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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
Please provide explanations.
Last edited by gluon on 01 Sep 2007, 22:01, edited 1 time in total.
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Manager
Status: Post MBA, working in the area of Development Finance
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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
Please provide explanations.
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Manager
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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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Senior Manager
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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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Director
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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 DJust 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.
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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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Manager
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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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Manager
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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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