DS #500 Test 8 - 24

Totally up to the individual as to whether to pick real numbers - whatever is easier for you. Just remember that we all make fewer mistakes with real numbers than we do with algebra.

For inequality problems like this one, you can add or subtract the inequalities (just like you can add or subtract equations). You just have to make sure that the greater than or less than signs point in the same direction

So, for (1), for example, you can do:
x + y + z > 0
- [x + y + 1 > 0]
= z - 1 > 0
= z > 1 Sufficient
[IGNORE the above calculations - I copied statement 2 here instead of statement 1 by accident - see later post]

(2):
x + y + z > 0
- [0 > x + y + 1]
= x + y + z > x + y + 1
= z > 1 Sufficient

D

Oops - that subtraction sign was supposed to be an addition sign. Thanks for the catch!

x + y + z > 0 (this statement is given in the question)
+ [0 > x + y + 1] (this statement is given as statement 2)
= x + y + z > x + y + 1
etc

Note that the second equation is statement 2, as written in the first post in this thread. I just wrote it the other way around (0 first instead of last) so that I could put the two inequality signs in the same direction.

24. If x + y + z > 0, is z > 1?
(1) z > x + y +1
(2) x + y + 1 < 0

Oh, I see - it wasn't statement 2 that I mis-read, it was statement 1 (or rather, I copied statement 2 twice by accident and never did statement 1 at all). Got it! Sorry about that!

Okay, let's try this again.

Statement 1:
Given x + y + z > 0
(1) z > x + y +1

Yes, if I add these, I get:
x + y + 2z > x + y + 1
2z > 1
z > 1/2

Which is a "maybe" so insufficient.