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

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

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New post 17 May 2008, 11:22
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18. Is x > y ?
(1) ax > ay
(2) a^2x > a^2 y?

Pls explain

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Re: DS-ax [#permalink]

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New post 17 May 2008, 11:37
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

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Re: DS-ax [#permalink]

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New post 18 May 2008, 06:02
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


GmatNub,can you please explain what is wrong about this approach

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 ?

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Re: DS-ax [#permalink]

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New post 18 May 2008, 06:23
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This post received
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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


GmatNub,can you please explain what is wrong about this approach

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

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Re: DS-ax [#permalink]

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New post 18 May 2008, 16:39
Thanks Itsme.

The OA is B

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Re: DS-ax [#permalink]

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New post 18 May 2008, 19:06
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
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Re: Is x > y ? (1) ax > ay (2) a^2x > a^2 y? Pls [#permalink]

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New post 19 Sep 2017, 08:05
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Re: Is x > y ? (1) ax > ay (2) a^2x > a^2 y? Pls   [#permalink] 19 Sep 2017, 08:05
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