Interesting problem - does anyone have an easier way to do it?
If zy < xy < 0, is |x-z| + |x| = |z|?
(1) z<x
(2) y>0
(1) Since z < x, then |x-z| = x-z
And since zy < xy < 0 and z < x, then x and z are negative (they must be of opposite sign of y since their product is less than zero, and knowing that z < x, then y must be positive)
so |x| = -x and |z| = -z
and the equality is therefore not correct x-z-x EQUAL to -z
So we can answer the question
SUFFICIENT
(2) y>0, then z and x must be negative for their product to be negative
And since zy < xy, than z < x (both negative) so their absolute values is their opposite.
In this case, we have: |x-z| = x-z, |x| = -x and |z| = -z
so, |x-z| + |x| = x-z-x EQUAL to -z
SUFFICIENT
So answer is D
If zy < xy < 0, is |x-z| + |x| = |z|?
(1) z<x
(2) y>0
(1) Since z < x, then |x-z| = x-z
And since zy < xy < 0 and z < x, then x and z are negative (they must be of opposite sign of y since their product is less than zero, and knowing that z < x, then y must be positive)
so |x| = -x and |z| = -z
and the equality is therefore not correct x-z-x EQUAL to -z
So we can answer the question
SUFFICIENT
(2) y>0, then z and x must be negative for their product to be negative
And since zy < xy, than z < x (both negative) so their absolute values is their opposite.
In this case, we have: |x-z| = x-z, |x| = -x and |z| = -z
so, |x-z| + |x| = x-z-x EQUAL to -z
SUFFICIENT
So answer is D
Last edited by fleurdelisse on Wed Apr 22, 2009 8:58 am, edited 1 time in total.
