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

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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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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 !

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

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.

lol yes , i meant median, edited it :D

[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.

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

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

[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
So answer is A

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

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!

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