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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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Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

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]

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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.

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.

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

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25 Nov 2016, 04:25

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

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12 Dec 2016, 09:37

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 _________________

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

gmatclubot

Re: If x is the average (arithmetic mean) of 5 consecutive even
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12 Dec 2016, 09:37

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