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

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

62% (02:18) correct
38% (01:14) wrong based on 189 sessions

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, 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

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.

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

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.

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

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) _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

gmatclubot

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

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