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Manager
Joined: 05 May 2005
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Problem Solving - Odds and Evens [#permalink]
 29 Jan 2007, 21:44
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
How would you solve:
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. 1 and III only
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SVP
Joined: 05 Jul 2006
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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. 1 and III only
2x+y+2 = 33 ie: 2x+y = 31
y is sure odd
my answer is B
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VP
Joined: 21 Mar 2006
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Just to add...
Once you have established that I need not be true, you can eliminate II as well because:
z = x + 2.
Unless you know whether x is even or odd, you cannot know whether z is even or odd.
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Director
Joined: 24 Aug 2006
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Location: Dallas, Texas
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Re: Problem Solving - Odds and Evens [#permalink]
02 Feb 2007, 00:18
X+Y+Z =33
Z=X+2
2X+2+Y=33
2(X+1)+Y=33
even + Y =odd
Therefore, Y must be odd
(B)
_________________
"Education is what remains when one has forgotten everything he learned in school."
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Re: Problem Solving - Odds and Evens
[#permalink]
02 Feb 2007, 00:18
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