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# Is x > y ? (1) ax > ay (2) a^2x > a^2 y? Pls

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VP
Joined: 03 Apr 2007
Posts: 1013
Is x > y ? (1) ax > ay (2) a^2x > a^2 y? Pls  [#permalink]

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17 May 2008, 12:22
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18. Is x > y ?
(1) ax > ay
(2) a^2x > a^2 y?

Pls explain

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Director
Joined: 23 Sep 2007
Posts: 693

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17 May 2008, 12:37
goalsnr wrote:
18. Is x > y ?
(1) ax > ay
(2) a^2x > a^2 y?

Pls explain

B

Because in statement 1, if a = a positive number, then x > y
But if a = a negative number, when you divide by a negative, you switch sign,
if a = negative, then x < y
VP
Joined: 03 Apr 2007
Posts: 1013

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18 May 2008, 07:02
gmatnub wrote:
goalsnr wrote:
18. Is x > y ?
(1) ax > ay
(2) a^2x > a^2 y?

Pls explain

B

Because in statement 1, if a = a positive number, then x > y
But if a = a negative number, when you divide by a negative, you switch sign,
if a = negative, then x < y

Stat 1:
(1) ax > ay
a(x-y) >0
a>0 or (x-y)>0
insuff

stat 2:
(2) a^2x > a^2 y?
a^2(x-y) >0

here too a^2 >0 or (x-y) >0
How can we assume a^2 >0 ?
Intern
Joined: 26 Apr 2008
Posts: 42

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18 May 2008, 07:23
1
goalsnr wrote:
gmatnub wrote:
goalsnr wrote:
18. Is x > y ?
(1) ax > ay
(2) a^2x > a^2 y?

Pls explain

B

Because in statement 1, if a = a positive number, then x > y
But if a = a negative number, when you divide by a negative, you switch sign,
if a = negative, then x < y

Stat 1:
(1) ax > ay
a(x-y) >0
a>0 or (x-y)>0
insuff

stat 2:
(2) a^2x > a^2 y?
a^2(x-y) >0

here too a^2 >0 or (x-y) >0
How can we assume a^2 >0 ?

First of all, your conclusion from statement 1 is wrong. The statement 1 gives -
a(x-y) > 0
=> { a>0 AND (x-y)>0 } or { a<0 AND (x-y)<0 }
Insuff

Coming to statement 2, same thing as above. But square of a number cannot be negative.
So (x-y)>0, means x>y
VP
Joined: 03 Apr 2007
Posts: 1013

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18 May 2008, 17:39
1
Thanks Itsme.

The OA is B
Director
Joined: 05 Jan 2008
Posts: 596

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18 May 2008, 20:06
1
B for me.

From Stmt 1, ax>ay -> if a is -ve then x<y, if a is +ve then x>y thus 1 is insufficient

from stmt 2-> a^2 is always positive thus x>y thus B
Non-Human User
Joined: 09 Sep 2013
Posts: 12068
Re: Is x > y ? (1) ax > ay (2) a^2x > a^2 y? Pls  [#permalink]

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18 Oct 2018, 17:15
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Re: Is x > y ? (1) ax > ay (2) a^2x > a^2 y? Pls   [#permalink] 18 Oct 2018, 17:15
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