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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]  02 Sep 2008, 05:45
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
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Re: time and work [#permalink]  02 Sep 2008, 06:00
arjtryarjtry 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

effective hours to fill the pool = 10 hours

when both are opened hours to fill the pool = 20 ( i.e 1/4 -1/5 = 1/20)
Let say x is the part of pool filled before opening the exit value.
1/4*x+1/20*(1-x)=1/10
x=1/4 --> 1/4 filled before opening exit value.
1/4 can be filled in 1 hour.

2.00 pm
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Last edited by x2suresh on 02 Sep 2008, 07:14, edited 1 time in total.
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Re: time and work [#permalink]  02 Sep 2008, 06:19
arjtryarjtry 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

I'm getting D. Here's how:

First of all, we should consider the formula work= rate * time, work is always 1.

We don't know how many hours the filling part lasted before the draining happened, so let's assume that the filling alone lasted for $$x$$ hours. Since the total time from 1 pm to 11 pm is 10 hours, the remaining hours that the draining came in is $$(10 - x)$$ hours. To fill the pool is 4 hours, so its rate is 1/4. The draining takes 5 hours, so its rate is 1/5 . So our formula should be:

$$1/4(x) + (1/4-1/5)(10-x) = 1$$

$$1/4(x) + (1/20)(10-x) = 1$$

$$1/4(x) + 1/2-1/20(x) = 1$$

x = 2.5 hours. So the filling went on by itself for 2.5 hours before the draining took place. Since the filling started at 1 pm, the draining started 2.5 hours later, which is 3:30 pm.

Last edited by tarek99 on 03 Sep 2008, 02:29, edited 1 time in total.
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Re: time and work [#permalink]  02 Sep 2008, 06:25
Let x be the pool capacity.
x/4 pool fills in 1 hour. Similarly, x/5 pool empties in 1 hour.
The pool gets filled for 10 hours and is simultaneously emptied for (10-y) hours where y is the time lag before the outlet drain is opened.
The equation then becomes:
(x/4)*10 - (x/5)*(10-y) = x
On solving, we get y= 2.50
Therefore, the outlet drain is opened 2.5 hrs. after the inlet drain i.e., at 1:00 + 2.50 = 3:30 pm.
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Re: time and work [#permalink]  02 Sep 2008, 06:26
x2suresh wrote:
arjtryarjtry 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

effective hours to fill the pool = 10 hours

when both are opened hours to fill the pool = 20 ( i.e 1/4 -1/5 = 1/20)
Let say x is the part of pool filled before opened the exit value.
1/4*x+1/20*(1-x)=1/10
x=1/4 --> 1/4 filled before opening exit value.
1/4 can be filled in 1 hour.

2.00 pm

I think the problem with your approach is that you've turned the time into work. Since our total is work, which is 1, you can only get work by multiplying time and rate together. So, the rate of 1/4 has to be multiplied by time in order to get a fraction of the total work of 1. So your 1/20, which is total rate when filling and draining are combined, needs to be multiplied by time so that you can convert it to a fraction of the total work.
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Re: time and work [#permalink]  02 Sep 2008, 07:16
tarek99 wrote:
x2suresh wrote:
arjtryarjtry 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

effective hours to fill the pool = 10 hours

when both are opened hours to fill the pool = 20 ( i.e 1/4 -1/5 = 1/20)
Let say x is the part of pool filled before opened the exit value.
1/4*x+1/20*(1-x)=1/10
x=1/4 --> 1/4 filled before opening exit value.
1/4 can be filled in 1 hour.

2.00 pm

I think the problem with your approach is that you've turned the time into work. Since our total is work, which is 1, you can only get work by multiplying time and rate together. So, the rate of 1/4 has to be multiplied by time in order to get a fraction of the total work of 1. So your 1/20, which is total rate when filling and draining are combined, needs to be multiplied by time so that you can convert it to a fraction of the total work.

Got it guys.
thanks!
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Re: time and work [#permalink]  02 Sep 2008, 17:35
tarek... your explanation is too good. i think that the formula u have mentioned is the key to gettign any work and time problem...

work is always 1. so dividing the total work to work done in unit time(rate) i.e hours/ min/ days etc will give the total time reqd.

its like saying total distance travelled/ dist travelled per hour = no. of hrs.

work and time is nothing but the next version of spped and distance problems.
the key in any work and time problem is to get the rate of work, i.e work done per hour.

sometimes it is given to fill a pool of 100 lts , to print 100 books etc..
my suggestion is to consider that irrespective of the no. given, to take it as 1 work.
i,e say 100lts can be filled by hose A in 5hrs
just take it as hose A takes 5 hrs to do 1 work(whatever it may be).
so 1/time to do the complete work, gives the rate of work.
1/ rate of work will give time to do the work AT THE MENTIONED RATE.

i will furnish one more example just to simplify things..
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Re: time and work [#permalink]  02 Sep 2008, 17:50
X and Y can print 60 books in 20 days and 12 days respectively.X started the work alone and after 4 days, Y joined him till the completion of work. how long did the work last?

here, take 60 books as 1 work.
X rate here is 1/20.
X works alone for 4 days. and X's rate is 1/20.
so in 4 days X will have done 4/20 work or 1/5 work.
now Y Joins in.
Y rate is 1/12.
now remaining work is 1-1/5 or 4/5 work.

so using our formula mentioned i.e work/ rate= time
work=1/5
rate now = rate x+ rate y.= 1/20+1/12= 2/15.

substituting , we get (4/5)/ ( 2/15) or 6 days.

so the work lasted for 4 +6 days or 10 days.
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Re: time and work [#permalink]  03 Sep 2008, 02:35
arjtryarjtry wrote:
X and Y can print 60 books in 20 days and 12 days respectively.X started the work alone and after 4 days, Y joined him till the completion of work. how long did the work last?

here, take 60 books as 1 work.
X rate here is 1/20.
X works alone for 4 days. and X's rate is 1/20.
so in 4 days X will have done 4/20 work or 1/5 work.
now Y Joins in.
Y rate is 1/12.
now remaining work is 1-1/5 or 4/5 work.

so using our formula mentioned i.e work/ rate= time
work=1/5
rate now = rate x+ rate y.= 1/20+1/12= 2/15.

substituting , we get (4/5)/ ( 2/15) or 6 days.

so the work lasted for 4 +6 days or 10 days.

exactly! you've demonstrated your understanding to this type of problem really well. i'm glad that I could help.

regards,
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Re: time and work [#permalink]  03 Sep 2008, 11:08
When ever possible avoid abstraction as much as possible. Since they didn't specify the size of the pool, it is most likely not relevant and therefore whatever answer we get using a specific value is most likely the answer we would get for any value of the pool size.

Let's use 20 gallons since this is the smallest number divisible by 5 and 4.

The valve fills the pool at 5 gallons/hr since it takes 4 hours to fill the pool

The drain empties the pool at 4 gallons/hr since it takes 5 hours to drain the pool.

In ten hours the valve will add 50 gallons but we only want 20 gallons to fill the pool.

Therefore the drain must be open for the amount of time it takes to empty 30 gallons (50 -20)

The drain takes 30/4=7.5 hours to drain 30 gallons

11pm - 7.5hrs = 3.5 or 3:30pm in the context of our problem
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Re: time and work [#permalink]  03 Sep 2008, 11:14
lsmv479 wrote:
When ever possible avoid abstraction as much as possible. Since they didn't specify the size of the pool, it is most likely not relevant and therefore whatever answer we get using a specific value is most likely the answer we would get for any value of the pool size.

Let's use 20 gallons since this is the smallest number divisible by 5 and 4.

The valve fills the pool at 5 gallons/hr since it takes 4 hours to fill the pool

The drain empties the pool at 4 gallons/hr since it takes 5 hours to drain the pool.

In ten hours the valve will add 50 gallons but we only want 20 gallons to fill the pool.

Therefore the drain must be open for the amount of time it takes to empty 30 gallons (50 -20)

The drain takes 30/4=7.5 hours to drain 30 gallons

11pm - 7.5hrs = 3.5 or 3:30pm in the context of our problem

good approach..!!
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Re: time and work   [#permalink] 03 Sep 2008, 11:14
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