<p>The average (arithmetic mean) of the 3 different positive integers x, y, and z is 50. If x<y<z and y=30, what is the least possible value of z?</p>
<p>A 50
B 51
C 59
D 90
E 91</p>
<p>Answer: B 51</p>
<p>Why?
I did:
(x+y+z)/3=50, thus x+y+z=150; y=30</p>
<p>Thus: x+z = 120
z must be greater than x as given by the inequality, so at the very least, z must be 61, and x 59. How is that given inequality reversed?</p>
<p>But the greatest possible value of x is 29. Since y=30, x<y<z means x<30<z.</p>
<p>If x takes its greatest possible value, the value of z will be the minimum possible. So let x be 29.</p>
<p>Then (29+30+z)/3 = 50.
29+30+z = 150
z = 91</p>
<p>To minimize the value of z, we must maximize the value of x.
x<30 and must be a positive integer;
The highest value that fulfills those criteria is x = 29;</p>
<p>Plug all this back into our equation to find the average:
(29 + 30 + z)/3 = 50;
29 + 30 + z = 150
z = 91;</p>
<p>The answer is FOR SURE (E) 91. Are you sure you’re not reading the answer key wrong or something?</p>