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undvaulter
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Post subject: Rate  Posted: Tue Apr 13, 2010 7:48 am |
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A certain liquid passes through a drain at a rate of w/20 gallons every x seconds. At that rate, how many minutes will it take y gallons to pass through the drain?
y/1200xy
20xy/w
xy/3w
w/3xy
3y/wx
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ps63739
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Post subject: Re: Rate Posted: Tue Apr 13, 2010 9:04 am |
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In one second w/(20*x) Gallons will pass or the rate is w/(20*x) gallons/second.
So y/(w/20x) = Number of seconds.
20xy/w = total number of seconds. For minutes devide by 3600,
xy/1200w?
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thoppae.saravanan
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Post subject: Re: Rate Posted: Tue Apr 13, 2010 7:08 pm |
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Quote: 20xy/w = total number of seconds. For minutes devide by 3600 From seconds to hours you need to divide by 3600. For this problem we need to convert seconds to minutes hence we have to divide the value by 60 => 20xy/60w = xy/3w. Option C should be the right answer.
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undvaulter
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Post subject: Re: Rate Posted: Tue Apr 13, 2010 7:56 pm |
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Im not understanding how you arrived at w/(20*x)
The question states "at a rate of w/20 gallons every x seconds"
Doesnt this mean the rate is w/20 and the time is x, thus the set up should be (w/20)*x, or rate*time (gallons*seconds)
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ps63739
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Post subject: Re: Rate Posted: Tue Apr 13, 2010 8:10 pm |
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Thanks thoppae.saravanan. I was taking in hours, sorry my mistake.
undvaulter,
In x seconds w/20 gallons,
so in 1 second w/(20*x) gallons.
In simple terms, suppose you can count till 120 in 30 seconds
So in 1 second, till how much can you count?
Assume x = 30. 120 = w/20. Makes sense?
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undvaulter
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Post subject: Re: Rate Posted: Wed Apr 14, 2010 12:44 am |
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Thank you for the reply, but im still a bit confused.
In the simple example provided above, 30seconds is the time, 120 is the distance, thus 4 would be the rate. R*T=D
This may be where my confusion lies. You stated the 120 is the w/20, however the 120(what you can count til) is the distance and the w/20 is the rate ("at a rate of w/20 gallons.")
The Rate in the original equation is w/20
The Time is x
so (w/20) * x or (w/20) * (x/1)
thus wx/20
this will give you an answer in seconds, so to convert that into minutes you get
wx/20=y/60
my final answer is 3wx/y
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douglas_longo
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Post subject: Re: Rate Posted: Fri Apr 23, 2010 1:05 am |
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it's easiest to pick numbers for each letter, plug them in and get a target number.
w=10
x=6
y=6
10/20= 1/2 gallon every 6 seconds = 6 gallons in 72 seconds
convert to minutes 72/60 = 6/5
now plug in the numbers you picked into each answer until one yields 6/5
C) 6(6)/3(10) 36/30 = 6/5
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Ben Ku
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Post subject: Re: Rate Posted: Wed Apr 28, 2010 11:50 pm |
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Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.
_________________
Ben Ku
Instructor
ManhattanGMAT
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akhan
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Post subject: Re: Rate Posted: Sat Sep 24, 2011 5:07 pm |
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The source of this question is Princeton Review, Online Test (1). (Comes via purchase of their PR GMAT book) The question, in its entirety is as follows:
A certain liquid passes through a drain at a rate of gallons every x
seconds. At that rate, how many minutes will it take y gallons to pass through the drain?
a) y/200xy
b) 20xy/w
c) xy/3w
d) w/3xy
e) 3y/wx
I chose "C" as the OA.
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jnelson0612
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Post subject: Re: Rate Posted: Fri Oct 07, 2011 11:02 pm |
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I concur with the poster who suggested choosing and plugging in real numbers. This problem is quite simple once that is done.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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