Is |x| = y - z ?

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Is |x| = y - z ?

by jsl » Thu Oct 16, 2008 11:56 pm
Is |x| = y - z ?

1) x + y = z
2) x < 0

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by stop@800 » Fri Oct 17, 2008 12:04 am

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by lunarpower » Fri Oct 17, 2008 2:52 am
statement 1:
we can rephrase this to x = z - y.
there are thus 3 possibilities for the absolute value |x| :
(a) if z - y is positive, then |x| = z - y, and will NOT equal y - z (which is a negative quantity).
(b) if z - y is negative, then |x| = y - z (the opposite of z - y).
(c) if z - y = 0, then |x| equals both y - z and z - y, since each is equal to 0.
TAKEAWAY: when you consider absolute value equations, you'll often do well by considering the different CASES that result from different combinations of signs.
notice that (a) and (b), or (a) and (c), taken together prove that statement 1 is insufficient.

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statement 2:
we don't know anything about y or z, so this statement is insufficient.**
if you must, find cases: say y = 2 and z = 1. if x = -1, then the answer is YES; if x is any negative number other than -1, then the answer is NO.

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together:
if x < 0, then this is case (b) listed above under statement 1.
therefore, the answer to the prompt question is YES.
sufficient.

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**note that, if i were particularly evil, i could craft a statement that doesn't mention all three of x, y, z and yet IS STILL SUFFICIENT.
here's one way i could do that:
(2) y < z
in this case, y - z is negative and therefore CAN'T equal |x| -- no matter what x is -- since |x| must be nonnegative.
so, this statement is a definitive NO, and is thus sufficient even though it doesn't mention x at all.
this is evil, but i see no reason why it wouldn't be on the test.
Ron has been teaching various standardized tests for 20 years.

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by stop@800 » Fri Oct 17, 2008 3:03 am
Thanks ron.

I did the most common mistake.
while solving for B, I kept A also in mind :(

and yes the last case u mentioned (the evil one) is wonderful :)

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by maihuna » Tue Apr 14, 2009 10:24 am
I have a simple logic it seems:

Given : |X| = y-z

if x<0, -x = y-z
if x>0 x = y-z

From 1: -x = y-z we dont know abt sign of x
From 2: We know x<0 so combine will give me Yes

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by mehravikas » Wed Jan 20, 2010 12:24 pm
Sorry for bumping into an old thread. I have a doubt...lets make the question simpler

Is |x| = 3?

1. x = -3
2. x < 0

from the question we get if x > 0 then x = 3 or if x < 0 then x = -3

Statement 1 - tells us that x = -3, why do we have to consider different values of x i.e. +ve or -ve. Isnt this statement sufficient to answer that x would be equal to -3 when x < 0
Statement 2 - if x < 0, then x will be equal to -3

Am I missing something, please explain.

maihuna wrote:I have a simple logic it seems:

Given : |X| = y-z

if x<0, -x = y-z
if x>0 x = y-z

From 1: -x = y-z we dont know abt sign of x
From 2: We know x<0 so combine will give me Yes

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by bigmonkey31 » Thu Mar 04, 2010 12:51 am
Yeah, I have the same question as the above poster. Don't both of these equations point the the negative solution for 'X'???

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by trzask.m » Wed Mar 31, 2010 2:56 pm
I have the same question as the above two posters and that is how I arrived at A for my answer. I wrote out the three cases for the stem, X=Y-Z, X=-y+Z, and 0=Y-Z=-y+Z.

If statement one states that X+Y=Z we can rephrase as X=-y+z, I assumed that this staement meant X was positive and therefore equal to Y-Z. I know that the OA is correct and I'm not arguing with the "bible", I just want to know the flaw in my thinking. Did I use cirular reasoning to deduce my answer?

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by pops » Wed Mar 31, 2010 10:31 pm
trzask.m wrote:I have the same question as the above two posters and that is how I arrived at A for my answer. I wrote out the three cases for the stem, X=Y-Z, X=-y+Z, and 0=Y-Z=-y+Z.

If statement one states that X+Y=Z we can rephrase as X=-y+z, I assumed that this staement meant X was positive and therefore equal to Y-Z. I know that the OA is correct and I'm not arguing with the "bible", I just want to know the flaw in my thinking. Did I use cirular reasoning to deduce my answer?
Is |x| = y - z ?

1) x + y = z
2) x < 0

Hey, how can you assume that X=-y+z means x was positive ?
here, if x is positive then |x|=z-y so |x|=y-z is false
but, if x is negative then |x|=y-z so |x|=y-z is true
i.e. it all depends on sign of x
hence, combining 2. will give a definite answer :)

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by lunarpower » Sat Nov 20, 2010 1:52 am
trzask.m wrote:I have the same question as the above two posters and that is how I arrived at A for my answer. I wrote out the three cases for the stem, X=Y-Z, X=-y+Z, and 0=Y-Z=-y+Z.

If statement one states that X+Y=Z we can rephrase as X=-y+z, I assumed that this staement meant X was positive and therefore equal to Y-Z. I know that the OA is correct and I'm not arguing with the "bible", I just want to know the flaw in my thinking. Did I use cirular reasoning to deduce my answer?
i kindly indicated the flaw in your thinking, in bright red. (:

basically -- at risk of giving advice that is essentially tautological -- you can't assume things that, well, you can't assume.
in this case,
* statement 1 just says "x + y = z"
* you assumed "x + y = z AND x is positive"
when this is spelled out explicitly, i think it should be pretty easy for you to see the problem -- there is no reason to assume that x is positive when you look at the statement.

also, i'm not sure if you're aware of this, but data sufficiency questions are NEVER written in such a way that the two statements contradict each other. therefore, if you think that statement #1 implies that x is positive, then, well, you're wrong ... because statement #2 says that x is negative.

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by lunarpower » Sat Nov 20, 2010 1:52 am
finally --
if anyone still has doubts about the insufficiency of statement #1, just plug in numbers to test it:
if x = 1, y = 2, z = 3, then these values satisfy statement #1, and the answer to the question is NO (since, no, |1| is not equal to 2 - 3).
if x = -1, y = -2, z = -3, then these values also satisfy statement #1, but the answer to the question is YES this time (since |-1| does, in fact, equal -2 - (-3)).

that's one "yes" and one "no", which together make an ironclad "insufficient".
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by gmatusa2010 » Fri Dec 17, 2010 3:57 am
Ron,

Can you discuss this a little further. I'm still confused about what this statement means "Is |X|=y-z?" because if we are given the expression |X|=y-z then I automatically think X=y-z or X=z-y THUS the question becomes "Is X=y-z or X= z-y"? It's obviously the wrong, but can you point out the logic error here?

Ex: Is |X|>1? means X>1 or X<-1? if we can prove one or the other statement then it is sufficient.

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by lunarpower » Fri Dec 17, 2010 8:15 am
gmatusa2010 wrote:Ron,

Can you discuss this a little further. I'm still confused about what this statement means "Is |X|=y-z?" because if we are given the expression |X|=y-z then I automatically think X=y-z or X=z-y THUS the question becomes "Is X=y-z or X= z-y"? It's obviously the wrong, but can you point out the logic error here?

Ex: Is |X|>1? means X>1 or X<-1? if we can prove one or the other statement then it is sufficient.
well ... the only reason that works is because you know that the number 1 is positive in the first place, so you don't have to think about the issue that is causing the complications in this thread.
for instance, if you just go around blindly applying these rules to other cases, you'll get in trouble:
* if you turn "|x| = 1" into "x = 1 or x = -1", then of course this is ok.
* on the other hand, if you try to use the same rule to turn "|x| = -2" into "x = -2 or x = 2", then that won't work.

taking the above examples, note how they would apply to the case of an even simpler equation, such as |x| = Q.
if the number Q happens to the positive (as in the case of 1 above), then you can simply split this into x = Q or x = -Q. on the other hand, if the number Q happens to be negative (as in the case of -2 above), then there are no solutions to this equation at all.
so, if you're given the equation |x| = Q, you can't write a general solution in one piece -- you have to split it up into two cases:
either
(1) Q is positive (or 0), and then x = Q or x = -Q,
or
(2) Q is negative, and there's no solution.

what's going on in the problem above is basically the same thing. just as you can't tell whether Q it is positive or negative in my example, you can't tell which of (y - z) and (z - y) is positive in the example here, and you must know which is which in order to translate the equation away from absolute value. since you don't know which is which at the outset of the problem, your only option is to split it into two cases, just as in the case of "Q" above.
Ron has been teaching various standardized tests for 20 years.

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by thebigkats » Tue Dec 21, 2010 4:07 pm
Somehow I got this practice Q today only and I can see that the thread is rather old. In any case, this is how I solved it:

1. x=z-y ==> x=+ve or x=-ve or X=0

If x=+ve ==> z>y ==> y-z <0 ==> |x| <0 which is not possible
if x=-ve ==> y>z ==> y-z > 0 ==> |x| = y-z
if x=0 ==> z=y ==> |x| = 0

Given that condition #1 fails - So INSUFFICIENT

2. x<0
doesn't tell us anything about y or z

2. both #1 and #2
x is fore sure negative from #2 ==> y >z (from #1) => y-z > 0 ==> |x| = y-z

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by emalby » Sat Mar 26, 2011 1:33 am
Is |x| = y - z ?

1) x + y = z
2) x < 0

I think this method can help:

1) x = z - y => - x = y - z => |x| = - x

therefore if x = 0 then equation |x| = y - z is TRUE, if x is any other number is FALSE since no absolute value of a number x can equal a negative number - x. So 1) is not sufficient because we can have two different cases.

2) is not sufficient alone since we don't have any information on y and z. They can be whatever number, so no way to tell whether they equal x.

1 & 2) is sufficient. 2) tells us that x is lower than 0, so we go in case 1) and se that x=0, which was the only case where equation was valid (TRUE), is excluded since x is lower than 0 based 2). So we have our answer that equation |x| = y - z is FALSE!