gmatdriller wrote:I missed this question because I made a wrong assumption; please clarify.

Is |x| = y - z ?

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

my approach:
If actually |x| = y-z
then x = y-z.................subject to x>0
OR x = z-y....................subject to x <0
my assumption is that, the question is asking for the sign of x.

(1) this implies that x < 0, isn't it? ....
(2) x<0.....meaning x=z-y....just as in (1)
I picked D as my answer.
Please correct me.

Well, what if z and y are equal, so that x=0? This particular end case falls outside of you algebraic analysis, and is the exact reason why the answer is C: you need stat. (2) to eliminate this possibility and get a "yes" answer".

Don't rely blindly on algebraic concepts - always be on the look out for weird "end cases" that you haven't thought of.

gmatdriller wrote:I missed this question because I made a wrong assumption; please clarify.

Is |x| = y - z ?

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

Question really is: y-z > 0 or y > Z?
Because |x| is always > 0

From 1) x = z-y; Now if x > 0 --> z>y BUT if x <0 then z < y therefore it is ambiguous; NOT SUFFI

From 2) x < 0 but |x| is always >0 not sufficient to make any conclusion

From 1) and 2) if x < 0 then we can conclude that z-y <0 or y > z Thus stmnt 1 and stmnt 2 together are suffi.

Ans:C

Geva@MasterGMAT wrote:

gmatdriller wrote:I missed this question because I made a wrong assumption; please clarify.

Is |x| = y - z ?

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

my approach:
If actually |x| = y-z
then x = y-z.................subject to x>0
OR x = z-y....................subject to x <0
my assumption is that, the question is asking for the sign of x.

(1) this implies that x < 0, isn't it? ....
(2) x<0.....meaning x=z-y....just as in (1)
I picked D as my answer.
Please correct me.

Well, what if z and y are equal, so that x=0? This particular end case falls outside of you algebraic analysis, and is the exact reason why the answer is C: you need stat. (2) to eliminate this possibility and get a "yes" answer".

Don't rely blindly on algebraic concepts - always be on the look out for weird "end cases" that you haven't thought of.

can i solve the problem as following
|x|=y-z
if x>0 so x=y-z
if x<0 x= z-y
state 1 and 2 gives only 1/2 sufficient condistion
(1+2) is sufficient. so pick C
can i solve like that?

+1 thanks to you Geva for the explanations.


we are asked, |x| = y-z ?
using number line,
consider: x(0) + y(0) = z(0)...answer is YES
x(2) + y(2) = z(4)...................answer is NO
x(-2) + y(-2) = z(-4) ..............answer is NO
INSUFFICIENT

(2) SAYS nothing about y and z,; so, INSUFFICIENT

combining, we know that x and y have opposite signs.
e.g x(-2) + y(2) = z(0)
we answer YES, since |-2| = 2 - 0

also, consider x(-2) + y(-2) = z(-4)
we answer YES, since |-2| = -2 - (-4)

also, x(-2) + y(0) = z(-2)
we answer YES, since |-2| = 0 - (-2)
SUFFICIENT

I found this option time wasting as I need to consider whether the
variables are either (+ve) - (+ve); (+ve) - (-ve); (-ve) - (-ve) or (-ve) - (+ve)

what is the shortest method, or rather, how do we approach it algebra?