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

Question

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

chetan2u · Answer

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

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

since 2xy>xy.....\(|x|y<x(|y|+y)\)
ans NO
b) case II - both x and y are < 0

therefore \(|x|y<x(|y|+y)\)
ans NO
sufficient

D

Kurtosis · Answer

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

tusharagarwal_02 · Answer

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

Kurtosis · Answer

Hi,

Revisit your calculations in the highlighted portions. You have taken '+' in all your calculations instead of '-' sign.

AbdurRakib · Answer

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

manishtank1988 · Answer

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

mdacosta · Answer

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

vard · Answer

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.

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