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mpopatia
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Post subject: Running at their respective constant rates, Machine X takes  Posted: Sat Jun 19, 2010 2:21 pm |
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Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if two machines together produce 5/4w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
A) 4
B) 6
C) 8
D) 10
E) 12
OA - E
I understand the OA method which uses algebra: 3(w/d + w/d+2) = 5/4w and solves to be 6, and then 2w = 12. However I attempted to use the RTW chart and got stuck! Is there a way to pick numbers for the variables and use the RTW chart for this type of problem?
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Viswanathan.harsha
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Thu Jul 01, 2010 12:49 am |
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Posts: 46
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Can you please work out the algebra for the given problem? I am having difficulty getting w=6. Is this problem from gmat prep?
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rash.patil
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Thu Jul 01, 2010 4:53 am |
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Let T total number of days taken by machine X to produce W widgets.
Then Work done by x is 1 day is W/T
Work done by machine y in 1 day = W/(T - 2).
Combined work done in 3 days = 3( W/T + W/(t-2) )
= 3W( 2T-1 ) / (T- 2) which is equal to 5W/4
So equating the two equations we get
12( 2T - 1 ) = 5T( T - 2)
24T - 12 = 5T2 - 10T
5T2 - 32T + 12 = 0
(T-6) (5T+2) = 0.
Which gives T = 6.
So no. of days to produce 2W widgets = 2T = 12.
Ans: E
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Viswanathan.harsha
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Fri Jul 02, 2010 3:32 am |
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I believe it is supposed to be 12 (2T-2)=5T(T-2)
This gives you 5T^2-34T+24=0
Giving you t=6
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mschwrtz
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Mon Jul 12, 2010 11:31 pm |
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| ManhattanGMAT Staff |
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Posts: 506
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That's right Viswanathan.harsha
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ayush.sharma
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Mon Feb 20, 2012 2:35 pm |
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Guys - I get stuck as well.
The difference is that you are taking one value to be
D days and then Y to be D-2.
But what if you take y to be D days and x to be D+2.
As given in the question ( Machine X takes 2 days longer)
So I put Machine Y takes D days
and Machine X takes 2+D days.........
By doing that I get stuck at a quadratic that does not make sense :
5d^2 -14d-24=0
I cant see why my approach is wrong.
Any help ?
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SRK
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Wed Feb 22, 2012 1:23 pm |
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Even i am facing the same issue as ayush. Could some one please help?
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ayush.sharma
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Thu Feb 23, 2012 1:21 am |
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Got it.
The equation is correct.
5t^2 -14t-24=0
When ever I see the co-efficient of t^2>1 - I just go for
roots of quadratic =
(-b +or - sqrt ( b^2-4ac))/2a
The mistake I was making before was that the value of c in this equation is not 24 but -24. and that will give one of the roots as 4 and that is what we were looking for.
Does anyone have a better way to deal with this revere foil quadratic eq with coeff x >1?
thanks- hope that helped
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tim
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Sun Feb 26, 2012 5:21 am |
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| ManhattanGMAT Staff |
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Posts: 2242
Location: Southwest Airlines, seat 21C
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cool. let us know if you have any other questions about this one..
_________________
Tim Sanders
Manhattan GMAT Instructor
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shubham_sagijain
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Mon Feb 27, 2012 3:21 pm |
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I think insted of solving the equation, it is much better to substitute the values in the eqn and check.
D --> Number of days
1/D - rate of Y
1/(D+2) - rate of X
we get 3 (1/D + 1/(D+2)) = 5/4.
Now, just substitute D = 4 and see whether the eqn is satisfied.
So, 4 + 2 = 6 days - No. of days taken by X to produce W widgets.
Just multiply by 2 to get the number of days to produce 2 W widgets.
2* 6 = 12..so E is the answer :)
Thanks,
Shubh
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RonPurewal
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Post subject: Re: Running at their respective constant rates, Machine X takes Posted: Sat Mar 03, 2012 8:34 am |
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Posts: 7146
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that's a nice way to do it.
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