A certain swimming pool of 10000 liters has three pipes;

Stolyar,
my answer comes to 15.01 hours

Sorry guys, it's late in the evening here and I needed to eat some water melon to reconsider my answer. The exam date is approaching and I often catch myself thinking about it even in the most peculiar places.

Unswer is 15 hours.

Logic is as follows:

A pours in 1000*55/60 liters an hour (l/h)
B pours out 200*50/60 l/h
C pours out 100*50/60 l/h

To get a total amount of water that is added into the swimming pool every hour, substract B&C from A. It's 666,(6)liters.
Time=work/rate => 10000/666(6)=15 hours.

We see that pipe A fills in 1000 gal in 65 minutes (1 hr filling and 5 min. rest).

We see that B & C drains out 300 gal. in 70 minutes (1/2 hr draining, 5 min rest, 1/2 hr draining and another 5 min. rest)

So, I suppose the total time taken is 910x. Why? Because, the LCM of 65 and 70 is 910. I used a factor x with it because I do not know yet whether 910 is the time taken in minutes, YET.

Now, A fills in 1000/65*910x gallons in 910x minutes
or, A fills in 14000x gallons

B&C drains 300/70*910x gallons in 910x minutes
or, B&C drains 3900x gallons

So, at the end of the total time (910x minutes) the pool will be filled.
=> 14000x - 3900x = 10000
=> x = 10000/10100 = 100/101

So, total time = 910x
= 910 * 100/101
= 15.01 hours

Becareful the way you word/interpret the problems. The problem actually states that A stops working AFTER working for one hour, and pipes B and C stop working AFTER working for 30 minutes. This is entirely different than if A stops working once every hour (i.e., after working for 55 minutes) and B and C stop working once every half hour (i.e., after working 25 minutes), which is the problem that everyone solved.

Stolyar, what is your intention here?

AkamaiBrah,
I don't agree. The problem states "A stops for 5 minutes after each hour of working. B and C stop for 5 minutes after each half an hour of working."

If "A stops working AFTER working for one hour, and pipes B and C stop working AFTER working for 30 minutes" then the pool will never be filled.

I'm sure he means 5 mnutes after each hour and half hour respectively, but your observation is besides my point. My point is that 5 minutes "after" each hour and half-hour is still different from 5 minutes "at the end" of each hour and half-hour respectively. The solution of 15 hours is the solution to the latter interpretation, which is not what Stolyar asked. The answer to the first interpretation is a few minutes less.

The answers are not different by much, BUT THEY ARE STILL DIFFERENT.

great problem...

the way i modeled this in order to be able to do this in about two minutes was to say that pipe A filled at a rate of 1000 * 11/12 lph and i mentally combined pipes B and C into one pipe emptying at 300 * 5/6 lph.

Then I set up the equation:

X (11000/12 - 1500/6) = 10000

By hand, X is about 15, a little more. Hope that is accurate enough for the GMAT!!!

AkamaiBrah,

Absolutely. You are correct. I see what you mean. Currently trying to formulate a way to see if I can solve that. Thanks

Sure you are correct! A works for a COMPLETE HOUR (60 minutes) then stops for 5 minutes. B and C work for HALF AN HOUR (30 min) and then stop for 5 min. And so on and so forth, until the pull is full. I invented the question occasionally yesterday, so figures can be raw. A hint: re-consider rates. My solution is to go public soon.

My solution:

1. Unite pouring pipes in one that works at the rate of 300 lph
2. Translate nominal rates into actual ones.

Since we deal with minutes, it is reasonable to get rid of hours

the nominal R(fill)=1000/60 lpm
the nominal R(pour)=300/60 lpm

the actual R(fill)=1000/65=200/13 lpm (because of a 5-minute stop after each 60 m)
the actual R(pour)=150/35=30/7 lpm (because of a 5-minute stop after each 30 m)

WORK=RATE*TIME

10000=(200/13-30/7)*X in minutes

X=910000/1010=901 munutes approximately, or 15 hours and 1 minute.