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Algebra Expressions- Not so simple!

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by Kemmy G » Wed Jun 15, 2011 12:39 pm
Can someone PLEASE explain question 148 of the 12 ed. OG quant section?
[x/x+y](10)+[y/x+y](20)= k.
What could be the value of k?

I've studied the explanation in the OG answer question and am still as stumped as ever. Can someone PLEASE explain it step by step, so it's really easy to follow and grasp?

Thanks, :)
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by Frankenstein » Wed Jun 15, 2011 1:23 pm
Hi,
Please post the complete question, including options. You haven't even posted the condition x<y and
x,y are positive numbers.

[x/x+y](10)+[y/x+y](20) = (10x+20y)/(x+y) = [10(x+y)+10y]/(x+y)
= 10[1 + (y/x+y)]
Given x < y => x+y < y+y
So, x+y < 2y => y/(x+y) > 1/2
and y/(x+y) < 1
So, 10[1 + (y/x+y)] > 10[1+ (1/2)] & 10[1 + (y/x+y)] < 10[1+ 1]
So, [x/x+y](10)+[y/x+y](20) > 15 & [x/x+y](10)+[y/x+y](20) < 20
So, any value between 15 and 20 could be the value of k.
Last edited by Frankenstein on Wed Jun 15, 2011 1:28 pm, edited 1 time in total.
Cheers!

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by pemdas » Wed Jun 15, 2011 1:27 pm
i'll try without looking up an official explanation
with our conditions given x<y and x,y,k>0 we set k=(10x+20y)/(x+y) and this is equivalent to 10+10y/(x+y); because 10y/(x+y) is more than 10/2 {x<y, hence x+y<2y} our answer can be in the interval (greater than 15 and less than 20), so it's d
Kemmy G wrote:Can someone PLEASE explain question 148 of the 12 ed. OG quant section?
[x/x+y](10)+[y/x+y](20)= k.
What could be the value of k?

I've studied the explanation in the OG answer question and am still as stumped as ever. Can someone PLEASE explain it step by step, so it's really easy to follow and grasp?

Thanks, :)
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by GMATGuruNY » Wed Jun 15, 2011 2:17 pm
I posted an easy solution here: https://www.beatthegmat.com/tough-algebr ... tml#292229
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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by Kemmy G » Tue Aug 02, 2011 3:28 am
Thank you so much guys, especially GMATGuruNY!!!

I thought I would receive alerts in my email box whenever anyone replied to my post, and since none came, I simply assumed the question wasn't answered. Will have to improve on checking BTG more regularly, :D. Thanks again everyone.

Off to study and understand the question again!
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