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positive integers

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Junior | Next Rank: 30 Posts
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by wishkaro » Tue May 05, 2009 9:01 am
Good one ... any short cut

If x, y, and k are positive integers such that 10 x/ (x+y) + 20 y /(x+y) = 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

OA D
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by sacx » Wed May 06, 2009 2:45 am
www.beatthegmat.com/tough-algebra-equation-t13568.html
SACX
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by sureshbala » Wed May 06, 2009 3:00 am
There's a nice explanation given by Ian.

Let's have a look from algebra point of view

We have

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

Given x<y

i.e x+y < 2y

i.e. x+y/y < 2

i.e. y/(x+y) > 1/2

So we have 1/2 < y/(x+y) < 1

Hence 10[(1 + y/(x+y)] must be in between 15 and 20.

So 18 is the possible answer
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by navalpike » Wed May 06, 2009 10:40 am
Suresh, this is probably obvious to you but how did you go from :

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

to

10[(1 + y/(x+y)]=k

Thanks,
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by sureshbala » Wed May 06, 2009 10:49 am
navalpike wrote:Suresh, this is probably obvious to you but how did you go from :

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

to

10[(1 + y/(x+y)]=k

Thanks,
Hi,

We know that (a+b)/c = a/c + b/c

Similarly [(x+y)+y]/(x+y) = (x+y)/(x+y) + y/(x+y) = 1 + y/(x+y)

(consider a= x+y and b = y in the above example)
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by navalpike » Wed May 06, 2009 10:56 am
Gotcha, thanks :)
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