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# If x is the average (arithmetic mean) of 5 consecutive even

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

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01 May 2012, 10:03
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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
[Reveal] Spoiler: OA

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Re: x is a nonzero integer [#permalink]

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01 May 2012, 11:28
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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01 May 2012, 14:41
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Expert's post
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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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink]

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19 Jun 2013, 04:52
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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14 Jul 2014, 21:59
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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14 Jul 2014, 22:19
Expert's post
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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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17 Jul 2014, 22:00
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 !

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

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17 Jul 2014, 22:06
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

Last edited by ayushee01 on 17 Jul 2014, 23:00, edited 1 time in total.

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

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17 Jul 2014, 22:58
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.

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

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17 Jul 2014, 23:03
lol yes , i meant median, edited it :D

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

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17 Jul 2014, 23:05
[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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23 May 2015, 01:08
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 ?

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

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23 May 2015, 08:33
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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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink]

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25 Nov 2016, 05:25
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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink]

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12 Dec 2016, 10:37
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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31 May 2017, 22:54
[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] 31 May 2017, 22:54
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