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by pradeepkaushal9518 » Sat Sep 11, 2010 4:10 am
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

I only
II only
III only
I and II only
I and III only
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by Rahul@gurome » Sat Sep 11, 2010 4:52 am
(x + y + z)/3 = 11 or x + y + z = 33
z = 2 + x
Then, x + y + (2 + x) = 33 or 2x + y = 31
If x is EVEN, y is ODD, then 2(EVEN) + ODD = ODD; this is true
If x is EVEN, y is EVEN, then 2(EVEN) + EVEN = EVEN; not true
If x is ODD, y is ODD, then 2(ODD) + ODD = EVEN + ODD = ODD; this is true.
If x is ODD, y is EVEN, then 2(ODD) + EVEN = EVEN + EVEN = EVEN; not true

Therefore, x may be even or odd, but y is odd is true. z may be even or odd, as z depends on x.
So, only II is true.

The correct answer is [spoiler](B)[/spoiler].
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by klmehta03 » Sat Sep 11, 2010 9:16 am
Agree the answer is B
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