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If x and y are positive. is y< 2 ?

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If x and y are positive. is y< 2 ? [#permalink] New post 03 May 2005, 07:51
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Please provide explanation for this soln.

If x and y are positive. is y< 2 ?

(1) x > 2y
(2) x < y + 2
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 [#permalink] New post 03 May 2005, 08:54
Alone (1) and (2) are obviously insufficient.

Let's take them together

x > 2y > 0
x < y + 2 > 0

0 < 2y < y + 2 => y < 2. Bingo
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 [#permalink] New post 03 May 2005, 12:46
i edited it, because it was wrong...

C)...

1) and 2) are each wrong

1) and 2) is sufficient because:

x > 2y
x < y + 2

=>

y + 2 > x > 2y => y - x + 2 > 2y - x => y + 2 > 2y => 2 > y
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 [#permalink] New post 05 May 2005, 08:10
OA is C.

christoph or anyone: could you please explain your soln.

y + 2 > x > 2y => y - x + 2 > 2y - x => y + 2 > 2y => 2 > y

y + 2 > x > 2y : okay

y - x + 2 > 2y - x : Here what happened to the middle " > x > " Have you subtracted "x" on all three terms in that case it should be
y +2- x > x-x > 2y - x , right. So this would become

y +2- x > 0 > 2y - x. Is it fair to say this is equivalent to y - x + 2 > 2y - x leaving behind the middle "> 0 >".

Please correct me if I am wrong.

This is true for any value just not 0 e.g: y +2- x > 23 > 2y - x is equivalent to y +2- x > 2y - x.
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 [#permalink] New post 05 May 2005, 11:09
TS wrote:
OA is C.

christoph or anyone: could you please explain your soln.

y + 2 > x > 2y => y - x + 2 > 2y - x => y + 2 > 2y => 2 > y

y + 2 > x > 2y : okay

y - x + 2 > 2y - x : Here what happened to the middle " > x > " Have you subtracted "x" on all three terms in that case it should be
y +2- x > x-x > 2y - x , right. So this would become

y +2- x > 0 > 2y - x. Is it fair to say this is equivalent to y - x + 2 > 2y - x leaving behind the middle "> 0 >".

Please correct me if I am wrong.

This is true for any value just not 0 e.g: y +2- x > 23 > 2y - x is equivalent to y +2- x > 2y - x.


yes, you are right ! it is true for any value. check this link: http://www.gmatclub.com/phpbb/viewtopic.php?t=14576
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 [#permalink] New post 05 May 2005, 19:12
Thanks christoph for the clarification.
  [#permalink] 05 May 2005, 19:12
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