<p>i hear a lot of people getting the answer "x+y+z=36, x+y-z=4", but from i remember, one of the requirements stated in the question was something like "z-x=4." so therefore, the only feasible solution would be x=10, y=12, and z=14. thus, z-x=4, and x+y+z=36 and x+y-z=8, right? (since 10+12= 22, then 22-14=8.)</p>
<p>it seems like i'm the only person who thought that z-x=4 was in the question.</p>
<p>it was 5940 or something,</p>
<p>wrong question... :)</p>
<p>No the only two reqs were the first 2 equations. You cant assume anymore. They did not state z - x = 4.</p>
<p>i'm almost positive that z-x was a requirement of some kind. i'm pretty sure it was number 12 in a 16-question math section. if the answer was x+y+z=36 and x+y+z=4, it wouldn't be that close to the end of the section, given that both answers are a bit easy</p>
<p>that is the answer because they only gave 2 requirements and nothing else. Average of XYZ=12 and z subtracted from the sum of x and y is 4.</p>
<p>Average of XYZ = 12. Therefore, x+y+z = 36. Additionally, x < y < z - therefore, z is the greatest. Lastly, the largest number subtracted from the sum of the other two is 4.</p>