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Is x < y? (I) 1/x < 1/y (2) x/y < 0

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Is x < y? (I) 1/x < 1/y (2) x/y < 0 [#permalink] New post 05 Jun 2009, 19:23
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A
B
C
D
E

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Is x < y?

(I) 1/x < 1/y

(2) x/y < 0

Last edited by skim on 06 Jun 2009, 06:03, edited 1 time in total.
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Re: Inequalities DS [#permalink] New post 06 Jun 2009, 00:26
I would have preferred 3 different threads for these 3 questions.

Anyway, picking the 1st one.

Is x<y ?
(I) 1/x < 1/y Insufficient
  • when both x and y are +ve
    1/x < 1/y => y < x
  • when both x and y are -ve
    1/x < 1/y => y < x
  • when x is -ve and y is +ve
    1/x < 1/y => y > x
  • when x is +ve and y is -ve
    Scenario not possible

(2) x/y < 0 Insufficient
Implies, both x and y have different signs
still we cannot conclude if x<y or y<x

(3) Together Sufficient
From 2nd we know both x and y have different signs, and
from 1st we know, if both have different signs, then x -ve and y +ve is only possible
=> x<y

Answer should be C
btw, whats the OA
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Re: Inequalities DS [#permalink] New post 06 Jun 2009, 06:02
bigoyal wrote:
I would have preferred 3 different threads for these 3 questions. - [Noted]

Anyway, picking the 1st one.

Is x<y ?
(I) 1/x < 1/y Insufficient
  • when both x and y are +ve
    1/x < 1/y => y < x
  • when both x and y are -ve
    1/x < 1/y => y < x
  • when x is -ve and y is +ve
    1/x < 1/y => y > x
  • when x is +ve and y is -ve
    Scenario not possible

(2) x/y < 0 Insufficient
Implies, both x and y have different signs
still we cannot conclude if x<y or y<x

(3) Together Sufficient
From 2nd we know both x and y have different signs, and
from 1st we know, if both have different signs, then x -ve and y +ve is only possible
=> x<y

Answer should be C
btw, whats the OA


OA is
[Reveal] Spoiler:
C
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Re: Inequalities DS [#permalink] New post 08 Jun 2009, 22:03
1) 1/x-1/y<0 => (y-x)/xy <0 =>{y>x and xy<0} or {y<x and xy>0} insuf
2) x/y<0 => {x>0 and y<0} or {x<0 and y>0} insuf
Both, we can ans YES
Hence C
Re: Inequalities DS   [#permalink] 08 Jun 2009, 22:03
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