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Is |x|*y - x*|y| > x*y, where x and y are non-zero integers

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Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 11 Jun 2016, 20:06
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  95% (hard)

Question Stats:

35% (02:19) correct 65% (01:28) wrong based on 100 sessions

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Is |x|*y - x*|y| > x*y, where x and y are non-zero integers?
(1) y < 0
(2) x*y > 0


Self Made
Slightly tricky

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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 11 Jun 2016, 20:06
OA and OE after some discussion
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 11 Jun 2016, 20:23
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Nice Question!!!

St1: y < 0 --> y is negative
y < 0 and x > 0 --> -xy - xy > -xy? --> -2xy > -xy ? No (-2xy is more negative than -xy)
y < 0 and x < 0 --> -xy + xy > xy --> 0 > xy ? No
Sufficient

St2: xy > 0 --> x and y have same signs --> 0 > xy ? No
Sufficient

Answer: D
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Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 11 Jun 2016, 23:06
St1: y<0

If x>0 then, suppose x=2, y=-3
=> 2.(-3) + 2.3 > 2.(-3)
=> Is 0>-6?, Yes

Now, if x<0 then, suppose x=-2, y=-3
=> 2.(-3) + (-2).3 > (-2).(-3)
=> Is -12>6?, No

Therefore, St.1 is not sufficient

St2: x*y>0

If x,y>0 then, suppose x=2, y=3
=> 2.(3) + 2.3 > 2.(3)
=> Is 12>6?, Yes

If x,y<0 then, suppose x=-2, y=-3
=> 2.(-3) + (-2).3 > (-2).(-3)
=> Is -12>6?, No

Therefore, St.2 is not sufficient

Ans=E
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Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 11 Jun 2016, 23:19
tusharagarwal_02 wrote:
St1: y<0

If x>0 then, suppose x=2, y=-3
=> 2.(-3) + 2.3 > 2.(-3)
=> Is 0>-6?, Yes

Now, if x<0 then, suppose x=-2, y=-3
=>2.(-3) + (-2).3 > (-2).(-3)
=> Is -12>6?, No

Therefore, St.1 is not sufficient

St2: x*y>0

If x,y>0 then, suppose x=2, y=3
=> 2.(3) + 2.3 > 2.(3)
=> Is 12>6?, Yes

If x,y<0 then, suppose x=-2, y=-3
=> 2.(-3) + (-2).3 > (-2).(-3)
=> Is -12>6?, No

Therefore, St.2 is not sufficient

Ans=E


Hi,

Revisit your calculations in the highlighted portions. You have taken '+' in all your calculations instead of '-' sign.
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Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 12 Jun 2016, 02:53
chetan2u wrote:
Is |x|*y - x*|y| > x*y, where x and y are non-zero integers?
(1) y < 0
(2) x*y > 0


Self Made
Slightly tricky



Statement(1): I think,Picking number is the best strategy here,

    If x=1 and y=-1 then -2>-1 ,and if x=-1 and x=-1 then 0>1
    If x=2 and y=-1 then -4>-2 ,and if x=-2 and x=-1 then 0>2.....Sufficient

Statement (2) :From Statement (1) I can say that same signs of x and y make the left hand side equation always 0 and right hand side positive. Sufficient


Correct Answer D
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Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 11 Sep 2016, 10:18
Hello Friends, i got this question right but i couldn't corelate my method with anything given here
(i did similar to what Vyshak's did but without some key steps i think it is difficult to understand). If i am wrong PLEASE correct me. Here it goes

[1] y < 0
[1.a] y < 0 and x > 0
|x|*y - x*|y| = x*(-y) - x*[-(-y)] = -xy - xy = -2xy
x*y = x*(-y) = -xy
-2xy >| -xy

[1.b] y < 0 and x < 0
|x|*y - x*|y| = [-(-x)]*(-y) - (-x)*[-(-y)] = -xy + xy = 0
x*y = (-x)*(-y) = xy
0 >| xy
Thus 1 Suff
[2] x*y > 0 => Both x and y have same sign
[2.a = 1.b] & [2.b] y > 0 and x > 0
|x|*y - x*|y| = x*y - x*y = 0
x*y = xy
0 >| xy
Thus 2 Suff

Hence D
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Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers [#permalink]

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New post 22 Feb 2017, 00:01
why for under (1) in the scenario y<0 & x>0 do you have -xy - xy ? -xy?

wouldn't it be: -xy + xy > xy? ==> 0>xy? ==> maybe/NS?

I understand the rest, but got lost at this part. you would only change the variables inside the abs to negatives, no? Thanks!

Michael

Vyshak wrote:
Nice Question!!!

St1: y < 0 --> y is negative
y < 0 and x > 0 --> -xy - xy > -xy? --> -2xy > -xy ? No (-2xy is more negative than -xy)
y < 0 and x < 0 --> -xy + xy > xy --> 0 > xy ? No
Sufficient

St2: xy > 0 --> x and y have same signs --> 0 > xy ? No
Sufficient

Answer: D
Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers   [#permalink] 22 Feb 2017, 00:01
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