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

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If x, y, and k are positive numbers such that (x/x+y)(10) + [#permalink]  01 Sep 2007, 20:25
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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

Last edited by gluon on 01 Sep 2007, 22:01, edited 1 time in total.
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Re: PS - set 1 [#permalink]  01 Sep 2007, 21:12
C. 15
eqn can be solved as
(10x+20y)/((x+y)=k
when x and y are 1, k = 15
The plugging-in yeilds the same result!

gluon 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

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I guess it is D.

As x<y, if we plug in X=1;Y=4.

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

Plugging in 18 will solve the problem.

OA?
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appuvar wrote:
I guess it is D.

As x<y, if we plug in X=1;Y=4.

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

Plugging in 18 will solve the problem.

OA?

OA is D

Just one more question... how did you think about trying out x=1 and y=4 for the equation? It would have taken me a long time to find this by trial and error. Is there any systematic way?
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gluon wrote:
appuvar wrote:
I guess it is D.

As x<y, if we plug in X=1;Y=4.

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

Plugging in 18 will solve the problem.

OA?

OA is D

Just one more question... how did you think about trying out x=1 and y=4 for the equation? It would have taken me a long time to find this by trial and error. Is there any systematic way?

I guess there is no specific way for trial and error. In this case it is pretty simple to try out values once you have the equation 10 - k = k -20. Keep multiplying the Right hand side with 2, 3,4 ... till you get an integer value for k.
Senior Manager
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I figured out a way to find values for x and y:

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

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

So know we have two equations:
x + 2y = k
x + y = 10

For k=10 we get x=10, y=0. But its given that x < y, hence incorrect.
For k=12 we get x=8, y=2. But its given that x < y, hence incorrect.
For k=15 we get x=5, y=5. But its given that x < y, hence incorrect.

For k=18 we get x=2, y=8. Finally we have x < y, hence correct answer is D.

But this is too slow. I wonder if there is a faster, systematic method we are missing.
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D

just simplify the equation u'll get

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

plug in any value wer x<y u'll get ans 50/3 nad the nearest value is 18
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D

just simplify the equation u'll get

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

plug in any value wer x<y u'll get ans 50/3 nad the nearest value is 18
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