A pool can be filled in 4 hours and drained in 5 hours. The : Quant Question Archive [LOCKED]
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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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07 Mar 2007, 21:09
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 later the drain that empties the pool was also opened. If the pool was filled by 11pm and not ealier, 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
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07 Mar 2007, 21:37

I will explain if it is right, but frankly speaking I don't know if my approach is right.
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07 Mar 2007, 21:52
(D)

Solution:

Fills 1/4 part in 1 hr and drains 1/5 part in 1 hr.

So when both runs together it fills (1/4 - 1/5) = 1/20 part in 1 hr.

Lets assume the valve which drains out opened after x hrs.

So, 1st valve ran for x hrs and filled x/4 part before 2nd valve opened.
Remaining 1-(x/4) part filled in between (1+x) pm and 11 pm and both valves were opened.

Together both valves ran for (10-x) hrs and filled (10-x)*1/20 part. Because difference between 1pm and 11 pm is 10 hrs.

So, (10-x)/20 = 1-(x/4)
=> (10-x)/20 = (4-x)/4
=> 10-x = 20-5x
=> 4x = 10
=> x = 10/4 = 2.5 hrs,

I guess, the answer should be at or before 1 pm + 2.5 hrs , i.e., at or before 3:30 pm., That is (D)
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08 Mar 2007, 01:01
Dear Candice,

You may use following shortcut to solve.

It fills tank in 4 hours that means 0.25X per hour (where X is capacity of tank)

It takes 5 hours to empty the tank, that means @ 0.2x per hour.

If both valves work together the speed of filling will be 0.25x-0.20x = 0.05x per hour. with this speed tank will take 20 hours to get filled.

In such questions, you may apply theory of elimination.

if you choose A which is 2pm you will have 9 hours left to fill 75% of the tank, which is not possible...

so you will immediately go to option C), which says 3 pm i.e. 2 hours from start, resulting in tank half filled. but we know that if both the valves will start it will take 10 hours to fill the remaining half of the tank, while we have only 8 hours left. so you can eliminate this too.

Now you have two options i.e. either 3:30 of 4:00 pm.
Lets see 4:00 pm, i.e. 3 hours from the begining. it will result in filling 75% of the tank. While we know that to fill 25% of the tank we need only 5 hours. but 3 hours + 5 hours = 8 hours and not 10 hours. So answer is 3:30pm.

Though, this appear long, but if you will get the logic, it can be resolved within 20 secs....
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08 Mar 2007, 02:12

Trying not to use fractions so as to make the solution quick and prolly more intuitive...

LCM of 4 and 5 is 20.
so assume that tank can hold 20 litres of water.

therefore, inlet fills at a rate 5 litres/hr and outlet empties at 4 litres/hr.

let outlet valve be open after t hours.

so we have the following eqn,
5t + (10-t) = 20
t = 2.5
hence, (D).

for solving the problem using this method, only the equation in italics needs to be written, everything else can be done mentally.
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08 Mar 2007, 18:28
I did it this way..

When inlet runs for 10 hours it can fill tank 2.5 times of capacity.
However at the end of 10 hours tank is full which is 1 times capacity

So outlet removed 1.5 times capacity at the rate of 5 hours for 1 full tank.
So it took 5 + 2.5 = 7.5 hours.

11 pm - 7.5 hours = 3:30 PM
08 Mar 2007, 18:28
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