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Four brothers Adam, Bill, Charles and David together [#permalink]
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01 Apr 2010, 03:37
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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 onefourth of the total contribution made by the other three brothers and the contribution made by Charles is twothird 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, 03:27, 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, 04:02
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, 04:40
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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, 05:04, edited 1 time in total.



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Re: Hard Ratio question [#permalink]
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01 Apr 2010, 05:43
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, 09:47
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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$. how did you get the contributed amounts? for example A contributed 1/3, B contributed 1/5th, etc?
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Re: Hard Ratio question [#permalink]
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08 Apr 2010, 13:18
azule45 wrote: sh00nya wrote: Given that A:(B+C+D)= 1:2. Hence A contributed 1/3rd of the total price.
Given B:(A+C+D)=1:4 Hence B contributed 1/5th of the total price.
Given C:(A+B+D)=2:3 Hence C contributed 2/5th of the total price.
Thus the contribution made by A,B and C= 1/3+1/5+2/5=14/15
So, the remaining 1/15th of the price is contributed by David. Hence David's contribution= 1/15×9900=660$. 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, 00:50
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there is another simple way 
a+b+c+d=t
a= .5(ta) b= .25(tb) c=2/3(tc)
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, 00:01
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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(9900C)/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, 03:30



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Re: Hard Ratio question [#permalink]
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04 Mar 2014, 21:13
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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26 Jan 2018, 09:48
hardnstrong wrote: 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 onefourth of the total contribution made by the other three brothers and the contribution made by Charles is twothird 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 ] We can let the contribution made by Adam, Bill, Charles and David be A, B, C, and D respectively. Since Adam contributes A dollars, the other three boys contribute 9900  A dollars; so we have: A = (1/2)(9900  A) 2A = 9900  A 3A = 9900 A = 3,300 Similarly, since Bill contributes B dollars, the other three boys contribute 9900  B dollars and we have: B = 1/4(9900  B) 4B = 9900  B 5B = 9900 B = 1,980 Finally, since Charles contributed C dollars, the other three boys contributed 9900  C dollars and thus: C = 2/3(9900  C) 3C = 2(9900)  2C 5C = 2(9900) C = 3,960 Now that we know A = 3,300, B = 1,980 and C = 3,960, we can easily find D by the following equation: 3,300 + 1,980 + 3,960 + D = 9,900 9,240 + D = 9,900 D = 660 Answer: D
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Re: Four brothers Adam, Bill, Charles and David together
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