It is currently 23 Aug 2017, 06:39

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

Hide Tags

Manager
Joined: 05 May 2005
Posts: 77

Kudos [?]: 57 [3] , given: 0

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

Show Tags

29 Jan 2007, 21:44
3
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

69% (02:14) correct 31% (01:18) wrong based on 910 sessions

HideShow timer Statistics

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, 22:54, edited 1 time in total.
Renamed the topic and added the OA.

Kudos [?]: 57 [3] , given: 0

SVP
Joined: 05 Jul 2006
Posts: 1742

Kudos [?]: 400 [2] , given: 49

Show Tags

30 Jan 2007, 01:12
2
KUDOS
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

Kudos [?]: 400 [2] , given: 49

VP
Joined: 21 Mar 2006
Posts: 1127

Kudos [?]: 46 [0], given: 0

Location: Bangalore

Show Tags

31 Jan 2007, 03:26
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.

Kudos [?]: 46 [0], given: 0

Director
Joined: 24 Aug 2006
Posts: 744

Kudos [?]: 186 [2] , given: 0

Location: Dallas, Texas
Re: Problem Solving - Odds and Evens [#permalink]

Show Tags

02 Feb 2007, 00:18
2
KUDOS
2
This post was
BOOKMARKED
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)
_________________

"Education is what remains when one has forgotten everything he learned in school."

Kudos [?]: 186 [2] , given: 0

Manager
Joined: 14 Jun 2011
Posts: 84

Kudos [?]: 38 [1] , given: 15

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

Show Tags

25 Aug 2013, 21:16
1
KUDOS
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
_________________

Kudos always encourages me

Kudos [?]: 38 [1] , given: 15

Math Expert
Joined: 02 Sep 2009
Posts: 41055

Kudos [?]: 119397 [0], given: 12020

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

Show Tags

25 Aug 2013, 23:00
Expert's post
5
This post was
BOOKMARKED
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.

_________________

Kudos [?]: 119397 [0], given: 12020

Director
Joined: 03 Aug 2012
Posts: 900

Kudos [?]: 837 [1] , given: 322

Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

Show Tags

19 Apr 2014, 23:41
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)
_________________

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

Kudos [?]: 837 [1] , given: 322

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17168

Kudos [?]: 263 [0], given: 0

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

Show Tags

29 Nov 2015, 07:42
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 263 [0], given: 0

Math Forum Moderator
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 2990

Kudos [?]: 1014 [0], given: 325

Location: India
GPA: 3.5
Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

Show Tags

29 Nov 2015, 10:52
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

_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Kudos [?]: 1014 [0], given: 325

Intern
Joined: 16 Jan 2016
Posts: 22

Kudos [?]: 5 [0], given: 29

Location: United States (CA)
Leonid: B
Concentration: Operations, General Management
GPA: 3.6
WE: Operations (Other)
Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

Show Tags

26 Feb 2016, 14:44
Could someone explain this? Any great breakdowns of the data-sufficiency type of questions?
_________________

I Write...

http://www.misterethoughts.blogspot.com

Kudos [?]: 5 [0], given: 29

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2680

Kudos [?]: 1636 [1] , given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
x, y, and z are positive integers. The average (arithmetic m [#permalink]

Show Tags

26 Feb 2016, 19:40
1
KUDOS
Expert's post
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.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1636 [1] , given: 792

BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2222

Kudos [?]: 761 [0], given: 595

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

Show Tags

18 Dec 2016, 14:35
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

_________________

Give me a hell yeah ...!!!!!

Kudos [?]: 761 [0], given: 595

Director
Joined: 02 Sep 2016
Posts: 671

Kudos [?]: 47 [0], given: 241

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

Show Tags

10 Apr 2017, 23:27
1
This post was
BOOKMARKED
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.
_________________

Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.

Kudos [?]: 47 [0], given: 241

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 1375

Kudos [?]: 724 [0], given: 2

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

Show Tags

18 Apr 2017, 16:32
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.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 724 [0], given: 2

Intern
Joined: 09 Mar 2017
Posts: 28

Kudos [?]: 19 [0], given: 51

Location: India
Concentration: Marketing, Organizational Behavior
WE: Information Technology (Computer Software)
Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

Show Tags

13 Aug 2017, 10:41
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.

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.
_________________

------------------------------
"Trust the timing of your life"

Kudos [?]: 19 [0], given: 51

Math Expert
Joined: 02 Sep 2009
Posts: 41055

Kudos [?]: 119397 [0], given: 12020

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

Show Tags

14 Aug 2017, 00:51
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.

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.
_________________

Kudos [?]: 119397 [0], given: 12020

Intern
Joined: 09 Mar 2017
Posts: 28

Kudos [?]: 19 [0], given: 51

Location: India
Concentration: Marketing, Organizational Behavior
WE: Information Technology (Computer Software)
Re: x, y, and z are positive integers. The average (arithmetic m [#permalink]

Show Tags

15 Aug 2017, 03:37
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.
_________________

------------------------------
"Trust the timing of your life"

Kudos [?]: 19 [0], given: 51

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1303

Kudos [?]: 707 [0], given: 5

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

Show Tags

18 Aug 2017, 09:09
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.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 707 [0], given: 5

Re: x, y, and z are positive integers. The average (arithmetic m   [#permalink] 18 Aug 2017, 09:09
Similar topics Replies Last post
Similar
Topics:
If the average (arithmetic mean) of distinct integers x, y, and z is 1 4 10 Aug 2017, 10:52
4 x, y, and z are positive integers such that x ≥ y ≥ z. If the average 5 22 May 2017, 18:34
4 If the average (arithmetic mean) of x, y, and z is 7x and x ≠ 0 5 22 Aug 2017, 12:09
4 x, y, and z are three positive integers whose average is 10 4 30 May 2016, 22:59
14 If the average (arithmetic mean) of positive integers x, y, 15 18 Dec 2016, 15:11
Display posts from previous: Sort by