On the number line the distance bet x & y is greater than the distance between x & z. Does Z lie between x & y on the number line.
a) xyz<0
b) xy<0

My approach:

stmt 1: x- y+ z+ xyz<0, z lies inside
x+ y- z+ xyz<0, y and z are on opp sides
x+ y+ z- xyz<0, y and z are on opp sides
x- y- z- xyz<0, z may lie or may not lie
insufficient

stmt 2: xy<0 x+ y- xy<0
x- y+ xy<0
no clue abt z.So, insufficient.

Both together: x- y+ z+ xyz<0, z lies inside
x+ y- z+ xyz<0, y and z are on opp sides
2 different answers so both not sufficient.
Ans E;
Is this correct?

given this statement, the number line can appear in one of exactly four ways:
case 1:--------------------x------z-------y---------
case 2: -------------z------x--------------y---------
case 3: -----y--------------x------z-----------------
case 4: -----y------z-------x------------------------

cases 1 and 4 are "yes" to the question; cases 2 and 3 are "no". since the GOAL is to TRY FOR "INSUFFICIENT" (as in just about any non-algebraic DS problem), you should try to get one instance of case 1 or 4, and one instance of case 2 or 3.


you can make any of the four cases with 3 negative numbers, so this statement is insufficient (you don't even have to consider the cases in which some of them are positive).
e.g.
x = -3, z = -2, y = -1 --> case 1 ("yes" to the question)
z = -4, x = -3, y = -1 --> case 2 ("no" to the question)



again, easy to get any of the three cases -- x and y just have to be one positive and one negative, and z can be, well, anything at all.
e.g.
x = -2, z = 1, y = 2 --> case 1 ("yes" to the question)
z = -3, x = -2, y = 2 --> case 2 ("no" to the question)

together:

if you have both statements, then you know:
x and y are one positive, one negative;
z is positive.
you can still get both answers:
x = -2, z = 1, y = 2 --> case 1 ("yes" to the question)
y = -2, x = 2, z = 3 --> case 2 ("no" to the question)
still insufficient.

answer = (e)