On the number line, the distance between x and y is greater

[angelapeltzer]

On the number line, the distance between x and y is greater than the distance between X and Z. Does Z lie between x and y on the number line?

(1) xyz<0
(2) xy <0

The answer is E.

I thought the answer was A. I see how this is a positive and negative integer problem.

From (2), I believe that X is negative and Y is positive. Given the explanation of the equation, and the description of (1), z could not lay to the left of x (z can't be negative). It seems that it would need to lay between X and Z.

Confused.

Thanks!

- Angie

[RonPurewal]

From (2), I believe that X is negative and Y is positive.

What about the other way around?

[angelapeltzer]

I'm sorry. That was typo. I thought the answer was C.

When I look at again. I'm thinking this:

(1) You know that either x, y, or z is negative or that all three are negative.

Insufficient.

(2) It doesn't tell us anything about z.

Insufficient.

From 1 and 2, you know that x or y need to be positive. If one of them is positive than you also know that z must be positive. What you don't know is where x or y is on the number line because you don't know which one is negative and which is positive. So C would be Insufficient.

[RonPurewal]

I'm sorry. That was typo. I thought the answer was C.

When I look at again. I'm thinking this:

(1) You know that either x, y, or z is negative or that all three are negative.

Insufficient.

(2) It doesn't tell us anything about z.

Insufficient.

From 1 and 2, you know that x or y need to be positive. If one of them is positive than you also know that z must be positive. What you don't know is where x or y is on the number line because you don't know which one is negative and which is positive. So C would be Insufficient.

This gets you the right answer, but it's not entirely a sound approach -- the problem doesn't boil down to just knowing the signs of the numbers.

E.g., even if I explicitly tell you that x = positive, y = negative, and z = positive, that's STILL insufficient. (Consider, say, x = 2, y = -2, z = 1, and then x = 2, y = -2, z = 3.)

Ultimately, the "I don't know the signs" approach gets lucky here. But, if you see a similar problem in the future, you're probably going to have to think about the literal distances on the number line.