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Source: — Problem Solving |
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by shankar.ashwin » Tue Nov 01, 2011 3:57 am
Substitute values of 'k' from answer choices. Only D satisfies the condition of x<y.
You could try it for all options
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by rijul007 » Tue Nov 01, 2011 4:00 am
10[x/(x+y) + 2y/x+y] = k
10[(x+2y)/(x+y)] = k
10[(x+y+y)/(x+y)] = k
10[1 + y(x+y)] = k

The above exprssion tells us that k is a multiple of 10 and greater than 10.

Only option E satisfies this condition.
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by neelgandham » Tue Nov 01, 2011 4:02 am
GmatKiss wrote:Will substitution work in this problem!!

Really hard to solve without sub.!
I am not a great fan of substitution, so here is my approach.

Exp :(10*(x/(x+y))) + (20*(y/(x+y))) = k
=> (10*((x+y)/(x+y)) + (10*(y/(x+y))) = k
=> 10 + (10*(y/(x+y)) = k (You can eliminate option A here as this expression is always > 10)

x,y>0 and y>x => 0<x<y.So the limits are 0 and y,

Let x = 0, then the expression 10 + (10*(y/(x+y)) = 20 > k
Let x = y, then the expression 10 + (10*(y/(x+y)) = 15 < k

=> 15<k<20

Answer : [spoiler]D-18![/spoiler]
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by neelgandham » Tue Nov 01, 2011 4:04 am
rijul007 wrote:10[x/(x+y) + 2y/x+y] = k
10[(x+2y)/(x+y)] = k
10[(x+y+y)/(x+y)] = k
10[1 + y(x+y)] = k

The above exprssion tells us that k is a multiple of 10 and greater than 10.

Only option E satisfies this condition.

10[1 + y/(x+y)] = k doesn't imply that k is a multiple of 10 !

Say y =8 and x =2 then the expression is 18(k=18). k is a multiple of 10, on an assumption that y/(x+y) is an integer, which isn't necessary.
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by rijul007 » Tue Nov 01, 2011 4:09 am
neelgandham wrote:
rijul007 wrote:10[x/(x+y) + 2y/x+y] = k
10[(x+2y)/(x+y)] = k
10[(x+y+y)/(x+y)] = k
10[1 + y(x+y)] = k

The above exprssion tells us that k is a multiple of 10 and greater than 10.

Only option E satisfies this condition.

10[1 + y/(x+y)] = k doesn't imply that k is a multiple of 10 !

Say y =8 and x =2 then the expression is 18(k=18). k is a multiple of 10, on an assumption that y/(x+y) is an integer, which isn't necessary.
sorry, i realized my mistake after posting
thnks for clarifying :)
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