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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 15:40

1

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

61% (02:22) correct
39% (01:31) wrong based on 699 sessions

I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.

If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of |x|-|y|?

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 17:07

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

xy not equal to 0 means neither of them is zero , the given equation can't hold if both are positive because in that case you can remove the mod sign and it will give you y=0 which is not possible

if both x,y are positive ,x - y = x + y

Same goes for x and y both being negative .

However, one of them can be neg and one positive , x=-2 and y = 1 satisfies the equation and hence ans is E

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 22:47

Expert's post

WholeLottaLove wrote:

Yes, but why does -2|xy| = 2xy?

and is this example posted by Bunuel... 2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

How does |x| - |y| = x+y? wouldn't it be x-y?

|x|-|y| will always be (x-y) ONLY for non-negative values of BOTH x and y. If even one of them is negative, then |x|-|y| will never (x-y). It will be either -x-y (In case x is negative and y positive) OR x-(-y)-->x+y(x positive and y negative).

When y is negative, as |y| is always a non-negative entity, thus we can't write |y| = y. Thus, we attach a negative sign to 'y' to make the entire term (-y) as positive.

Also, even though (x+y) does look like addition of two numbers, it actually isn't, as y is negative. It will be easier to understand this concept if you use valid nos. _________________

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
31 May 2013, 00:12

1

This post received KUDOS

WholeLottaLove wrote:

I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.

If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of |x|-|y|?

----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

x is greater than 0 => \(|x|=x\) y is less than 0 => \(|y|=-y\). You take -y from |y| because y<0 So \(|x|-|y|=x-(-y)=x+y\)

This is why \(|x|-|y|\) becomes \(x+y\). If you want you can try with real numbers, example: \(x=5\)>0 and \(y=-3\)<0 \(|x|-|y|=|5|-|-3|=5-3=2\) \(|x|-|y|=x+y=5-3=2\) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
31 May 2013, 02:05

2

This post received KUDOS

Expert's post

5

This post was BOOKMARKED

WholeLottaLove wrote:

How does x^2-2|xy|+y^2 become x^2+2xy+y^2? What happened to the -2|xy|? Even if xy were negative |xy| will be positive and pos. * neg (In this case, the -2) = negative, right?

Step 2: cancel x^2+y^2 in both sides to get \(-2|xy|=2xy\)

Step 3: reduce by -2 to get \(|xy|=-xy\)

Step 4: apply absolute value property, which says that the absolute value is always non-negative --> \(LHS=|xy|\) is always non negative, thus \(|xy|\geq{0}\), therefore RHS is also non-negative: \(|xy|=-xy\geq{0}\) --> \(-xy\geq{0}\).

Step 5: multiply by -1 and flip the sign of the inequality --> \(xy\leq{0}\)

Step 6: apply info given in the stem --> since given that \(xy\neq{0}\), then \(xy<0\).

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
10 Jul 2013, 09:10

If |x| - |y| = |x+y| and xy does not equal to 0, which of the following must be true?

|x| - |y| = |x+y| |x|-|y| = an absolute value which means that |x|-|y| is positive. Therefore, we can square both sides. (|x| - |y|)^2 = (|x+y|)^2 (|x| - |y|)*(|x| - |y|) = (|x+y|)*(|x+y|) x^2-2(xy|) + y^2 = x^2+2xy+y^2 -2|xy|=2xy (I take it we cannot add -2|xy| to 2xy?) |xy|=-xy -xy must be positive as it is equal to an absolute value. For that to be possible: |xy|=-xy |xy|=-(-xy) |xy|=xy xy=xy

So, xy < 0

(E)

A. x-y > 0 B. x-y < 0 C. x+y > 0 D. xy > 0 E. xy < 0

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
22 Jul 2013, 22:44

its better testing with some smart numbers.obviously will take some seconds more but worth it. As per this question there can be four cases: 1 )+ + 2)+ - 3)- + 4)- -. Check for the validity and you will see that choices 2 and 3 only suffice.So the product is going to be <0 with either case.Hence E

Answer which you 'll derive using smart numbers will be beyond doubt and perfect.Get used to smart numbers in areas like absolute values,percentages,ratios.They help a lot )

Re: PS - Number system [#permalink]
23 Apr 2014, 04:51

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.

Could you please explain following line as above defined in solution

Re: PS - Number system [#permalink]
23 Apr 2014, 06:15

Expert's post

PathFinder007 wrote:

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.

Could you please explain following line as above defined in solution

note that (|x+y|)^2=(x+y)^2

Thanks.

Generally, \(|x|^2=x^2\). For example, \(|-2|^2=4=2^2\) or \(|3|^2=9=3^2\). _________________

Re: PS - Number system [#permalink]
02 May 2014, 22:45

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 Bunnel,

3. --x------0--y----: x<0<y --> |x|-|y|=-x-y and |x+y|=-x-y --> -x-y={-x-y}

could you please clarify above statement. here y is positive so |x+y| should be -x+y

Re: PS - Number system [#permalink]
03 May 2014, 03:22

Expert's post

PathFinder007 wrote:

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 Bunnel,

3. --x------0--y----: x<0<y --> |x|-|y|=-x-y and |x+y|=-x-y --> -x-y={-x-y}

could you please clarify above statement. here y is positive so |x+y| should be -x+y

Please clarify

Thanks.

y is positive but x is negative and x is further from zero than y, so x+y is negative hence |x+y|=-(x+y). For example, x=-5 and y=1 --> x+y=-4=negative. _________________

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
06 May 2014, 20:00

My Analysis : xy # 0 mean x and y might have either positive or negative Numbers ( Integers) Concepts : Absolute value , Modules My Steps : As xy # 0 , So x and y have either positive or negative numbers By Plugging x and y as positive numbers |x| - |y| = |x+y| the Equation never becomes Equal So One Number has to be Negative Ex : x = -6 , y = 3 |-6| - |3| = |-6 + 3| 6 - 3 = |-3 | 3 = 3 So xy = -6*3 = -18

Re: PS - Number system [#permalink]
18 Jun 2014, 05:57

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 \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\)

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 ,

Can you please explain how \(|xy|=-xy\) --> \(xy\leq{0}\),

Re: PS - Number system [#permalink]
18 Jun 2014, 07:13

Expert's post

gauravsoni wrote:

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 \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\)

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 ,

Can you please explain how \(|xy|=-xy\) --> \(xy\leq{0}\),

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
18 Jun 2014, 11:41

Bunuel wrote:

gauravsoni 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 \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\)

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 ,

Can you please explain how \(|xy|=-xy\) --> \(xy\leq{0}\),

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
24 Sep 2015, 19:13

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