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pool and valves

by ace_gre » Mon Jan 25, 2010 5:58 pm
A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?
A) at 2:00 pm
B) at 2:30 pm
C) at 3:00 pm
D) at 3:30 pm
E) at 4:00 pm

I did it by backsolvling and ended up with D Does anyone have a better approach?
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by thephoenix » Mon Jan 25, 2010 8:16 pm
I got D ; my way

let after x hrs the drained valve was opened
in x hrs the pool will be filled x/4 @ 1/4
for rest 11-x hrs the pool will be filled @ 1/20----->(11-x)/20
eqn
(x/4)+(11-x)/20=1----->x=2.25----> 2 hrs 35-----> at 3:35 the drained valve is open( the only option nearest is D)

but i am not satisfied with my approach
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by ace_gre » Mon Jan 25, 2010 8:31 pm
After some contemplation, here is my way..

Let the capacity of the pool be x and the number of hours the inlet valve was open be n.
Then 10-n will be the no. of hours both inlet and drain valves were open.

When both inlet and outlet valve was open then net gain in the pool = x/4 - x/5 = x/20

According to the info given,
(x/4) * n + (x/20) * (10-n) = x
Cancelling x through out, and solving for n = 5/2 => 2 1/2 hrs.
Then the drain valve was opened 2 1/2 hrs after the inlet valve was opened. i.e 3.30PM :)
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by fibbonnaci » Mon Jan 25, 2010 8:41 pm
Hey thephoenix, your approach is rite. u made a simple error.

in one hour the inlet pipe fills 1/4th of the tank.
sat x hrs the inlet pipe alone functions. so the amount filled is x/4 th of the tank.

total time is 10 hrs.

the drain works for (10- x) hrs in combination with the inlet.
so in 1 hr (1/4 - 1/5) => 1/20th of the tank gets full
in (10 - x) hrs, (10-x)/20 th of the tank gets filled.

so total tank becomes full in x/4 + (10-x)/ 20 =1
x= 2.5hrs.

so the drain opens two and half hours after 1 PM. ie 3.30
- D
Hope this helps!
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by thephoenix » Mon Jan 25, 2010 8:52 pm
fibbonnaci wrote:Hey thephoenix, your approach is rite. u made a simple error.

in one hour the inlet pipe fills 1/4th of the tank.
sat x hrs the inlet pipe alone functions. so the amount filled is x/4 th of the tank.

total time is 10 hrs.

the drain works for (10- x) hrs in combination with the inlet.
so in 1 hr (1/4 - 1/5) => 1/20th of the tank gets full
in (10 - x) hrs, (10-x)/20 th of the tank gets filled.

so total tank becomes full in x/4 + (10-x)/ 20 =1
x= 2.5hrs.

so the drain opens two and half hours after 1 PM. ie 3.30
- D
Hope this helps!
THANKS I GOT IT
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by bigmonkey31 » Mon Jan 25, 2010 9:55 pm
This is how I did it.. a bit more conceptual...

Rate to fill = 1/4
Rate to drain = 1/5

Since you know the Fill is on for 10 hrs (1pm to 11pm), you can set up the equation to see how long the drain is on:

t = time the drain is on
(1/4)10 - (1/5)(t) = 1
solve for t.. t=7.5 hrs

Subtract 7.5 from 11pm...

=3:30
(D)
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by rahul.s » Sat Jan 30, 2010 9:57 am
bigmonkey31 wrote:This is how I did it.. a bit more conceptual...

Rate to fill = 1/4
Rate to drain = 1/5

Since you know the Fill is on for 10 hrs (1pm to 11pm), you can set up the equation to see how long the drain is on:

t = time the drain is on
(1/4)10 - (1/5)(t) = 1
solve for t.. t=7.5 hrs

Subtract 7.5 from 11pm...

=3:30
(D)
why does it = 1?
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by sars72 » Sat Jan 30, 2010 10:13 am
rahul.s wrote:
bigmonkey31 wrote:This is how I did it.. a bit more conceptual...

Rate to fill = 1/4
Rate to drain = 1/5

Since you know the Fill is on for 10 hrs (1pm to 11pm), you can set up the equation to see how long the drain is on:

t = time the drain is on
(1/4)10 - (1/5)(t) = 1
solve for t.. t=7.5 hrs

Subtract 7.5 from 11pm...

=3:30
(D)
why does it = 1?
The value 1 is used to represent the full tank i.e. complete task
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by bigmonkey31 » Sat Jan 30, 2010 11:02 am
^ exactly. You set up your equation (rate*time=work); and work in this case is 1 pool. Likewise, if the question were asking 1/2 the pool, then work=.5

Hope that helps.
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