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Re M2509
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16 Sep 2014, 00:23
Official Solution:A draining pipe can empty a pool in 4 hours. On a rainy day, when the pool is full, the draining pipe is opened and the pool is emptied in 6 hours. If rain inflow into the pool is 3 liters per hour, what is the capacity of the pool? A. \(9\) liters B. \(18\) liters C. \(27\) liters D. \(36\) liters E. \(45\) liters Let the rate of the draining pipe be \(x\) liters per hour. Then the capacity of the tank will be \(C=time*rate=4x\); Now, when raining, the net outflow is \(x3\) liters per hour, and we are told that at this new rate the pool is emptied in 6 hours. So, the capacity (C) of the pool also equals to \(C=time*rate=6(x3)\); Thus we have: \(4x=6(x3)\). Solving gives \(x=9\). Therefore \(C=4x=36\). Answer: D
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Re: M2509
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09 Oct 2014, 04:33
other method let x be the total capacity x/4 equal to the drain rate. 3 equals rain inflow 6 hours rain inflow equals 18 x/4=(x+18)/6 6x=4x+72 2x=72 x=36



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Here's how I approached it.
Rain inflow is 3l/hr After 6 hours it would have filled in 18l On top of this amount, the pipe would also have to drain the contents of the already filled pool.
We know that draining a filled pool takes 4 hours. But draining a filled pool + the 18l takes 6 hours, hence it takes 2 hours to drain 18l.
Then do a ratio comparison: If 18l takes 2 hrs, a filled pool takes 4 hrs. 18:2 = filled pool:4
Hence, filled pool = 36l
It's also easier to just draw a diagram that would look something like this: __+18___ 2 hrs  x  4 hrs ________
From the diagram, it is more obvious that x is 36.



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Re: M2509
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21 Oct 2014, 08:14
This is how I solved it. Recall rate eqn: 1/r = 1/r1 + 1/r2 In this case, the rate is negative, hence, 1/r = 1/r1  1/r2 i.e. 1/6 = 1/4  1/r2 1/r2 = 1/4  1/6 = 1/12 Hence r2 = 12 W = RT C = RT = 3 * 12 = 36
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Re: M2509
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18 Nov 2014, 20:58
dakrownie wrote: In this case, the rate is negative, hence, 1/r = 1/r1  1/r2 May I know why you considered rate to be negative, please? Any examples or resources to that extent is appreciated.



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Re: M2509
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18 Nov 2014, 21:22
Instead of solving question with conventional method, solve it by plugging in the answer choices and out of the given answer choices 36 is the best to pick up.



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Re: M2509
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30 Dec 2014, 02:37
I appreciate all the methods mentioned but they involved a 2step equation process.
I used this one involving only the rate equation :
Capacity of the pool : x litres Draining pipe rate : x/4  x litres per 4 hours Rain inflow rate : 3/1  3 litres per 1 hour (negative in this case as draining is emptying and rain is filling the pool) Group rate : x/6  x litres per 6 hours
Usual rate equation : 1/r1 + 1/r2 = 1/r ===> x/4  3/1 = x/6 ===> solve for x.



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I did it by calculating the reduction in rate, caused by the rain.
So, normally the drain pipe drains the full pool in 4 hours. This means that the draining rate is 1/4. That rainy day, because the rain was filling the pool while the drain was emptying it, the pool was drained in 6 hours. This means that the draining rate was 1/6.
Now, 1/41/6 = 1/12. This must be the rate of the rain inflow. This means that the pool will be totally full in 12 hours (only by the rain).
So, 3 liters per hour is 3*12 = 36. Sounds complicated even to me that I solved it in this way, but sometimes it happens... So, is this way indeed correct?



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Re: M2509
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22 Jul 2015, 03:32
If you follow % efficiency method it will solved in seconds. The outlet pipe can drain the water in 4 days which mean its efficiency is 25% (divide 100 by 4) and when it rains its efficieny is reduced to 16.66% (divide 100 by 6) that means the rain is filling the pool at an efficiency of 8.33% (25%16.66%). Now from the calculation we can deduce that the rain will take 12 hours to fill the pool (100/8.33%). So just multiply 12 with 3 and get the answer. This approach saves time but one needs to memorise the percentage to fraction conversion table. For example 12.5% =1/8.



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Re: M2509
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02 Jun 2017, 03:16
Tank Capacity=x Drain rate=x/4 l/hr Filling rate=3l/hr Capacity of Tank in 6 hrs= 0 0=x + (3*6)  6*(x/4) x=36 litres



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Re: M2509
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01 Aug 2017, 18:21
My 2 cents: Attachment:
my 2 cents.png [ 4.07 MiB  Viewed 6306 times ]
If you have any question, please let me know & i do like kudos



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Re: M2509
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08 Jun 2018, 06:20
Alternate approach:
Normal Time = 4 hours Because of rain the pipe had to work for extra 2 hours.
Extra inflow to the pool = 3 * 6 = 18 litres
The pipe took 2 hours extra to drain out the extra 18 litres > Rate = 18/2 = 9lt/hour
Total capacity = 9 * 4 = 36
Answer: D



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Re: M2509
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09 Jun 2018, 02:20
L/4 is drain rate.
for six hours it is 6*L/4
Total water drained in siz hours is 6*3 + L
Therefore, 18 + L = 6*L/4
=> L = 36 which is capacity of Tank



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Re: M2509
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09 Jun 2018, 09:18
I solved backwards from the answers: B  18 L / 4 hours = 4.5 L per hour drained. 4.5 * 6 hours total to drain = 27 liters total were drained. Subtract 18 (6 hours * 3 Liters per hour of rain) and you get 9. 9 does not equal 18 (because the pool started full), so try D. D  36 L / 4 hours = 9 L per hour drained. 9 * 6 = 54 liters total were drained. Subtract 18 and you get 36. This is the correct answer.



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Re: M2509
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23 Jun 2018, 05:43
Hours to drain with no rain = 4 hrs Hours to drain with rain = 6 hrs Hours to drain water from rain = 2 hrs Water from rain = 3L/hr * 6hrs = 18L Drain rate (I used only rain water) = 18L/2hrs = 9L/hr Capacity (I used without rain) = 4hours * 9L/hr = 36L



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Re: M2509
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16 Sep 2018, 02:41
Its a simple question, quite GMAT Like very easy to mess up in haste
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