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Manager  Joined: 05 May 2005
Posts: 71
x, y, and z are positive integers. The average (arithmetic m  [#permalink]

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19 00:00

Difficulty:   35% (medium)

Question Stats: 72% (01:50) correct 28% (02:07) wrong based on 1045 sessions

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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

Originally posted by above720 on 29 Jan 2007, 21:44.
Last edited by Bunuel on 25 Aug 2013, 22:54, edited 1 time in total.
Renamed the topic and added the OA.
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Re: x, y, and z are positive integers. The average (arithmetic m  [#permalink]

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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.

Answer: B.
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Re: Problem Solving - Odds and Evens  [#permalink]

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5
2
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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2
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
Director  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.
Manager  Joined: 14 Jun 2011
Posts: 66
Re: Problem Solving - Odds and Evens  [#permalink]

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1
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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1
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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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)
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Re: x, y, and z are positive integers. The average (arithmetic m  [#permalink]

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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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1
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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GRE 1: Q169 V154 Re: x, y, and z are positive integers. The average (arithmetic m  [#permalink]

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

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1
EVEN-ODD RULES are being tested here

Given: x,y,z>0
x+y+z=33 (Average*total no. of terms=11*3=33)
z=x+2

x+y+x+2=33
2x+y=31

EVEN+y= ODD
y= ODD-EVEN
y= ODD

x and z can be even or odd and its not possible to get a unique value for them.

Thus only statement 2 is correct.
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Re: x, y, and z are positive integers. The average (arithmetic m  [#permalink]

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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

We are given that the average (arithmetic mean) of x, y, and z is 11.

Using the formula average = sum/number, we have:

11 = (x + y + z)/3

33 = x + y + z

We are also given that z is two greater than x, thus we now have:

x + y + (2 + x) = 33

2x + y = 31

Since 31 is odd and 2x is even, and the only way to get an odd sum is even + odd (or vice versa), we see that y MUST be odd. However, we do not know anything about x or z.

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

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Bunuel wrote:
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.

Answer: B.

Hi Bunuel,
I have understood the solution. Just for getting better at concept, I would like to get my doubt clarified.
Since x,y and z are positive integers , 11 is the mean and z is two greater than x, why can't it imply that z is 12, x = 12-2= 10 and y=11(mean) ? In such case, x must be even and y is odd.

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

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TaN1213 wrote:
Bunuel wrote:
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.

Answer: B.

Hi Bunuel,
I have understood the solution. Just for getting better at concept, I would like to get my doubt clarified.
Since x,y and z are positive integers , 11 is the mean and z is two greater than x, why can't it imply that z is 12, x = 12-2= 10 and y=11(mean) ? In such case, x must be even and y is odd.

Thank you.

The point is that the question asks which of the following MUST be true, not COULD be true. So, while x = 10 = even, y = 11 = mean, and z = 12 = even is certainly possible, but this is only one of the cases, and x = even is true only for some of the cases (it COULD be true) but on the other hand y = odd is true for ALL cases (it MUST be true).

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

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Hi Bunuel,
I have understood the solution. Just for getting better at concept, I would like to get my doubt clarified.
Since x,y and z are positive integers , 11 is the mean and z is two greater than x, why can't it imply that z is 12, x = 12-2= 10 and y=11(mean) ? In such case, x must be even and y is odd.

Thank you.[/quote]

The point is that the question asks which of the following MUST be true, not COULD be true. So, while x = 10 = even, y = 11 = mean, and z = 12 = even is certainly possible, but this is only one of the cases, and x = even is true only for some of the cases (it COULD be true) but on the other hand y = odd is true for ALL cases (it MUST be true).

Hope it helps.[/quote]

Yes! I get it now.. Thanks Bunuel. I went wrong in assuming x,y,z to be consecutive integers which is why I thought 10,11,12 is the only possible case.
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Re: x, y, and z are positive integers. The average (arithmetic m  [#permalink]

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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

Since the average (arithmetic mean) of x, y, and z is 11:

x + y + z = 33

Since z = x + 2:

x + y + x + 2 = 33

2x + y = 31

y = 31 - even number, so y must be odd. Thus, we only can determine that y must be odd.

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

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