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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink]
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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]
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]
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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]
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]
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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]
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
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink]
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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]
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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]
Expert Reply
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]
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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]
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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]
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Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y [#permalink]
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