x, y, and z are positive integers. The average (arithmetic mean) of x, y, and z is 11. If z is two greater than x, which of the following must be true?
I. x is even
II. y is odd
III. z is odd
A)I only
B)II only
c)III only
D)I and II only
E)I and III only
OA is B)
Explanation given is :
We know that the average of x, y, and z is 11. We can therefore set the up the following equation:
(x + y + z)/3 = 11
Cross-multiplying yields
x + y + z = 33
Since z is two more than x, we can replace z:
x + y + x + 2 = 33
2x + y + 2 = 33
2x + y = 31
Since 2x must be even and 31 is odd, y must also be odd (only odd + even = odd). x and z can be either odd or even. Therefore, only statement II (y is odd) must be true.
The correct answer is B.
I understand the above solution, but my logic was slightly different. I figured if 11 was the average, and the z = x+2, the only solution was
x y z
10 11 12
Using this logic, I thought the right answer was x= even and y = odd. Where am I making a mistake? i.e I and II Option D)