## Is absvalue (x - y) > absvalue(x) - absvalue(y)?

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Anne1276

### Is absvalue (x - y) > absvalue(x) - absvalue(y)?

From GMAT Prep Practice CAT:

Is absvalue (x - y) > absvalue(x) - absvalue(y)?

1) y<x

2) XY< 0

I have a tough time with absolute value problems. How do you approach this - and other, similar absolute value problems? Thanks.
Jeff

### Absolute value problems

Is abs(x-y)> abs(x) - abs(y)

the key to this is a good reprasing of the problem. When could this be true? Only in two cases 1) X and y have opposite signs or 2) x>y and they are both negative

Example of the first case: x=5,y=-2. Then abs(x-y) = 7 and the abs(x)-abs(y) =3.
Example of the second case: x=-3, y=-7 then abs(x-y) =4 and the abs(x)-abs(4) =-4

So let's consider each statement in turn:

1) y<x. Does this meet either of the two conditions above? It could meet either one, but doesn't have to, so 1) is insufficient and you can rule out answer A and D

2) xy<0. This tells you that they have different signs (one positive, one negative) so this meets the first condition we discussed above and 2) is sufficient by itself so the answer is B.

Jeff
laiusergiu
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### Re: Is absvalue (x - y) > absvalue(x) - absvalue(y)?

I am never comfortable with solutions that resort to number picking.

Let's start with the fact that |a|+|b|>= |a+b|. This is always the case, with equality occurring when a>=0 and b>=0.

Our question asks us to determine whether |x-y|>|x|-|y|.
|x-y|>|x|-|y| rearrange the terms
|x-y|+|y|>|x| expand x
|x-y|+|y|>|x-y+y| we are now in the |a|+|b|>= |a+b| scenario because the sum of the expressions within the individual absolute values equals the expression under the collective absolute value. Therefore we conclude that this inequality always holds weakly (i.e. >=), with equality occurring when the expressions within the individual absolute values are non-negative (i.e. x-y>=0 and y>=0). In other words
CASE 1: if x>=y>=0 then the left and right sides are equal.
CASE 2: for all other relations between x, y, and 0, the left side will be strictly greater than the right side

Conceptually, this question amounts to us deciding whether we have enough information to unequivocally determine in which of the 2 possible scenarios we are: 1) equality of the two sides or 2) left side is strictly greater than the right side.

Now, let's go through the answer choices:
(1) y<x
This is not enough for us to know whether x>=y>=0 holds. If we also knew that y>=0 then we'd have enough information to state that we are in CASE 1 (i.e. the two sides are equal). In the absence of that piece of information, however, we cannot determine whether we're in CASE 1 or CASE 2.

(2) xy<0
This tells us that x and y are of opposite signs, which implies that x>=y>=0 does not hold, and that puts us in CASE 2 and allows us to conclude that the left side is unequivocally greater than the right side. To be more specific, this allows us to conclude that:
|x-y|+|y|>|x-y+y|
therefore
|x-y|+|y|>|x|
or if you prefer
|x-y|>|x|-|y|

Cheers,
Sergiu
messi10
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### Re: Is absvalue (x - y) > absvalue(x) - absvalue(y)?

Hi,

That is a good solution.

However, you mention that you spent some time coming up with this. On the GMAT, if a solution is not found within 2 mins, then we at risk of messing up the timing. Unless this is one of the last few questions and you have gained time over the course of the quant section, I don't see whether this is as a practical solution.

But you may be good at this so I am sure this is a case of personal opinion. However, you will find that picking numbers on these kind of absolute value and inequality problems is the quicker method.

Regards

Sunil
RonPurewal
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### Re: Is absvalue (x - y) > absvalue(x) - absvalue(y)?

varun_783 wrote:But you may be good at this so I am sure this is a case of personal opinion. However, you will find that picking numbers on these kind of absolute value and inequality problems is the quicker method.

yes.
you should definitely not hesitate to pick numbers if you find that using theory is just not working for you.

remember, the issue is not trying to figure out ahead of time what will be fastest -- the issue is just trying to collect as many methods as possible. this way, if you are stuck on any one particular method, you can just say "oh well", abandon that method, and start trying something else.
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### Re: Is absvalue (x - y) > absvalue(x) - absvalue(y)?

Hi. Though I am comfortable with the above mentioned solutions, I want to know if my approach is correct as well.

Is |x-y| > |x| - |y| ?
Squaring both sides,
(|x-y|)^2 > (|x| - |y|)^2
x^2 + y^2 - 2xy > |x|^2 + |y|^2 - 2|x||y|
As |x|^2 = x^2 and |y|^2 = y^2, the equation finally comes down to,
Is -2xy > -2 |x||y| ?

Statement (1) : Insufficient as we do not know the values/signs of x and y.

Statement (2) :
If xy < 0, then -2xy is > 0 (xy is negative and multiplying it with -2 results in a positive value).
But -2|x||y| is actually negative always. Hence, statement 2 is sufficient as -2xy > -2|x||y| when xy < 0.

Appreciate your help ! Cheers !
tim
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### Re: Is absvalue (x - y) > absvalue(x) - absvalue(y)?

no. squaring both sides will not work unless you can be sure both sides are positive to begin with..
Tim Sanders
Manhattan GMAT Instructor