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Try this

by Bidisha800 » Tue Feb 03, 2009 12:27 am
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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Re: Try this

by Vemuri » Tue Feb 03, 2009 1:58 am
The answer should be 'D'. I attempted this question from the answer choices. Started with C by substituting k with 15. On solving the equation, I got x=y, which cannot be true since x<y. Then tried option D. On solving the equation, got 4x=y. This satisfies the condition x<y
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more detailed

by valentindima » Tue Feb 03, 2009 4:19 am
I simplified the equation to (10x)/(x+y)=20-k.
Now we should start plugging in the possible values:
A. k=10 --> x=x+y --> y=0: x can be anything >0; This is not an option though, since y=0, not positive;
B. k=12 --> 2x=8y --> x=4y: This is where we use the fact that x<y, to weed out options. This not an option.
C. k=15 --> x=y, not an option;
D. k=18 --> x=y/4, correct;
E. k=30 --> x=-y/10, this means that either x or y must be negative. Not an option

Too bad that in the exam you don't get the time to be that analytical.
If these are the options, then I recommend plugging in like Vemuri did and moving on.
If this is a I, II and III, etc kind o f question, then no luck, you have to compute all options.
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by krisraam » Tue Feb 03, 2009 5:17 am
(x/(x+y) )(10) + ( y/(x+y))(20) = k

simplifying the equation you get

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

we know that x<y ==> x+y <2y ==> 1/2 < y/x+y

substitute y/(x+y) = 1/2 we get the minimum value for k.

15< k.

The answer should be greater than 15

As x,y,k are positive the maximum value of y/x+y is 1 ie when x = 0

so k <= 20.

the possible values of k are 15<k<=20

D is the answer
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