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

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If x is the average (arithmetic mean) of 5 consecutive even [#permalink] New post 01 May 2012, 09:03
00:00
A
B
C
D
E

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  35% (medium)

Question Stats:

56% (01:48) correct 44% (01:00) wrong based on 139 sessions
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] New post 01 May 2012, 10:28
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] New post 01 May 2012, 13:41
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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

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}.

Answer: A.

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] New post 19 Jun 2013, 03:52
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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink] New post 14 Jul 2014, 20:59
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] New post 14 Jul 2014, 21:19
Expert's post
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] New post 17 Jul 2014, 21:00
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}.

Answer: A.

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] New post 17 Jul 2014, 21:06
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, 22:00, edited 1 time in total.
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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink] New post 17 Jul 2014, 21:58
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. :-D
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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink] New post 17 Jul 2014, 22:03
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] New post 17 Jul 2014, 22:05
[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

Hence I is the answer.
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Re: If x is the average (arithmetic mean) of 5 consecutive even   [#permalink] 17 Jul 2014, 22:05
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