IMO D

given: zy<xy<0---equ 1

zy and xy are negative.

Either x and z are negative or y is negative.

St 1: z<x

Let x = -1, z = -3 and y = 2

xy = -2; zy = -6

-6<-2; statisfies equ 1

Let x = 3, z = 1, y = -2

xy = -6, zy = -2
zy>xy, this violates equ 1.

Hence x and z are negative.

Hence |x - z| is positive ( |-1+3| = 2)

We have to show if |x - z| + |x| = |z|
LHS
(x-z)+(-x)
= -z

RHS
(-z)

LHS = RHS Hence st 1 is SUFF

St 2 y>0
This gives the same information as st 1.---SUFF

Another shortcut would be you know St 1 and St 2 give you the same information since xz and y are opposite signs. So that brings down the options to either D or E. So in case your running out of time you can use a little bit of logic and bring down the options to D or E and make a guess.

If zy < xy < 0 is | x - z | + | x | = | z | ?

zy < xy < 0
z +ve y -ve or y +ve z -ve
x +ve y -ve or y +ve x -ve
x and z are both +ve or -ve since y is the same


Basically the question boils down to if the distance between x and z and distance between x and 0 put together is equal to distance between z and 0. The only time this can happen is if x > z or z < x.Picture it on a number line

If both x and z are postive

0--------x------------z

or

If both x and z are negative

z----------x------------0

Stmt I

z<x Exactly what we want from the explanation above

| x - z | + | x | = | z | TRUE

SUFF

Stmt II

y> 0

Therefore x and z must be negative for the inequality of zy < xy < 0 to hold good and x must be greater than z (eg: z= -10 x=-5)

| x - z | + | x | = | z | TRUE

SUFF

Done D)

NICEEE

Vishu