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If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
 13 Jul 2010, 23:00
Question Stats:
57% (02:25) correct
42% (01:41) wrong based on 50 sessions
If |x| - |y| = |x+y| and xy does not equal to 0, which of the following must be true? A. x-y > 0 B. x-y < 0 C. x+y > 0 D. xy > 0 E. xy < 0
Last edited by Bunuel on 04 Dec 2012, 02:11, edited 1 time in total.
Renamed the topic and edited the question.
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Re: PS - Number system [#permalink]
14 Jul 2010, 05:56
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divakarbio7 wrote: if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0 |x|-|y|=|x+y| --> square both sides --> (|x|-|y|)^2=(|x+y|)^2 --> note that (|x+y|)^2=(x+y)^2 --> (|x|-|y|)^2=(x+y)^2 --> x^2-2|xy|+y^2=x^2+2xy+y^2 --> |xy|=-xy --> xy\leq{0}, but as given that xy\neq{0}, then xy<0. Answer: E. Another way:Right hand side, |x+y|, is an absolute value, which is always non-negative, but as xy\neq{0}, then in this case it's positive --> RHS=|x+y|>0. So LHS must also be more than zero |x|-|y|>0, or |x|>|y|. So we can have following 4 scenarios: 1. ------0--y----x--: 0<y<x --> |x|-|y|=x-y and |x+y|=x+y --> x-y\neq{x+y}. Not correct. 2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct. 3. --x------0--y----: x<0<y --> |x|-|y|=-x-y and |x+y|=-x-y --> -x-y={-x-y}. Correct. 4. --x----y--0------: x<y<0 --> |x|-|y|=-x+y and |x+y|=-x-y --> -x+y\neq{-x-y}. Not correct. So we have that either y<0<x (case 2) or x<0<y (case 3) --> x and y have opposite signs --> xy<0. Answer: E. Hope it helps.
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Re: PS - Number system [#permalink]
15 Jul 2010, 05:12
divakarbio7 wrote: if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0 Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E.
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Re: PS - Number system [#permalink]
01 Sep 2010, 05:03
SnehaC wrote: divakarbio7 wrote: if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0 Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E. the only way lxl - lyl can equal lx+yl is when one number is positive and one number is negative or both are zero. So then xy must be negative. I think I read if somewhere in Bunuel's post
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If |x| - |y| = |x+y| and xy does not equal to 0, which of the fo [#permalink]
03 Dec 2012, 20:27
When you see |x| or absolute values being tested in the GMAT, this means testing
(1) positive and negative signs
(2) zeroes
(3) nature of |x| and |y| against each other
Looking at the equation we know |x| - |y| = |x + y| where xy is not 0.
(1) We figure that |x| and |y| are non-zeroes
(2) We figure that |x| - |y| > 0. This means |x| > |y|
Now, we figure the signs of the two variables by lining up possibilities
(1) both negative ==> x=-5,y=-4 ==> |-5| - |-4| = |-9| FALSE
(2) both positive ==> x=5,y=4 ==> |5| - |4| = |5 + 4| FALSE
(3) x is positive, y is negative ==> |5| - |-4| = |5-4| TRUE
(4) y is positive, x is negative ==> |-5| - |4| = |-5+4| TRUE
Now, let's find the answer
A) x-y > 0 ==> -5-(-4) = -1 FALSE
E) xy <0 ==> This is exactly what we need. x and y have both different signs.
Answer: E
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Re: I need a strategy for this one. [#permalink]
04 Dec 2012, 00:46
I think the best strategy is to square up the equation.
|x| - |y| = |x+y|
> (|x| - |y|)^2 = (|x+y|)^2
> x^2 + y^2 - 2|x|.|y| = x^2 + y^2 + 2xy
> |x|.|y| = - xy
'Cause xy#0 and |x|.|y|>=0 --> xy<0
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
06 Jan 2013, 19:22
this is specific to Eden,
For your rule. what if X = -1 and Y = 2 then it would be l - 1 l - l 2 l = l -1 + 2 l ... 1 - 2 = 1 incorrect. is this correct?
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Re: PS - Number system [#permalink]
10 Feb 2013, 10:31
Bunuel wrote: divakarbio7 wrote: if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0 |x|-|y|=|x+y| --> square both sides --> (|x|-|y|)^2=(|x+y|)^2 --> note that (|x+y|)^2=(x+y)^2 --> (|x|-|y|)^2=(x+y)^2 --> x^2-2|xy|+y^2=x^2+2xy+y^2 --> |xy|=-xy --> xy\leq{0}, but as given that xy\neq{0}, then xy<0. Answer: E. Another way:Right hand side, |x+y|, is an absolute value, which is always non-negative, but as xy\neq{0}, then in this case it's positive --> RHS=|x+y|>0. So LHS must also be more than zero |x|-|y|>0, or |x|>|y|. So we can have following 4 scenarios: 1. ------0--y----x--: 0<y<x --> |x|-|y|=x-y and |x+y|=x+y --> x-y\neq{x+y}. Not correct. 2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct. 3. --x------0--y----: x<0<y --> |x|-|y|=-x-y and |x+y|=-x-y --> -x-y={-x-y}. Correct. 4. --x----y--0------: x<y<0 --> |x|-|y|=-x+y and |x+y|=-x-y --> -x+y\neq{-x-y}. Not correct. So we have that either y<0<x (case 2) or x<0<y (case 3) --> x and y have opposite signs --> xy<0. Answer: E. Hope it helps. Hi Bunuel, I thought an absolute value of a product can never be a negative. Could you please explain in your equation how did you progress after getting this negative?
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
10 Feb 2013, 18:03
Basically, remember in these exercises to do square root in both sides to simplify: (|x+y|)^2 = (x+y)^2 or (|x|)^2 = x^2 And also remember that: |x||y|=|xy| Therefore, with: -|x||y| = xy I would say: xy is always equal to something negative. Solution: E.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
12 Mar 2013, 05:44
Let us separate LHS and RHS :
1] |x| - |y| =0 It follows the curve x=y in 1st and 2nd quadrant
2] |x+y|=0. It follows he curve x=-y in 4rth and 2nd quadrant
The common set is 2nd quadrant which implies E with the exception of origin as xy<>0
Am I wrong in my interpretation ? But this is the way I visualize the problem.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
09 May 2013, 07:15
Interpreting the given
the distance between x and 0 is larger than that between y and 0 and this difference in distance of each of x and y from zero is equal to the distance between x and -y
draw this on a number line
..................x..............-y...............0.............y
this means that x and y has to be on opposite sides from zero , i.e. different signs thus xy<0
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
12 May 2013, 07:56
can we do the following
Generally /x/ - /-y/ <= /x-(-y)/ and the equality hold true only when -xy>o or xy<0 and /x/>/-y/ or /x/>/y/
given xy not = 0 , and given /x/-/y/ = /x+y/ , i.e. /x/ - /-y/ = /x-(-y)/ therefore /x/> /-y/ and -xy>o ,i.e xy<0
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of
[#permalink]
12 May 2013, 07:56
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