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Four brothers Adam, Bill, Charles and David together [#permalink]
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 01 Apr 2010, 04:37
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Question Stats:
73% (03:55) correct
27% (03:51) wrong based on 144 sessions
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Four brothers Adam, Bill, Charles and David together contributed certain amount of money and purchased a car. The contribution made by Adam is half that of the total contribution made by the other three brothers, contribution made by Bill is one-fourth of the total contribution made by the other three brothers and the contribution made by Charles is two-third of the total contribution made by the other three brothers. If the total price of the car is $9900, find the contribution made by David. A. $540 B. $580 C. $600 D. $660 E. $680
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Last edited by Bunuel on 02 Dec 2012, 04:27, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Hard Ratio question [#permalink]
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01 Apr 2010, 05:02
Hmm, if you write it down you get literally: \(A = \frac{1}{2}(B + C + D)\) \(B = \frac{1}{4}(A + C + D)\) \(C = \frac{2}{3}(A + B + D)\) \(A + B + C + D = 9900\) Therefore you have a system of 4 unknowns and 4 equations, so it's solvable (too bad it's not a DS problem we could just stop here  ). Now I guess the easiest way to solve it is to use Gauss' Pivot Method? Remember you are only interested in D, so no need to calculate A, B and C, that would be a waste of time 
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Re: Hard Ratio question [#permalink]
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01 Apr 2010, 05:40
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Given that A:(B+C+D)= 1:2.
Hence A contributed 1/3rd of the total price.
Given B:(A+C+D)=1:4
Hence B contributed 1/5th of the total price.
Given C:(A+B+D)=2:3
Hence C contributed 2/5th of the total price.
Thus the contribution made by A,B and C= 1/3+1/5+2/5=14/15
So, the remaining 1/15th of the price is contributed by David.
Hence David's contribution= 1/15×9900=660$.
Last edited by sh00nya on 01 Apr 2010, 06:04, edited 1 time in total.
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Re: Hard Ratio question [#permalink]
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01 Apr 2010, 06:43
sh00nya wrote: Given that A:(B+C+D)= 1:2.
Hence A contributed 1/3rd of the total price. Hi, I don't understand where you get the 1/3 from?
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Re: Hard Ratio question [#permalink]
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06 Apr 2010, 10:47
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sh00nya wrote: Given that A:(B+C+D)= 1:2.
Hence A contributed 1/3rd of the total price.
Given B:(A+C+D)=1:4
Hence B contributed 1/5th of the total price.
Given C:(A+B+D)=2:3
Hence C contributed 2/5th of the total price.
Thus the contribution made by A,B and C= 1/3+1/5+2/5=14/15
So, the remaining 1/15th of the price is contributed by David.
Hence David's contribution= 1/15×9900=660$. how did you get the contributed amounts? for example A contributed 1/3, B contributed 1/5th, etc?
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Re: Hard Ratio question [#permalink]
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08 Apr 2010, 14:18
azule45 wrote: sh00nya wrote: Given that A:(B+C+D)= 1:2.
Hence A contributed 1/3rd of the total price.
Given B:(A+C+D)=1:4
Hence B contributed 1/5th of the total price.
Given C:(A+B+D)=2:3
Hence C contributed 2/5th of the total price.
Thus the contribution made by A,B and C= 1/3+1/5+2/5=14/15
So, the remaining 1/15th of the price is contributed by David.
Hence David's contribution= 1/15×9900=660$. how did you get the contributed amounts? for example A contributed 1/3, B contributed 1/5th, etc? Add the portions of the ratio. 1:2 => 3 parts total, A is 1/3 1:4 => 5 parts total, B is 1/5 2:3 => 5 parts total, C is 2/5
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Re: Hard Ratio question [#permalink]
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01 May 2010, 01:50
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there is another simple way ----
a+b+c+d=t
a= .5(t-a)
b= .25(t-b)
c=2/3(t-c)
so t = 9900
so a=3300 , b= 1980 , c= 3960
so d= 660
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Re: Hard Ratio question [#permalink]
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02 Dec 2012, 01:01
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(1) A = (B + C + D)/2 (2) B = (A + C + D)/4 (3) C = 2(B + D + A)/3 (4) A + B + C + D = 9900 Combine (1) and (4) A = (9900 - A)/2 ==> A=3300 Combine (2) and (4) B = (9900 - B)/4==>B=1980 Combine (3) and (4) C = 2(9900-C)/3==>C=3960 D = 9900 - (3300 + 1980 + 3960) D = 660 Answer: 660
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Re: Four brothers Adam, Bill, Charles and David together [#permalink]
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02 Dec 2012, 04:30
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Re: Four brothers Adam, Bill, Charles and David together [#permalink]
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25 Feb 2014, 05:15
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Hard Ratio question [#permalink]
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04 Mar 2014, 22:13
sh00nya wrote: Given that A:(B+C+D)= 1:2.
Hence A contributed 1/3rd of the total price.
Given B:(A+C+D)=1:4
Hence B contributed 1/5th of the total price.
Given C:(A+B+D)=2:3
Hence C contributed 2/5th of the total price.
Thus the contribution made by A,B and C= 1/3+1/5+2/5=14/15
So, the remaining 1/15th of the price is contributed by David.
Hence David's contribution= 1/15×9900=660$. Nice method to solve this kind of problems, when only 1 variable value is required
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Re: Four brothers Adam, Bill, Charles and David together [#permalink]
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Re: Four brothers Adam, Bill, Charles and David together [#permalink]
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21 Apr 2017, 11:59
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Re: Four brothers Adam, Bill, Charles and David together
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21 Apr 2017, 11:59
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