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

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Jivana
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Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   Updated on: 28 Jun 2021, 00:11
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Is \(x > y\) ?

(1) \(ax > ay\)
(2) \(a^2 * x > a^2 * y\)
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Official Answer
B

Originally posted by Jivana on 10 Aug 2009, 18:21.
Last edited by Bunuel on 28 Jun 2021, 00:11, edited 2 times in total.
Edited the question and added the OA
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Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   01 Jul 2012, 02:12
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Is x>y?

Question: is \(x-y> 0\)

(1) \(ax>ay\) --> \(a(x-y)>0\), two cases:
A. \(a>0\) and \(x-y>0\);
OR
B. \(a<0\) and \(x-y<0\);
Depending on \(a\), \(x-y\) may or may not be greater than zero. Not sufficient.

(2) \(a^2x > a^2y\). Since \(a^2>0\) (\(a^2\) can not be zero, because if \(a=0\), then the inequality won't hold true) we can safely reduce by it: \(x>y\). Sufficient.

Answer: B.

Hope it's clear.
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   10 Aug 2009, 18:57
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1. ax > ay. The inequality between x and y depends on a, which can be either positive or negative.

2. a^2.x > a^2.y
Though a is positive or negative a^2 will always be positive.
So we can conclude x > y.
2 alone is sufficient, where as 1 alone is not sufficient.
So answer is B
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   10 Aug 2009, 23:09
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Is x > y ?

(1) ax > ay
(2) a^2 . x > a^2 . y

from 1

ax-ay>0

a(x-y)>0 we dont know whether a is -ve or +ve...insuff

from 2

a^2(x-y)>0 a^2 is +ve thus x-y is +ve thus x>y...suff

B
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   10 Aug 2009, 23:14
B it is ...

by squaring 'a' you are making it +ve ....
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   08 Oct 2009, 14:04
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Is x > y ?

Is x > y? --> is x - y > 0?

(1) ax > ay --> ax-ay>0 --> a(x-y)>0 this equation holds true when both a and (x-y) has the same sign, meaning that they are both positive or both negative. So x-y can be negative or positive. Not sufficient.

(2) (a^2)*x > (a^2)*y --> a^2x-a^2y>0 -> a^2(x-y)>0 -> as a^2 is always positive (it cannot be zero as equation>0) --> (x-y) also must be positive for equation to hold true. Hence x-y>0. Sufficient

Answer: B.
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   08 Oct 2009, 14:07
do we not need to consider complex values for a? i mean its not been specified that a is real.
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   08 Oct 2009, 14:20
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jax91 wrote:
do we not need to consider complex values for a? i mean its not been specified that a is real.


On the GMAT all numbers are real, so x in even power is greater than or equal to zero.
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   13 Apr 2012, 20:52
AshForGMAT wrote:
B it is ...

by squaring 'a' you are making it +ve ....



What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   14 Apr 2012, 02:02
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Avh86 wrote:
AshForGMAT wrote:
B it is ...

by squaring 'a' you are making it +ve ....



What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit


Welcome to GMAT Club. Below is an answer to your question.

First of all, the GMAT is dealing only with Real Numbers, so even roots from negative numbers are not defined. So, \(a^{even}\geq{0}\).

Is x>y?

Question: is \(x-y> 0\)

(1) \(ax>ay\) --> \(a(x-y)>0\), two cases:
A. \(a>0\) and \(x-y>0\);
OR
B. \(a<0\) and \(x-y<0\);
Depending on \(a\), \(x-y\) may or may not be greater than zero. Not sufficient.

(2) \(a^2x > a^2y\) --> \(a^2(x-y)>0\) --> since \(a^2>0\) (\(a^2\) cannot be zero, since the product is more than zero) we can safely reduce by it: \(x-y>0\). Sufficient.

Answer: B.

Hope it's clear.
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   25 Oct 2013, 00:29
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Bunuel wrote:
Avh86 wrote:
AshForGMAT wrote:
B it is ...

by squaring 'a' you are making it +ve ....



What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit


Welcome to GMAT Club. Below is an answer to your question.


First of all, the GMAT is dealing only with Real Numbers, so even roots from negative numbers are not defined. So, \(a^{even}\geq{0}\).

Is x>y?

Question: is \(x-y> 0\)

(1) \(ax>ay\) --> \(a(x-y)>0\), two cases:
A. \(a>0\) and \(x-y>0\);
OR
B. \(a<0\) and \(x-y<0\);
Depending on \(a\), \(x-y\) may or may not be greater than zero. Not sufficient.

(2) \(a^2x > a^2y\) --> \(a^2(x-y)>0\) --> since \(a^2>0\) (\(a^2\) can not be zero, since the product is more than zero) we can safely reduce by it: \(x-y>0\). Sufficient.

Answer: B.

Hope it's clear.



Hi Bunuel

What if question is

Is x = y

A) ax = bx
B ) a^2x = a^2y

Would then the answer be D

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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   25 Oct 2013, 01:55
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DivyanshuRohatgi wrote:
Bunuel wrote:
Avh86 wrote:


What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit


Welcome to GMAT Club. Below is an answer to your question.


First of all, the GMAT is dealing only with Real Numbers, so even roots from negative numbers are not defined. So, \(a^{even}\geq{0}\).

Is x>y?

Question: is \(x-y> 0\)

(1) \(ax>ay\) --> \(a(x-y)>0\), two cases:
A. \(a>0\) and \(x-y>0\);
OR
B. \(a<0\) and \(x-y<0\);
Depending on \(a\), \(x-y\) may or may not be greater than zero. Not sufficient.

(2) \(a^2x > a^2y\) --> \(a^2(x-y)>0\) --> since \(a^2>0\) (\(a^2\) can not be zero, since the product is more than zero) we can safely reduce by it: \(x-y>0\). Sufficient.

Answer: B.

Hope it's clear.



Hi Bunuel

What if question is

Is x = y

A) ax = bx
B ) a^2x = a^2y

Would then the answer be D

Regards


No, in this case the answer would be E. Consider: x=y=0 or a=b=0.
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   20 Sep 2016, 08:04
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Stiv wrote:
Is x > y?

(1) ax > ay
(2) a^2x > a^2y


Responding to a pm:

Is x > y?

(1) ax > ay

We don't know whether a is positive or negative.

Say a is positive.
Divide both sides by a. You get x > y

Say a is negative.
Divide both sides by a. You get x < y (the inequality sign will flip)

Is x > y? We cannot say. Not sufficient.

(2) a^2x > a^2y
a^2 cannot be negative since its the square of a real number. a^2 must be positive only.
So when you divide both sides by a^2, you get x > y
Sufficient alone.

Answer (B)
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   28 May 2018, 20:14
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shruti.shambhavi wrote:
Is x>y?

A) ax>ay
B) a^2x>a^2y



A) ax>ay...
If a is positive, yes otherwise no
Insufficient
B) \(a^2x>a^2y....x>y\)
Sufficient

B
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink] New post   10 Jan 2023, 21:09
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink]
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