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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]

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New post 14 Jun 2012, 01:18
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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, 01:22, 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]

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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]

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New post 16 Jun 2012, 15:02
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]

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New post 07 Oct 2013, 22:47
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]

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Re: If y is not equal to 0 and y is not equal to 1, which is [#permalink]

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New post 16 Feb 2017, 22:56
Q. Is x/Y >X/(Y+1)

X/Y - X/(Y+1)>0
XY+X-XY>0
So the question is
IS X>O?

statement 1. X≠0
Not sufficient as X>0 or X<0

Statement 2. X>Y
Not sufficient as we do not know the sign of Y and the distance between X and Y.

Combine...not sufficient as can't find the sign of x

Ans E

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Re: If y is not equal to 0 and y is not equal to 1, which is   [#permalink] 16 Feb 2017, 22:56
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