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

Question

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

Bunuel · Accepted Answer

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.

LalaB · Answer

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

Bunuel · Answer

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

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ayushee01 · Answer

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

KarishmaB · Answer

Check out the blog posts on the link given in my signature below.

parul1591 · Answer

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 !

ayushee01 · Answer

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

parul1591 · Answer

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

ayushee01 · Answer

lol yes , i meant median, edited it :D

PerfectScores · Answer

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.

shallow9323 · Answer

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 ?

Bunuel · Answer

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

stonecold · Answer

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

tareqobaida · Answer

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
So answer is A

Funsho84 · Answer

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

Answer A

BrushMyQuant · Answer

Given that x is the average (arithmetic mean) of 5 consecutive even integers and we need to find 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.

============================================================

Theory
 ‣‣‣ Mean = (Sum Of All The Numbers) / (Total Number Of Numbers).  
============================================================

Let the 5 consecutive even numbers be y - 4, y - 2 , y , y + 2, y + 4
=> Mean =  \frac{Sum}{5}  =  \frac{y - 4 + y - 2 + y + y + 2 + y + 4 }{ 5}  =  \frac{5y}{5}  = y

=> Mean = y = x
So, the numbers are x - 4, x - 2 , x , x + 2 , x + 4

I. x is an even integer. 
This is true as x is the middle term and we know that all numbers are even

II. x is a nonzero integer. 
This doesn't have to be true as x can very well be 0 and the set will become
0 -4, 0 -2 , 0 , 0 +2 , 0 + 4 or -4, -2, 0, 2, 4

III. x is a multiple of 5. 
This doesn't have to be true. Same example as II proves this one wrong too
-4, -2, 0, 2, 4

So,  Answer will be A .
Hope it helps!

Watch the following video to Learn the Basics of Statistics

https://www.youtube.com/watch?v=Cvy_HHw4KIs

bumpbot · Answer

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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

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