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# If x, y, and k are positive numbers such that ( x/x+y)(10) +

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Manager
Joined: 03 Oct 2008
Posts: 61
If x, y, and k are positive numbers such that ( x/x+y)(10) + [#permalink]

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06 Oct 2008, 07:34
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 ( x/x+y)(10) + (y/x+y )(20) = 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
Retired Moderator
Joined: 05 Jul 2006
Posts: 1741

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06 Oct 2008, 08:02
albany09 wrote:
If x, y, and k are positive numbers such that ( x/x+y)(10) + (y/x+y )(20) = 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

k = 10 (x+2y/x+y) ,

A is ruled out , the numinator > doniminator question becomes can x+2y/x+y = (6/5 or 3/2 0r 9/5 0r 3)

D is my answer ( x= 1, y = 4)
Intern
Joined: 21 Aug 2008
Posts: 23

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06 Oct 2008, 08:04
let x=1, y=4

1/5(10) + 4/5(20) = 18

D.
Senior Manager
Joined: 04 Jan 2006
Posts: 276

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06 Oct 2008, 12:36
albany09 wrote:
If x, y, and k are positive numbers such that ( x/x+y)(10) + (y/x+y )(20) = 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

10x + 20y = k (x + y)
10x + 20y = kx + ky
10x - kx = ky - 20y
x(10 - k) = y(k - 20)

y = x (10 - k) / (k - 20)

Since x < y,
y = x * [some positive integers more than 1].
y = x * [(10 - k)/ (k - 20)]

Therefore, (10 - k) / (k - 20) must be a positive integer more than 1. Check with the answer choice.

E. k = 30, (10 - 30) / (30 - 20) = -20/10 = -2. Wrong, because this make y a negative number.
D. k = 18, (10 -18) / (18 - 20) = -8 / -2 = 4. Correct.
I will skip C, B, and A. You can check the answer yourself.

Re: math problem   [#permalink] 06 Oct 2008, 12:36
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# If x, y, and k are positive numbers such that ( x/x+y)(10) +

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