DS #500 Test 8 - 24

[f2001290]

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

Please explain the methodology to solve these type of problems.

[mschling52]

Is the answer B?

I don't usually try to pick numbers for a problem like this, but prefer to work with the equations instead.

(1) gives us z > x + y + 1. Rearranging, we have z - x - y > 1. From the stem, x+y+z>0, or z > -x - y. Therefore,

2z > z - x - y > 1. So, 2z>1 implies z>(1/2). Not Sufficient.

(2) gives us x+y < -1, so for x+y+z to be positive, z must be greater than 1. Sufficient.

[Stacey Koprince]

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

[mschling52]

Hi Stacey,

Can you expand a bit on how you manipulated statement (2) in your response (I think this is actually statement (1) in the problem as given). From how you did the subtraction of the inequalities in the first part of your response, I would think that

x + y + z > 0
- [0 > x + y + 1]
= x + y + z > -x - y - 1

[Stacey Koprince]

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

[mendiratta]

(1) gives us z > x + y + 1. Rearranging, we have z - x - y > 1. From the stem, x+y+z>0, or z > -x - y. Therefore,

2z > z - x - y > 1. So, 2z>1 implies z>(1/2). Not Sufficient.

Stacey Koprince, can you pls explain what is wrong with above explanation?..because I'm also getting B as solution.
anyway, what is OA?

[mschling52]

Stacey,

In your original response aren't you using Statement (2) twice? Just adding it in one case and subtracting it in another? If you use your method and statement (1), I think you come up with

x+y+z>0 (from stem)
+[z>x+y+1] (from Stmt 1)

Adding these gives,

x+y+2z > x+y+1
2z>1
z>(1/2),

which is the same thing I came up with from Statement 1 in my original response.

[Stacey Koprince]

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.

[gviren]

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


(1) ---- z - x - y - 1 > 0 Adding it with x + y + z > 0

2z - 1 > 0 ; Z > 1/2 hence not true

(2) -( x + y + 1 ) > 0 ; Adding it with x + y + z > 0

z - 1 > 0 ; z > 1 true