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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]

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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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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

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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]

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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]

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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]

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25 Aug 2013, 22:00
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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]

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19 Apr 2014, 22:41
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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]

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29 Nov 2015, 06:42
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Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

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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

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

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26 Feb 2016, 13:44
Could someone explain this? Any great breakdowns of the data-sufficiency type of questions?
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x, y, and z are positive integers. The average (arithmetic m [#permalink]

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26 Feb 2016, 18:40
leonidbasin1 wrote:
Could someone explain this? Any great breakdowns of the data-sufficiency type of questions?

You need to note what is given to you and what is getting asked. Additionally, this is a MUST BE TRUE type of a question ---> correct option will be true for ALL cases.

Given: x,y,z are POSITIVE INTEGERS and the average of x,y,z=11

From average of x,y,z = 11----> (x+y+z)/3 = 11 ---> x+y+z=33 (odd)

Also, z=x+2 ---> if x=odd, then z=odd+even=odd (think of 2+3=5, odd) and if x=even, then z=even+2=even. (Remember that odd+odd=even, even + even = even, odd +even =odd)

Also as x+y+z=33, odd ---> you need to have 1 odd number to make the sum of 3 integers = odd. Thus, the only case possible is

x=even, y=odd and z=even.

Lets analyse the options:

I. x is even , not possible. Eliminate.
II. y is odd, yes, keep.
III. z is odd, not possible. Eliminate.

Thus, only II is must be true and hence B is the correct answer.

As for your DS related question, it is a very broad question that can not be answered in simpler terms. You need to practice to understand how to tackle DS problems. The concept remain the same as those for PS only the application part differs. Take a look at Manhattan GMAT's DS book and practice questions with DS tags at viewforumtags.php

It will be of help to you if you ask specific questions rather than asking such broad ended questions.

Hope this helps.
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Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

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18 Dec 2016, 13:35
Here is my take on this one->

x+y+z=33
z=x+2=> z-x = 2 => Even

Hence x and z will have the same Even/Odd nature.
Hence y=> odd-eve=> odd
y is always odd
x and y can be both even or both odd

Hence B

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Mock Test -2 (Evens and Odds Basic Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768023

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Re: x, y, and z are positive integers. The average (arithmetic m   [#permalink] 18 Dec 2016, 13:35
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