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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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17 Oct 2011, 20:13
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A certain sum of money is divided among A, B and C such that A gets onethird of what B and C together get and B gets twoseventh 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
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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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17 Oct 2011, 21:29
stoy4o wrote: A certain sum of money is divided among A, B and C such that A gets onethird of what B and C together get and B gets twoseventh 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 shortcut on how to get to an answer? A = 1/3 (B+C) => C = 3A  B (1) B = 2/7 (A+C) => C = 3.5 B  A (B) AB = $12.4 A = 12.4+B (1)===> C = (37.2)+3B  B = 2B+37.2 ==> 2BC = 37.2 (3) (2)===> C = 3.5 B  B12.4 = 2.5B12.4==>2.5BC = 12.4 (4) from (4) and (3) 0.5B = 49.6 B = $99.2 A= $111.6 C =334.899.2=$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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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 \(\frac{(98)}{36}\)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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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(AB) = 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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01 Jul 2017, 19:42
stoy4o wrote: A certain sum of money is divided among A, B and C such that A gets onethird of what B and C together get and B gets twoseventh 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 timeB 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
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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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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
\(\frac{(98)}{36}\)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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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
\(\frac{(98)}{36}\)x = 12.4
\(\frac{1}{36}\)x = 12.4
x= 12.4*36=446.4
Hope that helped Just a small correction... 1/4x2/9x=12.4




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