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Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
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Updated on: 27 Jun 2021, 23:11
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Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
01 Jul 2012, 01:12
01 Jul 2012, 01:12
7
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Expert Reply
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.
_________________
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]
10 Aug 2009, 17:57
10 Aug 2009, 17:57
4
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2
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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
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
General Discussion
Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
10 Aug 2009, 22:09
10 Aug 2009, 22:09
2
Kudos
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
(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]
10 Aug 2009, 22:14
10 Aug 2009, 22:14
B it is ...
by squaring 'a' you are making it +ve ....
by squaring 'a' you are making it +ve ....
Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
08 Oct 2009, 13:04
08 Oct 2009, 13:04
1
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2
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Expert Reply
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.
_________________
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]
08 Oct 2009, 13:07
08 Oct 2009, 13:07
do we not need to consider complex values for a? i mean its not been specified that a is real.
Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
08 Oct 2009, 13:20
08 Oct 2009, 13:20
2
Kudos
Expert Reply
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]
13 Apr 2012, 19:52
13 Apr 2012, 19:52
AshForGMAT wrote:
B it is ...
by squaring 'a' you are making it +ve ....
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
Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
14 Apr 2012, 01:02
14 Apr 2012, 01:02
2
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2
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Expert Reply
Avh86 wrote:
AshForGMAT wrote:
B it is ...
by squaring 'a' you are making it +ve ....
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 ToolKitWelcome 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]
24 Oct 2013, 23:29
24 Oct 2013, 23:29
1
Bookmarks
Bunuel wrote:
Avh86 wrote:
AshForGMAT wrote:
B it is ...
by squaring 'a' you are making it +ve ....
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 ToolKitWelcome 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
Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
25 Oct 2013, 00:55
25 Oct 2013, 00:55
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 ToolKitWelcome 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]
20 Sep 2016, 07:04
20 Sep 2016, 07:04
2
Kudos
4
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Expert Reply
Stiv wrote:
Is x > y?
(1) ax > ay
(2) a^2x > a^2y
(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]
28 May 2018, 19:14
28 May 2018, 19:14
1
Kudos
Expert Reply
shruti.shambhavi wrote:
Is x>y?
A) ax>ay
B) a^2x>a^2y
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
_________________
Re: Is x > y ? (1) ax > ay (2) a^2*x > a^2*y
[#permalink]
10 Jan 2023, 20:09
10 Jan 2023, 20:09
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