f2001290 wrote:is |x| = y-z
(1) x+y = z
(2) x<0

Is it C ?

1 - insufficient. y-z = -x. Now, if x is positive, y-z = -ve and so y-z != |x|.
If x is negative, y-z = +ve and so y-z = |x|

2 - insufficient. Just knowing x<0 does not help.

1 and 2 together sufficient since y-z = -x and x is negative.

and whats problem with A ?

AND what this all to do with subject line? yahoo group?
have you guys started a yahoo group or what?
was not active in sectional sub forums lately.

is |x| = y-z
(1) x+y = z
(2) x<0

Statement II : Clearly insufficient. Does not say anything about y and z...Hence rule out choices B and E

Statement I : x+y = z...thus x = z-y
If z = 2 and y = 2 ; then z-y = 0 and y-z = 0...here IxI = y-z
If z = -4 and y = 6 then z-y = x = -10 and y-z = -10 and here IxI is not equal to -10; hence insufficient

statement I and II : together sufficient

One more for C

|x| = y-z
x= y-z or -x=y-z

1) x = z-y or -x = y-z which is true only in case of x<0
2) insuff

1+2) suff