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Manager  Joined: 10 Nov 2010
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If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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10 00:00

Difficulty:   55% (hard)

Question Stats: 57% (01:28) correct 43% (01:39) wrong based on 436 sessions

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If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

x= n+(n+2)+(n+4)+(n+6)+(n+8)/5

x = n+4

IMO I & II is correct

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Originally posted by GMATD11 on 01 May 2012, 10:03.
Last edited by pushpitkc on 30 Oct 2018, 01:09, edited 1 time in total.
Edited question
Math Expert V
Joined: 02 Sep 2009
Posts: 59129
Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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4
1
If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

First of all notice that we are asked which of the following MUST be true, not COULD be true.

Also notice that the average of 5 consecutive even integers equals to the median, so it's just a middle term. So, I must always be true: {even, even, even, even, even}

Next:
II. x is a nonzero integer. Not necessarily true, consider: {-4, -2, 0, 2, 4};
III. x is a multiple of 5. Not necessarily true, consider: {0, 2, 4, 6, 8}.

Hope it's clear.

P.S. In your example if n=-4 then x=n+4=0, so II is not always true.
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GMAT 1: 690 Q47 V38 Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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2
İİ is not correct, since nowhere mentioned that even integers MUST BE positive
I is true, since the avrg sum= (sum of even integers)/ 5 (odd integer)=even/odd=even

III is also out. u can check it , if u pick some integers
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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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GMAT 1: 660 Q49 V31 Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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consider any even integer , a
we can form a series as a - 4 , a -2 , a , a + 2 , a + 4

average will be 5a/5 = a

so, i) always true , since a is integer

iii) not always true,a can be any integer as we considered, not necessarily multiple of 5

ii) not always true, for -4 -2 0 2 4 , average 0

ans - A
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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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GMATD11 wrote:
If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

x= n+(n+2)+(n+4)+(n+6)+(n+8)/5

x = n+4

IMO I & II is correct

Here are some posts on arithmetic mean that you might find helpful.

http://www.veritasprep.com/blog/2012/04 ... etic-mean/
http://www.veritasprep.com/blog/2012/04 ... questions/
http://www.veritasprep.com/blog/2012/05 ... eviations/
http://www.veritasprep.com/blog/2012/05 ... tic-means/
http://www.veritasprep.com/blog/2012/05 ... on-median/
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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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Bunuel wrote:
If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

First of all notice that we are asked which of the following MUST be true, not COULD be true.

Also notice that the average of 5 consecutive even integers equals to the median, so it's just a middle term. So, I must always be true: {even, even, even, even, even}

Next:
II. x is a nonzero integer. Not necessarily true, consider: {-4, -2, 0, 2, 4};
III. x is a multiple of 5. Not necessarily true, consider: {0, 2, 4, 6, 8}.

Hope it's clear.

P.S. In your example if n=-4 then x=n+4=0, so II is not always true.

Hello Bunuel

I have a dout in this statement that you made :the average of 5 consecutive even integers equals to the median.

How did you establish this ?

Thank you !
Intern  Joined: 06 May 2014
Posts: 6
Concentration: Marketing, Technology
GMAT 1: 660 Q49 V31 Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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for evenly spaced numbers , average is same as median

consider 3 integers having common difference d,

a , a+d , a +2d --> average 3a + 3d /3 = a+d
--> median = a+d

hope it helps Originally posted by ayushee01 on 17 Jul 2014, 22:06.
Last edited by ayushee01 on 17 Jul 2014, 23:00, edited 1 time in total.
Intern  Joined: 12 May 2014
Posts: 10
Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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ayushee01 wrote:
for evenly spaced numbers , average is same as mean

consider 3 integers having common difference d,

a , a+d , a +2d --> average 3a + 3d /3 = a+d
--> median = a+d

hope it helps Hello Ayushee

If I infer correctly, you mean to say that for evenly spaced numbers the average is the same as the median ?

Thank you for the explanation. It's clear now. Intern  Joined: 06 May 2014
Posts: 6
Concentration: Marketing, Technology
GMAT 1: 660 Q49 V31 Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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lol yes , i meant median, edited it :D
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Posts: 116
Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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[quote="GMATD11"]If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

x-4, x - 2, x, x+2, x+4

Average = x.

I. x will always be even as they are consecutive even integers.
II. if x = 0 then -4, -2, 0, 2, 4
III. x = 10 is a possibility

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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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Hi Bunuel,

First if all i'd like to Thank you for all the help.
I am deeply appreciate your efforts.

Is zero an even integer ?
Math Expert V
Joined: 02 Sep 2009
Posts: 59129
Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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shallow9323 wrote:
Hi Bunuel,

First if all i'd like to Thank you for all the help.
I am deeply appreciate your efforts.

Is zero an even integer ?

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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GRE 1: Q169 V154 Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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Great Question.
Here is my solution to this one ->
Since the consecutive integer set is an arithmetic progression with common difference =2
Mean =Median=Average of the first and the last term.
Clearly 5 is odd
Hence median will be the 3rd term itself.
And since all elements are even => x must be even integer.

Statement 1->
True
Statement 2->
May or may not be true.
E.g => -4,-2,0,2,4 => Mean=0
and 0,2,4,6,8 => Mean ≠0
Hence False

Statement 2->
x Again w can use the some examples =>
6,8,10,12,14 => x=10 i.e multiple of 5
0,2,4,6,8 => Mean = 4 => non multiple of 5
Once False
Hence only 1 must be true.

Thus A

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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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[quote="GMATD11"]If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

Let n is the first integer of the series. So x = (n+n+2+n+4+n+6+n+8)/5 = (5n+20)/5=n+4
Since n is an integer, n+4 must be an integer. Option I is true. It’s never mentioned that the numbers are positive. In that case n might be -4 that will bring x = 0. So option II cannot be true. Option III cannot be true because n+4 is not divisible by 5
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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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Nice Question

X = avg of 5 consecutive even integers
*no restriction on sign

Pick numbers to try.

-4 + -2 + 0 + 2 +4 -> fits question

X= 0

Statements
1. True -> 0 is neither positive nor negative but it's even
2. False - could be true. But we proved wrong with statement 1
3. False - could be true. Try a different set such as -8 + -6+ -4 + -2 +0 /5 = 4

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Re: If x is the average (arithmetic mean) of 5 consecutive even  [#permalink]

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_________________ Re: If x is the average (arithmetic mean) of 5 consecutive even   [#permalink] 11 Oct 2019, 09:01
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