Problem Solving

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

Answer: A

if you take 2,4,6,8,10
average is 6
eliminate D and E

if you take -4,-2,0,2,4
average 0
eliminate B and C
A is left

Hence A is the answer.

irock 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

Answer: A

Picking numbers and elimination is a great way to tackle roman numeral questions - amar covered that approach very well. On test day, that would likely be the best approach.

Let's take a step back, however, and go into "review mode", something you should do for every question you attempt, whether you get it right or wrong. Remember, it doesn't matter how you do on practice questions - what matters is whether you learn something that will make you quicker and/or more accurate on test day.

A useful math fact is that the average of a set of consecutive numbers (be they integers, odds, evens, multiples of 4, etc...) is equal to the middle number if there's an odd number of terms or the average of the two middle terms if there's an even number of terms. The average of the set is also equal to the average of the first and the last terms.

On this question we have an odd number of terms (5), so the average is simply the middle term. Since we have 5 consecutive even integers, all that we know about the middle term is that it's even.

So:

I) always true;
II) 0 is even, so could be true or false; and
III) some even integers are multiples of 5 and some aren't, so could be true or false.

Only (I) is a "must be true": choose (A).

thanks both of you.