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Re: Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
shridhar786 wrote:
Is xy > x^2 ?

(1) y > x

(2) y > 0


Stem: xy > x^2

Statement 1. y>x
y x
3 2 True
2 -3 False

Statement 2. y>0
y x
3 2 True
2 -3 False

1+2
y x
3 2 True
2 -3 False

E, Insufficient.
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Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
rishabhshreyas14 wrote:
xy > x^2

Can't we rewrite this statement as xy>xx -->>>> (dividing by x both sides) y>x ?


rishabhshreyas14
We can't; check the video



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Re: Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
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rishabhshreyas14 wrote:
xy > x^2

Can't we rewrite this statement as xy>xx -->>>> (dividing by x both sides) y>x ?



rishabhshreyas14, I believe the video below is the one LeoN88 meant to reference:
The part about dividing by variables starts around 2:30

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Re: Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
GMATPrepNow wrote:
rishabhshreyas14 wrote:
xy > x^2

Can't we rewrite this statement as xy>xx -->>>> (dividing by x both sides) y>x ?



rishabhshreyas14, I believe the video below is the one LeoN88 meant to reference:
The part about dividing by variables starts around 2:30


I HAVE A POSSIBLY SIMPLER APPROACH IF IT STRIKES YOU

WE CAN REWRITE THE TERMS I SUPPOSE LIKE THIS
x(y-x)>0

what does this mean?
neg x neg
or
pos x pos

if x = -ve
y= smaller negative: then yes

if x= +ve
y= bigger positive: for inequality to be true

looking at the statements, y>x : we can get both no/yes
y>0 again two cases

combining: we still cannot say about x.
E
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Re: Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
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shridhar786 wrote:
Is xy > x^2 ?

(1) y > x

(2) y > 0



Can anyone tell me if I am doing the right way ?

given
xy > x^2
xy-x^2>0
x(y-x) >0
there exist two cases now for the question now
either x > 0
or y-x >0 or y>x

Now lets look at the statements

1) says y>x but doesn't tell us anything about x>0 or x <0
therefore not sufficient

2) says y>0 but we don't know if x > 0 or not..
therefore, not sufficient

Combining 1 and 2
y>x from 1
and y> 0 from 2
but what about x > 0 ?

therefore, not sufficient
hence E
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Re: Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
Expert Reply
shridhar786 wrote:
Is xy > x^2 ?

(1) y > x

(2) y > 0



Do we know whether x is positive or negative?…..No
So, bring it to one side

\(xy-x^2>0……….x(y-x)>0\)
two cases
If x>0, then y>x. => y>x>0
If x<0, then y<x. => y<x<0

(1) y>x
Nothing about x.

(2) y>0
Nothing about x


Combined
Nothing about x.
We still do not know whether x is positive or negative.

E
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Re: Is xy > x^2 ? (1) y > x (2) y > 0 [#permalink]
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