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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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14 Jun 2012, 01:18
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55% (hard)

Question Stats:

60% (02:23) correct 40% (01:38) wrong based on 178 sessions

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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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14 Jun 2012, 01:34
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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.

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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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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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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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16 Feb 2017, 19:27
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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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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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