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A sum of money was divided between John and Bob so that the ratio of

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A sum of money was divided between John and Bob so that the ratio of [#permalink]

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New post 30 Jun 2017, 02:30
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A sum of money was divided between John and Bob so that the ratio of John’s share to Bob’s share was 5 : 3. If John’s share exceeded (5/9)th of the total sum of money by $50, what was Bob’s share?

(A) $180
(B) $270
(C) $340
(D) $450
(E) $720
[Reveal] Spoiler: OA

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A sum of money was divided between John and Bob so that the ratio of [#permalink]

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New post 30 Jun 2017, 02:54
Let John's share be x and Bob's share be y

From question stem,

\(\frac{x}{y} = \frac{5}{3}\) -> 3x - 5y = 0 ->1
\(x = \frac{5*(x+y)}{9} + 50\)
9x = 5x + 5y + 450 -> 4x - 5y = 450 ->2

Substracting 1 from 2 gives x(John's Share) = 450

Hence, John's share is 450$
Since the ratio's of their shares are 3:5, Bob's share is 270$(Option B)
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Re: A sum of money was divided between John and Bob so that the ratio of [#permalink]

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Bunuel wrote:
A sum of money was divided between John and Bob so that the ratio of John’s share to Bob’s share was 5 : 3. If John’s share exceeded (5/9)th of the total sum of money by $50, what was Bob’s share?

(A) $180
(B) $270
(C) $340
(D) $450
(E) $720


Let's assume John:Bob = 5x:3x

Since John’s share exceeded (5/9)th of the total sum of money by $50, we may write -
5x = \(\frac{5}{9}\) 8x + 50

Solving the equation, we get x = 90

Bob's share = 3x = 90 x 3 = 270

Hence Answer B
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Re: A sum of money was divided between John and Bob so that the ratio of [#permalink]

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Bunuel wrote:
A sum of money was divided between John and Bob so that the ratio of John’s share to Bob’s share was 5 : 3. If John’s share exceeded (5/9)th of the total sum of money by $50, what was Bob’s share?

(A) $180
(B) $270
(C) $340
(D) $450
(E) $720


\(John : Bob = 5:3\)

Total amount \(= 5x + 3x = 8x\)

John’s share exceeded (\(\frac{5}{9}\))th of the total sum of money by \($50\).

\(\frac{5}{9}*x = 50\)

\(x = 90\)

Bob's Share \(= 3x = 3*90 = $270\)

Answer (B)...
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Re: A sum of money was divided between John and Bob so that the ratio of [#permalink]

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Bunuel wrote:
A sum of money was divided between John and Bob so that the ratio of John’s share to Bob’s share was 5 : 3. If John’s share exceeded (5/9)th of the total sum of money by $50, what was Bob’s share?

(A) $180
(B) $270
(C) $340
(D) $450
(E) $720


We are given that the ratio of John’s money to Bob’s money is 5 : 3 or 5x : 3x. Thus, the total is 8x.
Since John’s share exceeded 5/9 of the total sum of money by $50:

5x = (5/9)(8x) + 50

45x = 40x + 450

5x = 450

x = 90

So, Bob’s share was 3 x 90 = 270.

Answer: B
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Re: A sum of money was divided between John and Bob so that the ratio of [#permalink]

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New post 14 Jan 2018, 23:07
Given ratio: 5:3 = 8 parts

also it is mentioned john's share in 5/9 th + 50, total sum could be divisible by 9. Also total sum must be divisible by 8 (8 parts).

LCM (8, 9) = 72, but choices are multiple of 10s.

so let total sum = 720, then john's share = (5/9) * 720 + 50 = 450, bob's share = 270 => (B)
Re: A sum of money was divided between John and Bob so that the ratio of   [#permalink] 14 Jan 2018, 23:07
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