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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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30 Jun 2017, 03:30
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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

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

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30 Jun 2017, 03:54
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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30 Jun 2017, 04:03
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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 _________________ If you like the post, show appreciation by pressing Kudos button Director Joined: 04 Dec 2015 Posts: 700 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) Re: A sum of money was divided between John and Bob so that the ratio of [#permalink] ### Show Tags 30 Jun 2017, 04:08 1 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)... Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2758 Location: United States (CA) Re: A sum of money was divided between John and Bob so that the ratio of [#permalink] ### Show Tags 04 Jul 2017, 07:50 3 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.

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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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15 Jan 2018, 00:07
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] 15 Jan 2018, 00:07
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# A sum of money was divided between John and Bob so that the ratio of

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