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Re: Is xy > y^2? (1) yx < 1 (2) |x| > y [#permalink]
1
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The expression is = \(xy > y^2\)

Statement 1

xy < 1

if x = 1 y= -2 => -2< 1

-2 is not greater than 4..............Expression doesn't hold good

if x =1/2; y = 1/3 => xy = 1/6 < 1

but 1/6 is greater than 1/9 ............Expression holds good.

Statement 1 is insufficient

Statement 2

|x| > y

=> x = -2; y =1

xy = -2 > 1 ...................Expression doesn't hold good

If x = 2; y = 1

xy = 2 > 1 ....................Expression holds good

Statement 2 is insufficient


Statement 1 + Statement 2

xy < 1 & |x| > y

If x =1; y = 1/2 => xy = 1/2 > 1/4 ..............Expression holds good

If x = -2; y = 1=> xy = -2 > 1 ..................Expression doesn't hold good

Both statements are also insufficient


Hence OA should be E
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Re: Is xy > y^2? (1) yx < 1 (2) |x| > y [#permalink]
Archit3110 wrote:
is xy>y^2
y*(x-y)>0

is y>0 and x>y
#1
yx < 1
possible when either of x or y is -ve
insufficient
#2
lxl>y
y is always +ve or 0 , but x can be +ve or -ve
we get yes & no to target insufficient
from 1 &2
y is +ve or y = 0 and x is -ve
in that case y*( x-y) can be >0 or =0
insufficient
option E is correct

Bunuel wrote:
Is xy > y^2 ?

(1) yx < 1

(2) |x| > y



Hey Archit3110

In statement 1, it need not be -ve, right?

If x & y are fractions then also xy will satisfy the condition. It's not given that x & y are integers.

Moreover, in statement 2 y can be -ve when x = 0. Why did you write that y is always +ve or 0? Just wanted to confirm the reasoning here.
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Re: Is xy > y^2? (1) yx < 1 (2) |x| > y [#permalink]
sj296
Well for DS question the objective while solving should be try and make statement stand invalid/insufficient for target..
Which has been done for both cases in my solution..
I could make the #2 stand invalid by the reasoning I mentioned so I did not pick y as -ve also x is given in mod so it's better to test cases Of non zero inside mod..

Your reasoning for #1 is right as well, fraction values can also be taken to make statement invalid but then testing it would be a task for target .. so go for whats quick and gets you correct answer. :)


sj296 wrote:
Archit3110 wrote:
is xy>y^2
y*(x-y)>0

is y>0 and x>y
#1
yx < 1
possible when either of x or y is -ve
insufficient
#2
lxl>y
y is always +ve or 0 , but x can be +ve or -ve
we get yes & no to target insufficient
from 1 &2
y is +ve or y = 0 and x is -ve
in that case y*( x-y) can be >0 or =0
insufficient
option E is correct

Bunuel wrote:
Is xy > y^2 ?

(1) yx < 1

(2) |x| > y



Hey Archit3110

In statement 1, it need not be -ve, right?

If x & y are fractions then also xy will satisfy the condition. It's not given that x & y are integers.

Moreover, in statement 2 y can be -ve when x = 0. Why did you write that y is always +ve or 0? Just wanted to confirm the reasoning here.


Posted from my mobile device
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Re: Is xy > y^2? (1) yx < 1 (2) |x| > y [#permalink]
Archit3110 thanks, man. Agree with 'whats quick and gets you correct answer'. :thumbsup:
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Re: Is xy > y^2? (1) yx < 1 (2) |x| > y [#permalink]
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Re: Is xy > y^2? (1) yx < 1 (2) |x| > y [#permalink]
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