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VP
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If x, y, and k are positive numbers such that [#permalink]
05 Jan 2008, 13:13
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
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CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Dk=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
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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
Answer D
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Manager
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D is the answer, by trial and error method. only D satisfies x<y
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