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Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Is |x|*y - x*|y| > x*y, where x and y are non-zero integers  [#permalink]

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3
5 00:00

Difficulty:   95% (hard)

Question Stats: 32% (02:38) correct 68% (02:20) wrong based on 136 sessions

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

Slightly tricky

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Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Is |x|*y - x*|y| > x*y, where x and y are non-zero integers  [#permalink]

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OA and OE after some discussion

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

$$|x|*y - x*|y| > x*y..............|x|*y >x*|y| + x*y.............|x|y>x(|y|+y)$$

(1) $$y < 0$$
$$|x|y>x(|y|+y)$$
since y<0
# |x|*y<0
# x(|y|+y)=x*0=0
therefore $$|x|y<x(|y|+y)$$
ans NO
sufficient

(2) $$x*y > 0$$
this means x and y are of same sign
$$|x|y>x(|y|+y)$$
a) case I - both x and y are >0
# |x|y=xy
# x(|y|+y)=2xy
since 2xy>xy.....$$|x|y<x(|y|+y)$$
ans NO
b) case II - both x and y are < 0
# |x|y<0
# x(|y|+y)=x*0=0
therefore $$|x|y<x(|y|+y)$$
ans NO
sufficient

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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1
1
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

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GMAT 1: 710 Q50 V34 Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers  [#permalink]

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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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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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chetan2u wrote:
Is |x|*y - x*|y| > x*y, where x and y are non-zero integers?
(1) y < 0
(2) x*y > 0

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

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Md. Abdur Rakib

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Posts: 160
Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers  [#permalink]

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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

 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
 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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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

Intern  B
Joined: 01 Sep 2017
Posts: 1
Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers  [#permalink]

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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

The question can be re-written as:
is |x|*y - x*|y| - x*y > 0
= |x|*y - x(|y| + y) > 0 ...........(A)

Statement 1: y < 0
therefore |y| = -y (substituting in (A) we get :
is |x|*y - x(-y + y) > 0 ?
= is |x|*y > 0 ?
Since y is negative, |x|y can never be greater than 0. Therefore the statement is sufficient.

Statement 2: x*y > 0
this means either both x and y are positive or both are negative.

Case I: Both are positive.
The question becomes x*y - x(y + y) > 0
= is xy - 2xy>0 or is xy > 2xy?
which is not possible, therefore NO.

Case II: Both are negative.
The question becomes
is -x*y - x(-y + y) > 0 ?
= is -xy > 0? since both x and y are negative, product of -x(positive) and y(negative) will be negative. Therefore the answer is NO.

Statement 2 is also sufficient. Re: Is |x|*y - x*|y| > x*y, where x and y are non-zero integers   [#permalink] 11 Aug 2018, 23:35
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