<p>If the average(arithmetic mean) of x,y,z,5, and 7 is 8, which of the following must be ture?
one:the median of the five numbers cannot be 5
two:at least one of x,y, and z is greater than 9
Three:the range of the five numbers is 2 or more.
A.one only
B.two only
C.three only
D.one and two
E.two and three</p>
<p>I got the first and the second statement. As for the third one, according to ETS, it is right.
I've seen some explanation elsewhere which said that whatever nubmers you use to substitute x,y and z, the range will always be bigger than 2. Well, I get that, but the original statment also contains the the hypothesis that it could be 2 which is not possible. Thus I think the right answer is B instead of E.</p>
<p>The first statement is wrong.
The second statement is right! ;
(x+y+z+5+7)/5 = 8
40=x+y+z+12
28=x+y+z
therefore,one of those terms must be greater than 9- their average is 9.3</p>
<p>The third statement is also right and no need to explain it.
It is E</p>
<p>This is a GRE Math question written by ETS. There is a mistake on the part of ETS. The range must be greater than two, it cannot be equal to 2. Because that would require x, y, and z to fall between 5 and 7(inclusive), but there sum needs to be 28, which is not possible. The correct answer is indeed B. </p>
<p>They should have written option III as: The range of the five numbers is more than 2.
Then the answer would be E.</p>
<p>The first statement is wrong (counterexample: 5 5 5 7 18).</p>
<p>The second statement is true. (x+y+z+5+7)/5 = 8 → x+y+z = 28. At least one of x,y,z is greater than 9.</p>
<p>For the third statement, the phrase “two or more” includes the “more” case. For example, the statement “16 is greater than or equal to 15” is technically a true statement. The third statement is true.</p>