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# x, y, and z are positive integers. The average (arithmetic m

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x, y, and z are positive integers. The average (arithmetic m [#permalink]  29 Jan 2007, 20:44
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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. I and III only
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Aug 2013, 21:54, edited 1 time in total.
Renamed the topic and added the OA.
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[#permalink]  30 Jan 2007, 00:12
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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[#permalink]  31 Jan 2007, 02:26
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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Re: Problem Solving - Odds and Evens [#permalink]  01 Feb 2007, 23:18
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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]  25 Aug 2013, 20:16
Hi Guys,

I got a very good qns. Try it.

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

OA is B
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Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]  25 Aug 2013, 22:00
Expert's post
above720 wrote:
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

The average (arithmetic mean) of x, y, and z is 11 --> $$x+y+z=3*11=33$$

z is two greater than x --> $$z=x+2$$ (from this we have that either both x and z are odd or both are even) --> $$x+y+(x+2)=33$$ --> $$2x+y=31$$ --> $$even+y=odd$$ --> $$y=odd-even=odd$$.

As for x and z: if $$y=1=odd$$, then $$x=15=odd$$ and $$z=17=odd$$ BUT if $$y=3=odd$$, then $$x=14=even$$ and $$z=16=even$$.

Therefore only II must be true.

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Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]  19 Apr 2014, 22:41
1
KUDOS
Given that:

x+y+z = 33 and z=x+2

using both equations

2x(Even) + y = 31 (Odd)

Since the sum is Odd 'y' has to be odd. Hence , only options (B) and (D) to be verified.

To check for 'x'.

x+y+z=33

x+z= 33- Odd

x+z= Even

Even + Even = Even
Odd+Odd = Even

So (B)
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Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]  29 Nov 2015, 06:42
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Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]  29 Nov 2015, 09:52
above720 wrote:
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

x + y + z = 33

x + y + (x+2) = 33

2x + y + 2 = 33

2x + y = 31

2x = 31 -y

x = $$\frac{31-y}{2}$$

So, we can say y must be an odd no and x can be even / odd.

Further we have z = x + 2

Since x can be odd/even ; z will be odd if x is odd and z will be even if x is even

So, we can be definite only about y as ODD

Hence answer is (B)
Re: x, y, and z are positive integers. The average (arithmetic m   [#permalink] 29 Nov 2015, 09:52
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# x, y, and z are positive integers. The average (arithmetic m

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