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A certain sum of money is divided among A, B and C such that A gets on

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A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 17 Oct 2011, 20:13
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Difficulty:

  85% (hard)

Question Stats:

62% (02:31) correct 38% (02:55) wrong based on 90 sessions

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A certain sum of money is divided among A, B and C such that A gets one-third of what B and C together get and B gets two-seventh of what A and C together get. If the amount received by A is $12.4 more than that received by B, find the total amount shared by A, B and C.

A. $345.20
B. $386.40
C. $520.30
D. $446.40
E. None
[Reveal] Spoiler: OA

Kudos [?]: 132 [1], given: 165

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Re: A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 17 Oct 2011, 21:29
stoy4o wrote:
A certain sum of money is divided among A, B and C such that A gets one-third of what B and C together get and B gets two-seventh of what A and C together get. If the amount received by A is $12.4 more than that received by B, find the total amount shared by A, B and C.

A. $345.20
B. $386.40
C. $520.30
D. $446.40
E. None

Can someone give me a short-cut on how to get to an answer?

[Reveal] Spoiler:
D


A = 1/3 (B+C) => C = 3A - B ---(1)
B = 2/7 (A+C) => C = 3.5 B - A --(B)
A-B = $12.4
A = 12.4+B

(1)===> C = (37.2)+3B - B = 2B+37.2 ==> 2B-C = -37.2 ---(3)
(2)===> C = 3.5 B - B-12.4 = 2.5B-12.4==>2.5B-C = 12.4 ---(4)

from (4) and (3) 0.5B = 49.6
B = $99.2
A= $111.6
C =334.8-99.2=$235.6


Ans $446.4 (D)
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A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 01 Jul 2017, 14:16
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D
The problem is not hard at all, if you get the logic of it
If A has \(\frac{1}{3}\) of B and C, this means that B and C together have \(\frac{3}{3}\) . So A has 1 peace, B and C have 3 peaces, overall A has \(\frac{1}{4}\) of the total share. If we apply similar logic to B, B has\(\frac{2}{9}\). Now we can make an equality.

\(\frac{1}{4}\)x - \(\frac{1}{9}\)x = 12.4

\(\frac{(9-8)}{36}\)x = 12.4

\(\frac{1}{36}\)x = 12.4

x= 12.4*36=446.4

Hope that helped
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A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 01 Jul 2017, 19:02
D
Just need to get an equation.

A = \(\frac{(B+C)}{3}\) = \(\frac{(total - A)}{3}\)
B = \(\frac{2(A+C)}{7}\) = \(\frac{2(total - B)}{7}\)

A - B = 12.4
\(\frac{(total - A)}{3}\) - \(\frac{2(total - B)}{7}\) = 12.4
\(\frac{(7Total - 7A - 6Total + 6B)}{21}\) = 12.4
(Total - 7A + 6B)= 21*12.4
A + B + C - 7A + 6B = 21*12.4
B + C - 6A + 6B = 21*12.4
B + C - 6(A-B) = 21*12.4
B + C = 21*12.4 + 6*12.4
B + C = 27*12.4 => A = 9*12.4
A + B + C = 36*12.4 = 446.4

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Re: A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 01 Jul 2017, 19:42
stoy4o wrote:
A certain sum of money is divided among A, B and C such that A gets one-third of what B and C together get and B gets two-seventh of what A and C together get. If the amount received by A is $12.4 more than that received by B, find the total amount shared by A, B and C.

A. $345.20
B. $386.40
C. $520.30
D. $446.40
E. None



Hi,

Since the method with equation is given, I will touch on an alternate method.

A bit of quick thinking at times can save you a lot of time
B gets 2/7 of (B+C), so total= A+B+C=2/7*(B+C)+B+C=9/7*(B+C)..
Here B+C=TOTAL*7/9..
So answer should be div by 9, only D is left.

If there are two three choices fitting in, one more calculation can get you to answer.
Best if you are clueless on making equations and the choices does not have None in choices
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 28 Sep 2017, 07:18
Vardan95 wrote:
D
The problem is not hard at all, if you get the logic of it
If A has \(\frac{1}{3}\) of B and C, this means that B and C together have \(\frac{3}{3}\) . So A has 1 peace, B and C have 3 peaces, overall A has \(\frac{1}{4}\) of the total share. If we apply similar logic to B, B has\(\frac{2}{9}\). Now we can make an equality.

\(\frac{1}{4}\)x - \(\frac{1}{9}\)x = 12.4

\(\frac{(9-8)}{36}\)x = 12.4

\(\frac{1}{36}\)x = 12.4

x= 12.4*36=446.4

Hope that helped


This is brilliant. Thanks! Kudos to you.

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Re: A certain sum of money is divided among A, B and C such that A gets on [#permalink]

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New post 01 Oct 2017, 22:37
Vardan95 wrote:
D
The problem is not hard at all, if you get the logic of it
If A has \(\frac{1}{3}\) of B and C, this means that B and C together have \(\frac{3}{3}\) . So A has 1 peace, B and C have 3 peaces, overall A has \(\frac{1}{4}\) of the total share. If we apply similar logic to B, B has\(\frac{2}{9}\). Now we can make an equality.

\(\frac{1}{4}\)x - \(\frac{1}{9}\)x = 12.4

\(\frac{(9-8)}{36}\)x = 12.4

\(\frac{1}{36}\)x = 12.4

x= 12.4*36=446.4

Hope that helped


Just a small correction...

1/4x-2/9x=12.4

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Re: A certain sum of money is divided among A, B and C such that A gets on   [#permalink] 01 Oct 2017, 22:37
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