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# Problem Solving - Odds and Evens

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Manager
Joined: 05 May 2005
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Problem Solving - Odds and Evens [#permalink]  29 Jan 2007, 21:44
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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
SVP
Joined: 05 Jul 2006
Posts: 1565
Followers: 4

Kudos [?]: 63 [0], given: 35

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

VP
Joined: 21 Mar 2006
Posts: 1138
Location: Bangalore
Followers: 2

Kudos [?]: 14 [0], given: 0

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.
Director
Joined: 24 Aug 2006
Posts: 758
Location: Dallas, Texas
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Kudos [?]: 9 [0], given: 0

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)
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Re: Problem Solving - Odds and Evens   [#permalink] 02 Feb 2007, 00:18
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