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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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Economist wrote:
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:
10/4 - x/5 = 1/10 ( x = number of hours drain pipe is opened ). But I cant find the answer.


m20 q32


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

When only the valve is open, the pool is filling at a rate of \(\frac{1}{4}\) an hour. When both the valve and the drain are open, the pool is filling at a rate of \(\frac{1}{4} - \frac{1}{5} = \frac{1}{20}\) an hour. Let \(x\) denote the time when only the valve was open. Then both the valve and the drain were open for \(11 - 1 - x = 10 - x\)hours. Now we can compose the equation \(\frac{1}{4}x + \frac{1}{20}(10 - x) = 1\) which reduces to \(\frac{x}{5} = 0.5\) from where \(x = 2.5\). Thus, the drain was opened at \(1:00 + 2:30 = 3:30\)pm.

Answer: D.
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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Economist wrote:
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?
*at 2:00 pm
* at 2:30 pm
* at 3:00 pm
* at 3:30 pm
* at 4:00 pm

Don't have the OA :( I am interested in the approach.
I did:
10/4 - x/5 = 1/10 ( x = number of hours drain pipe is opened ). But I cant find the answer.


Imagine both the drain and the valve are on at the same time. You can use a formula here; I just convert each to the same amount of time (20 hours):

valve will fill 5 pools in 20 hours
drain will empty 4 pools in 20 hours
valve+drain will fill 1 pool in 20 hours

So we really have two 'workers' here; the valve, which fills one pool in 4 hours, is on for the first t hours, and the valve+drain combo, which fills one pool every 20 hours, which was on for the remaining 10 - t hours. While the valve was on, t/4 of the pool was filled, and while the valve+drain combo were on, (10-t)/20 of the pool was filled. Adding these, we must get 1 pool:

t/4 + (10-t)/20 = 1
5t + 10 - t = 20
4t = 10
t = 2.5

So the drain was turned on 2.5 hours after 1pm, or at 3:30pm.
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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A different approach :

So in one hour V1 will fill 1/4th of the pool
V2 will empty 1/5th of the pool.

If the valves were opened at the same time: 1/20th of the pool will be full in 1 hour ( Difference of 1/4 and 1/5 )

So if both the valves were opened at 1 PM it would take 20 hours to fill the pool. That's not the case. Work from the answer choices.

If V2 was opened at 3 PM. In 2 hours V1 would have filled : 1/2 of the tank. To fill the remaining half it would take 10 hours. 3 + 10 = 13. So the time should be after 3.

If V2 was opened at 3.30 : in 2.5 hours, V1 would have filled 1/4*2.5 : 5/8th of the tank. From 3.30 to 11 : V1 and V2 have 7.5 hours to fill the remaining 3/8th of the tank.

Operating together : 1/20 can be filled in 1 hour
3/8 can be filled in 3/8 *20 = 7.5 hours. Bingo :)

-pradeep
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We know that Rate x Time = Work done

Here,
Ri = Rate of the inlet valve
Ro = Rate of the outlet valve
Ti = Time for which the inlet valve was open
To = Time for which the outlet valve was open
x = Time after which the outlet valve is open

The time taken for pool to fill to capacity is 10 hours.

RiTi - RoTo = 1
(1/4)(10) - (1/5)(10-x) = 1

Solving the above equation gives x = 2.5 hours

Hence the outlet valve was opened 2.5 hours after 1pm i.e., 3:30pm
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
Ok.....why is it that i can't understand any of the explanations??????? Guys, i am really weak with these kind of work problems. Would anyone care to provide a solution to this problem once again??? please :-(
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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bibha wrote:
Ok.....why is it that i can't understand any of the explanations??????? Guys, i am really weak with these kind of work problems. Would anyone care to provide a solution to this problem once again??? please :-(


You can fill 1 pool in 4 hours. However we have a drain that drains at a rate of 1 pool in 5 hours.

Lets set up a basic rate equation.

we know the rate of the pump is 1pool/4hours. We also know this pump operated from 1PM to 11PM, thus 10 hours. Thus the pump filled 1pool/4hours*10hours or 10/4pools.

But the drain drained the pool at 1pool/5hours * t hours. t is our unknown value and is the number of hours the drain was on.

The total work was one pool, or represented as 10/4+1/5t = 1. Now we just solve for t.

t=15/2 = 7.5 hours or the number of hours the drain was on. Thus we have to make one final step of 10-7.5hrs = 2.5hrs, which is the number of hours the drain was turned on after the pump.

1PM +2.5hrs = 3:30PM

D
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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Filled rate= 1/4, Drained rate = 1/5
Total hours took = 10 [Given]
Let drained valve work for x hours
so, (1/4)(10) - (1/5)(10-x) = 1
x = 2.5 hours

Thus, the outlet valve was opened 2.5 hours after 1pm i.e., 3:30pm
Ans. D
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
D is the answer.
Nice problem. Thanks a ton. I got the solution but it took me about 2.5 minutes to get to the correct answer. Is that reasonable for such problems or is it a little over the limit?
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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siddharthmuzumdar wrote:
D is the answer.
Nice problem. Thanks a ton. I got the solution but it took me about 2.5 minutes to get to the correct answer. Is that reasonable for such problems or is it a little over the limit?


This is not an official question, so you shouldn't be so concerned about solving it in under 2 minutes. I think it is a bit of a lengthy problem anyway. On the real test, you need to average 2 minutes per question - if you can solve some questions in 1.5 minutes, then it's okay to spend 2.5 minutes sometimes. Normally I wouldn't recommend ever spending more than 3 minutes on a question (and if you do, you better be sure you'll get the right answer) unless you know you have extra time.
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
Bunuel

If the value was open for x hours, then why is it (11-1-x) hours when both the valve and the drain were open and not just (11−x)hours?

Thanks!
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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mahrah wrote:
Bunuel

If the value was open for x hours, then why is it (11-1-x) hours when both the valve and the drain were open and not just (11−x)hours?

Thanks!


The valve was opened from 1:00 pm to 11:00 pm, so for 11 - 1 = 10 hours. If x is the amount of time when only the valve was open, then 10 - x is the amount of time when both the valve and the drain were open.
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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Solution



Given:
    • The pool can be filled in 4 hours
    • The pool can be drained in 5 hours.
    • Filling valve was opened at 1:00 pm
    • The Pool was filled at 11:00 pm

Working out:

We need to find out the time when the drain valve was opened.

Per our conceptual knowledge, we know that the total volume of the pool can be taken as the LCM of 4 and 5.
    Thus, let’s assume that the total volume of the pool = 20 Liters.

Since the pool can be filled in 4 hours, in 1 hour, the amount of water filled in it = 5 litres.
Similarly, the amount of water drained in 1 hour = 4 litres.

Let’s also assume that the time after which the drain valve was opened = x hours.
So, from 1 pm to (1+x) pm, the pool was being filled at the rate of 5 liters per hour.
And from (1+x) pm to 11 pm, the pool was filled at a rate of (5-4) = 1 liter per hour.

Thus,
    • 5*x+ 1(10-x) = 20
    • Or, 5x + 10 -x = 20
    • Or, 4x = 10
    • Or, x =2.5 hours.

Thus, the drain valve was opened at 1+ 2:30 hours = 3:30 pm.

Answer: Option D
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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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 sometime 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?.
Considering the total work required to fill the pipe = W.
The rate of work done only when the inlet is working =\( \frac{W}{4}\)
The rate of work done when both the inlet and the outlet are working = \(\frac{W}{4}-\frac{W}{5}=\ \frac{W}{20}\)

The total time taken to get the work done is equal to 10 hours.

For the 10 hours complete period the inlet pipe was working. The total work done by the inlet = \(\frac{10W}{4}\)

For a certain period of time assuming to be x, the outlet was draining : = \(-x\left(\frac{W}{5}\right)\)

The work combination amounts to total work done: \(\frac{10W}{4}-\ \frac{xW}{5}=\ W\)
\(\frac{3W}{2}=\ \frac{xW}{5}\)
x = 7.5 hours.
Hence the outlet pipe started 2 and half hours later the inlet pipe started working.
Hence it was opened at 3: 30 pm.

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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
Rate for filling

1/4

Rate for draining

1/5

Effective rate

1/4 - 1/5

= 1/20

Pool was filled for some hours before draining pipe opened.

Total time 10 hours.

Let a be time filling pipe operated and b when draining pipe opened.

a + b = 10.

So b = 10-a


Let y be fraction of pool filled before drain was opened.

1/4 = y/a

y = a/4


And let z be remained work done after drain was opened.

1/20 = z / 10-a

20z = 10-a

z = 10-a / 20

Therefore total work done

a / 4 + 10-a/ 20 = 1

5a + 10 - a = 20

4a = 10

a = 2.5

Which means drain was opened after 2 hours 30 minutes.

Answer is 3:30pm.

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Re: A pool can be filled in 4 hours and drained in 5 hours. The valve that [#permalink]
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A filling pipe can fill an empty swimming pool in 4 hours, while a drain valve can empty the full pool in 5 hours. To fill the empty pool, the filling pipe was opened at 1:00 pm, and after some time, a drain valve was also opened. If the pool was filled at 11:00 pm, when was the drain valve 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


The rate of the filling pipe is given as \(\frac{1}{4}\) pool/hour and the rate of drain valve is given as \(\frac{1}{5}\) pool/hour. The filling pipe was opened for 10 hours, from 1:00 pm to 11:00 pm. Let x denote the time for which the drain valve was opened, then we'd have \(10*\frac{1}{4} - x*\frac{1}{5} = 1\), which gives \(x = 7.5\) hours. Hence, the drain valve was opened for 7.5 hours, and therefore was opened at 11:00 pm - 7.5 hours = 3:30 pm.


Answer: D
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