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If y is not equal to 0 and y is not equal to 1, which is

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If y is not equal to 0 and y is not equal to 1, which is [#permalink] New post 14 Jun 2012, 00:18
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E

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64% (02:20) correct 36% (01:36) wrong based on 100 sessions
If y is not equal to 0 and y is not equal to 1, which is greater, x/y or x/(y+1)

(1) x is not equal to 0
(2) x > y
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Jun 2012, 00:22, edited 1 time in total.
Edited the question.
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Re: If y is not equal to 0 and y is not equal to 1, which is [#permalink] New post 14 Jun 2012, 00:34
Expert's post
If y is not equal to 0 and y is not equal to 1, which is greater, x/y or x/(y+1)

Is \frac{x}{y}>\frac{x}{y+1}? --> is \frac{x}{y}-\frac{x}{y+1}>0? --> is \frac{xy+x-xy}{y(y+1)}>0? --> is \frac{x}{y(y+1)}>0?

(1) x is not equal to 0. Not sufficient.
(2) x > y. Not sufficient.

(1)+(2) Still not sufficient, for example: if x>0>(y=-1/2) answer is NO but if x>y>0 answer is YES.

Answer: E.
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Re: If y is not equal to 0 and y is not equal to 1, which is [#permalink] New post 16 Jun 2012, 14:02
Bunuel, is there any easier method to solve this problem? Maybe, a non-algebraic or lesser-algebraic method? :-/
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Re: If y is not equal to 0 and y is not equal to 1, which is [#permalink] New post 07 Oct 2013, 21:47
You can plug numbers. Based on the question stem, it seems pretty easy to plug and play with some numbers. \frac{x}{y} or \frac{x}{(y+1)}

(1) x is not equal to 0 ----> I think from a quick glance, this is not sufficient because we don't know any thing about y. But just for kicks work through the process. Pick numbers: y= 2 , x=2

\frac{2}{2}=1 and \frac{2}{2+1}=\frac{2}{3} Therefore, \frac{x}{y} 1 > 2/3 \frac{x}{(y+1)}

but if we pick y= -2 , x=-2 then \frac{-2}{-2} = 1 and \frac{-2}{-2+1}=\frac{-2}{-1}= 2, Therefore, \frac{x}{y} 1 < 2 \frac{x}{(y+1)}

So based on the results, we have two answer, therefore the statement N/S

(2) x > y ----> same thing for this statement. Try using a variation of the same number. Make the both positive and both negative but satisfying the parameters of the statement. Pick numbers: y= 2 , x=4

\frac{4}{2}=2 and \frac{4}{2+1}=\frac{4}{3} or 1 \frac{1}{3}. Therefore, \frac{x}{y} 2 >1 \frac{1}{3} \frac{x}{(y+1)}

but if we pick y= -4 , x=-2 then \frac{-2}{-4} =\frac{1}{2}and \frac{-2}{-4+1}=\frac{-2}{-3}= \frac{2}{3}, Therefore, \frac{x}{y} .50 < .666 \frac{x}{(y+1)}

So we get two different results from picking numbers that satisfied the parameters in Stmt 2. So this is N/S. When you put the two statements together its also I/S. From Stmt 1, we don't if x is positive or negative. That will have a bearing on the results of the numbers as noted above.

I know its been awhile since you made this post, but the time it took me rationalize this correct answer helped me understand why i got the question wrong. Hope this explanation helps someone else where an algebra equation is not intuitive.
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Re: If y is not equal to 0 and y is not equal to 1, which is   [#permalink] 07 Oct 2013, 21:47
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