is |x-y| > |x| - |y|?

[inetcleopatra]

Is |x-y| > |x| - |y|?

1) y< x
2) xy< 0

Anyone know how to solve this algebraically?
Source: GMATPREP1

OA: B

[sanidhya510]

is OA : C

1) y< x

y could be -ive and + ive. insufficient on plugging in.

2) xy< 0

x or y is negative.

Combined y is -ive and x +ive. and thus proven by pluggging in figures.

[klinh.vu]

Hello !

Is C the right answer ?

I would have said B.

ex: x and y have different signs

x= -1 and y= 3
then [-1-3] = 4 and [-1] - [3] = -2
So statement is true.

or
x= 3 and y=-1
then [3+1]= 4 and [3] - [-1] = 2
So statement is true also.

What do you think ? is there another way to approach that problem ?

Thanks.

[RonPurewal]

Anyone know how to solve this algebraically?

heh.
actually, you can't.

seriously -- a HUGE part of the mission of this exam is to create problems/equations/inequalities that CANNOT be solved with routine algebra.
remember that this is a standardized test. one of the key directives of standardized test ands, especially in math, is to create problems that are deliberately UNLIKE "school" problems.

TAKEAWAY:
if you absolutely can't get anywhere with algebra on a given equation, it's possible that the equation CAN'T be solved with algebra.
if this happens, you should QUICKLY ABANDON THE ALGEBRA and go with PLUG-IN METHODS.

--

LET'S PLUG IN

statement (1)
let's try x = 2, y = 1
is |2 - 1| > |2| - |1| ?
no.
let's try x = -1, y = -2
is |-1 - (-2)| > |-1| - |-2| ?
yes.
insufficient.

statement (2)
this means we have to pick OPPOSITE SIGNS. therefore, there are basically 6 cases to try:
* x is negative "bigger", y is positive "smaller"
* x is negative, y is positive, same magnitude
* x is negative "smaller", y is positive "bigger"
* x is positive "bigger", y is negative "smaller"
* x is positive, y is negative, same magnitude
* x is positive "smaller", y is negative "bigger"

if you try all of these --
-2, 1 --> is |-3| > 2 - 1? YES
-1, 1 --> is |-2| > 0? YES
-1, 2 --> is |-3| > 1 - 2? YES
2, -1 --> is |3| > 2 - 1? YES
1, -1 --> is |2| > 0? YES
1, -2 --> is |3| > 1 - 2? YES
the pattern is pretty clear: in each of these cases, the two numbers' magnitudes are working together on the left, but against each other on the right. therefore, the left-hand side is always going to be bigger.
sufficient.

[RonPurewal]

incidentally, if you're the type who is absolutely obsessed with having a "textbook solution" for EVERYTHING under the sun, then
"|x - y| > |x| - |y|"

can actually be rephrased to
"EITHER x and y have opposite signs, OR x = 0 and y ≠ 0, OR x and y have the same sign but x has a smaller magnitude than y"

...and that is absolutely the best you're going to do with that inequality.

--

alternatively, you can break statement (2) into two cases:
(a) x is positive, y is negative.
in this case, let's call x = "A" and y = "-B", so that A and B are both positive.
then this becomes
is |A - (-B)| > |A| - |-B| ?
--> is |A + B| > A - B ?
--> is A + B > A - B ?
YES

(b) x is negative, y is positive.
in this case, let's call x = "-A" and y = "B", so that A and B are both positive.
then this becomes
is |-A - B| > |-A| - |B| ?
--> is |-A - B| > A - B ?
--> is A + B > A - B ?
YES

SUFFICIENT

of course, any of these approaches is INSANELY difficult.

my goal here, in providing algebraic solutions, is to prove the value of NOT using algebra on this problem!

there are some problems on which plugging-in-numbers is the ONLY sensible method, unless you are a mathematician by trade.

another such problem is Problem Solving #164 in the quant OG supplement (either edition).
on that problem, the algebra solution is INSANE. i have a grad-school mathematics education, and there's no way i could come up with that sort of solution in two minutes.
on the other hand, if you just plug the answer choices in and see whether they work, it reduces to grade-school arithmetic.

don't be married to "textbook" solutions. if they don't work, start plugging in numbers!

[kevinmarmstrong]

Looking at examples is definitely the easiest approach. If you like to think of absolute values as distances, the following may help:

if x and y have opposite signs, x and - y have the same sign. x + (- y) will thus be farther from 0 than is x

Thus |x - y| > |x| > |x| - |y| since |y| > 0
SUFF

[akhp77]

Statement 1:
x > y
x = 5, y = 1, |x-y| = 4, |x| - |y| = 4 violated
x = 5, y = -1, |x-y| = 6, |x| - |y| = 4 Not violated
x = -1, y = -5, |x-y| = 4, |x| - |y| = -4 Not voilated

Not sufficient

Statement 2:
xy < 0
x = 5, y = -1, |x-y| = 6, |x| - |y| = 4 Not Violated
x = 5, y = -10, |x-y| = 15, |x| - |y| = -5 Not Violated
x = -1, y = 5, |x-y| = 6, |x| - |y| = -4 Not Violated
x = -10, y = 5, |x-y| = 15, |x| - |y| = 5 Not Violated

Only one opinion

sufficient

Ans B

[StaceyKoprince]

nice work, guys!