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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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Manager
Joined: 20 Sep 2008
Posts: 80

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

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

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17 Oct 2011, 20:13
1
KUDOS
5
This post was
BOOKMARKED
00:00

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 Manager Joined: 21 Aug 2010 Posts: 186 Kudos [?]: 135 [0], given: 141 Location: United States GMAT 1: 700 Q49 V35 Re: A certain sum of money is divided among A, B and C such that A gets on [#permalink] ### Show Tags 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) _________________ ------------------------------------- Kudos [?]: 135 [0], given: 141 Manager Joined: 23 Dec 2016 Posts: 65 Kudos [?]: 40 [4], given: 41 Schools: Fuqua GMAT 1: 720 Q49 V40 GPA: 3.21 A certain sum of money is divided among A, B and C such that A gets on [#permalink] ### Show Tags 01 Jul 2017, 14:16 4 This post received KUDOS 2 This post was BOOKMARKED 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 _________________ If you find my solution useful, hit the "Kudos" button Kudos [?]: 40 [4], given: 41 Intern Joined: 20 Jan 2017 Posts: 13 Kudos [?]: 8 [0], given: 9 Location: Armenia Concentration: Finance, Statistics GMAT 1: 730 Q51 V38 GPA: 3.93 WE: Consulting (Accounting) A certain sum of money is divided among A, B and C such that A gets on [#permalink] ### Show Tags 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 Kudos [?]: 8 [0], given: 9 Math Expert Joined: 02 Aug 2009 Posts: 5347 Kudos [?]: 6125 [0], given: 121 Re: A certain sum of money is divided among A, B and C such that A gets on [#permalink] ### Show Tags 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

Kudos [?]: 6125 [0], given: 121

Intern
Joined: 29 May 2012
Posts: 38

Kudos [?]: 5 [0], given: 11

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

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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.

Kudos [?]: 5 [0], given: 11

Manager
Joined: 19 Aug 2016
Posts: 64

Kudos [?]: 3 [0], given: 1

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

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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

Kudos [?]: 3 [0], given: 1

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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