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# If x, y, and k are positive numbers such that

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VP
Joined: 22 Nov 2007
Posts: 1079
If x, y, and k are positive numbers such that [#permalink]

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05 Jan 2008, 13:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x, y, and k are positive numbers such that 10x/(x+y)+20y/(x+y) = k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30
CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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05 Jan 2008, 13:18
D

k=10x/(x+y)+20y/(x+y)=(10x+10y+10y)/(x+y)=10+10*y/(x+y)

x<y and x,y are positive: 1/2<y/(x+y)<1 --> 15<k<20 --> k=18
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Director
Joined: 12 Jul 2007
Posts: 858

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05 Jan 2008, 13:29
10x + 20y = k(x+y)

10 = (x+2y) = (x+y) = impossible
12 = (5/6x + 5/3y) = (x+y) = possible
15 = (2/3x + 4/3y) = (x+y) = possible
18 = (5/9x + 10/9y) = (x+y) = possible
30 = (1/3x + 2/3y) = (x+y) = impossible

so now we have three possibilities to narrow down.

for 12: (x-5/6x) = (5/3y-y)
1/6x = 2/3y = impossible since x is smaller than y

for 15: (x-2/3x) = (4/3y-y)
1/3x = 1/3y = impossible since x is smaller than y

for 18: (x-5/9x) = (10/9y -y)
4/9x = 1/9y = possible since x is smaller than y

Manager
Joined: 04 Jan 2008
Posts: 83

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05 Jan 2008, 13:34
D is the answer, by trial and error method. only D satisfies x<y
Re: algebra   [#permalink] 05 Jan 2008, 13:34
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