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by saurabhmahajan » Wed Aug 11, 2010 1:38 am
How to solve this is more easy way ?

If x, y, and k are positive numbers such that (x/x+y)(10)+(x/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
Thanks and regards,
Saurabh Mahajan

I can understand you not winning,but i will not forgive you for not trying.
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by kvcpk » Wed Aug 11, 2010 1:53 am
saurabhmahajan wrote:How to solve this is more easy way ?

If x, y, and k are positive numbers such that (x/x+y)(10)+(x/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
x/x+y = k/30
When k=30, x= x+y -> y=0 NOT possible
When k = 18, 5x = 6x+6y -> x= -6y .. Not possible because x,y are positive
When K=15, 2x=x+y -> x=y .. not possible because given x<y
when k=12, 5x = 2x+2y -> 3x = 2y.. When y is integr, x need not be an integer. Because x= 2/3 Y
when k=10, 3x = x+y -> 2x = y when y is integer, x is always integer... POSSIBLE

pick A
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by saurabhmahajan » Wed Aug 11, 2010 1:58 am
sorry...its not the correct answer
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by kvcpk » Wed Aug 11, 2010 2:03 am
saurabhmahajan wrote:sorry...its not the correct answer
Are you sure that you have copied the question correctly?

I dont see any option better here. I chose A just thru elimination as stated in my above post.
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by kevincanspain » Wed Aug 11, 2010 2:12 am
saurabhmahajan wrote:How to solve this is more easy way ?

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
Note the correction!
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