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If x, y and z are integers and x – y – z < 0

Expert replies
by Brent@GMATPrepNow » Tue Mar 21, 2017 6:47 am
If x, y and z are integers and x - y - z < 0, is z > 1?

(1) x - y > 1 - z
(2) y - x < -2

Difficulty level: 800
Source: www.gmatprepnow.com

Answer: D
Last edited by Brent@GMATPrepNow on Wed Mar 22, 2017 9:50 am, edited 2 times in total.
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Source: — Data Sufficiency |

by DavidG@VeritasPrep » Tue Mar 21, 2017 10:52 am
[redacted]
Last edited by DavidG@VeritasPrep on Thu Mar 23, 2017 11:05 am, edited 2 times in total.
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by DavidG@VeritasPrep » Tue Mar 21, 2017 10:53 am
For more delightful inequality fun in which it's important to remember that we can sum our inequalities, so long as the inequality signs are facing the same direction, see here: https://www.beatthegmat.com/is-m-z-0-t13539.html
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by Brent@GMATPrepNow » Tue Mar 21, 2017 10:55 am
Solid solution, David.
One small error: x cannot equal 3/4 (since z must be an integer)

Cheers,
Brent
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by DavidG@VeritasPrep » Tue Mar 21, 2017 11:03 am
Brent@GMATPrepNow wrote:Solid solution, David.
One small error: x cannot equal 3/4 (since z must be an integer)

Cheers,
Brent
Dang. You win this round. Editing now.
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by Mo2men » Wed Mar 22, 2017 7:53 am
DavidG@VeritasPrep wrote:
Brent@GMATPrepNow wrote:Solid solution, David.
One small error: x cannot equal 3/4 (since z must be an integer)

Cheers,
Brent
Dang. You win this round. Editing now.
Hi David,

I committed the same error with same number (3/4) like you. Brent caught me too :) So I challanged myself to find any arrangement with based on:

z=1 and both conditions in the stem x - y - z < 0 & FACT 1 ) x - y > 1 - z and x, y and z are Integers

Look to my work below:

If z=1 & x-y=3

3 - 1 < 0........Not valid

3 + 1> 1 ........Valid

If z=1 & x-y=- 3

-3 -1 <0....valid

- 3 + 1 > 1...Not valid

Z can't be viable number to test as it violates either conditions in the stem of fact 1

More trials

If z=4 & x-y=3

3 - 4 < 0........valid

3 + 4> 1 ........Valid

If z=5 & x-y=-3

- 3 - 5 < 0........valid

- 3 +5> 1 ........Valid

If z=-5 & x-y=-4

-5 + 4 < 0........valid

-5 - 4 > 1........Not valid

Your comment You or Brent or any expert are welcome.

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by Brent@GMATPrepNow » Wed Mar 22, 2017 9:49 am
Mo2men has raised a very interesting point. It has caused me to change the official answer to D, unless I'm missing something.

I have also upgraded the difficulty level to 800!!!!!

Given: x - y - z < 0

Target question: Is z > 1

(1) x - y > 1 - z

From statement 1 (and the given inequality), we learn that z > 1/2. So, we might (incorrectly) conclude that it could be the case that z = 1 or z = 2, in which case, we get different answers to the target question.
HOWEVER, if we try to come up with values for x, y and z that demonstrate this, we find that we have a problem.

If z = 1, then we can plug this value into our two inequalities.
For the statement 1 inequality, we get x - y > 1 - 1
Simplify to get: x - y > 0

For the given inequality, we get x - y - 1 < 0
Simplify to get: x - y < 1

When we combine the two inequalities, we get: 0 < x - y < 1
In other words, the difference between x and y is a fractional value BETWEEN 0 and 1.
This is IMPOSSIBLE, since it's given that x and y are integers.
So, it cannot be the case that z = 1

Since we already know that z > 1/2, we CAN conclude that it's possible that z = 2, z = 3, z = 4, etc.
For example, consider these situations:
Case a: x = 0, y = 0 and z = 2. In this case, z IS greater than 1
Case b: x = 0, y = 0 and z = 3. In this case, z IS greater than 1
Case c: x = 0, y = 0 and z = 4. In this case, z IS greater than 1
Case d: x = 0, y = 0 and z = 5. In this case, z IS greater than 1
etc..

So, it turns out that the correct answer is actually D (both statements are sufficient)

Sorry for not knowing the correct answer when I first posted the question. It's even harder than I first imagined!!

Cheers,
Brent
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by DavidG@VeritasPrep » Thu Mar 23, 2017 11:05 am
I have also upgraded the difficulty level to 800!!!!!


A momentous moment for the community indeed. (And a humbling, humbling thread.)

Kudos to Mo2men
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