is the start time.
my way of doing it,
let x be the volume of the tank
speed at which it is filled =x/4
speed at which it is drained= x/5
time for which it was filled is 10 hrs (11-1)
and let y be the time after which drain was open, then

(x/4)10-(x/5)y = x
solve and we get y=2.5 hrs
so, 2.5 hrs from 1 pm is 3.30 p.m.

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Manager
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05 Jan 2008, 06:07
Dellin wrote:
3.30 is the start time.
my way of doing it,
let x be the volume of the tank
speed at which it is filled =x/4
speed at which it is drained= x/5
time for which it was filled is 10 hrs (11-1)
and let y be the time after which drain was open, then

(x/4)10-(x/5)y = x
solve and we get y=2.5 hrs
so, 2.5 hrs from 1 pm is 3.30 p.m.

nice way of solution too) but y is not 2.5 rather y is 7.5 and x is 2.5 am I right?

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Manager
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05 Jan 2008, 06:16
Dellin wrote:
3.30 is the start time.
my way of doing it,
let x be the volume of the tank
speed at which it is filled =x/4
speed at which it is drained= x/5
time for which it was filled is 10 hrs (11-1)
and let y be the time after which drain was open, then

(x/4)10-(x/5)y = x
solve and we get y=7.5 hrs
so, x= 2.5 hrs from 1 pm is 3.30 p.m.

Last edited by Dellin on 05 Jan 2008, 11:17, edited 1 time in total.

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Director
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05 Jan 2008, 06:25
bmwhype2 wrote:
A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1pm and some time alter the drain that empties the pool was also opened. If the pool was filled by 11pm and not earlier, when was the drain opened?

2 pm
2:50 pm
2:30 pm
3:30 pm
5pm

Let Pool = 20
Filled at the rate of: 20/4 = 5 pool/hr
Drained at the rate of: 20/5 = 4 pool/hr
When both valves are open, the pool is filled at the rate of: 5-4 = 1 pool/hr

Let x be the amount that is filled when only filling valve is open.
Therefore, 20-x is the amount that is filled when both valves are open.
Total time to fill the pool = 10 hrs

x/5 + (20-x)/1 = 10 --> (time taken when only filling valve is open = x/5; time taken when both valves are open = 20-x)
x = 12.5 --> (this is the amt. of pool filled when only the filling valve is open, i.e., when the draining valve is shut)
x/5 = 12.5/5 = 2.5 hrs --> (this is the duration for which only the filling valve was open)
Therefore, the draining valve was opened at 1 pm + 2.5 hrs = 3:30 pm

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06 Jan 2008, 07:51
very good Q

It takes 10 hours ie the valve that fills exactly fills the tank 2.5 times

=>at the end of the effort the tank is one time filled

=> 1.5 times the effort is gone down the drain
=> 5X1.5=7.5 hrs the rain is open
as the tank fills exactly at 11 pm ie both events happen simultaneously
11pm-7.5 hrs = 3 30 pm

ans is D

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Re: A pool can be filled in 4 hours and drained in 5 hours. The [#permalink]

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01 Jan 2017, 00:58
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Re: A pool can be filled in 4 hours and drained in 5 hours. The   [#permalink] 01 Jan 2017, 00:58
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