For a set X containing n integers, is the mean even?

The mean of a set = The sum of the elements / number of elements, so the question is whether mean=sum/n=even.

(1) n is even --> mean=sum/even. Not sufficient.

(2) All of the integers in set X are even --> so the sum of the elements is even --> mean=even/n. Not sufficient.

(1)+(2) The question becomes whether mean=even/even=even, which can not be determined as even/even could be even (for example 4/2=2=even), could be odd (for example 6/2=3=odd) or could be non-integer (for example 6/4=3/2). Not sufficient.

Answer: E.
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This one is tricky.

1. Insufficient
n = 2
X = {0,2} mean = 1. The answer is NO
X = {3,5} mean = 4. The answer is YES

2. Insufficient
X = {0,2} mean = 1. The answer is NO
X = {2,2} mean = 2. The answer is YES

1) + 2)
X = {0,2} mean = 1. The answer is NO
X = {2,2} mean = 2. The answer is YES

Hence E.

From (1), 1,2 mean = 1.5 or 4,8 mean = 6, so (1) is not enough


Again, 2,4 mean = 3 or 4,8 mean = 6, so (2) is not enough

From (1) and (2), there is no definitive conclusion, so answer is E.
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For a set X containing n integers, is the mean even?

(1) n is even.

(2) All of the integers in set X are even.

Doesn't both the statements combined give an odd answer? Thus Statement (C) is correct but the MGMAT says that (E) is correct. Can someone explain the answer.

take Set S as {2,4} then you will get the mean as 3 (odd)
take Set S as {4,4,4,4} then you will get the mean as 4 (even)

so answer will be E
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Trick here is that integers in Set S is NOT UNIQUE EVEN INTEGER. Hence Ans. E. If integers were unique EVEN integers then Ans. would have been C.
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Not true, for instance.

If set {2,4}, then Avg. = 3
If set {2,6}, then Avg. = 4

Since both sets are unique even integers, the average can still be odd of even, and therefore whatever is above is not valid.

Hi Bunuel,

I still do not understand how to solve these kind of questions methodically because everytime we have to consider cases I forget one or other cases to consider. So instead of trying numbers can you please describe a methodical approach to test this.

For eg when we consider cases how many cases are possible and how to check against them. Such as we know that for the mean to be even the sum mandatorily has to be even however the base could be even or odd and can yield even or odd mean in both the cases i.e when N= even or when n=odd as non integers are also included.

One doubt - if the question had mentioned n positive integers, how does the answer varies ? My doubt is - whether the addition of "zero" to the set changes the even\odd nature of the mean ?

good question

take 2,4,6,8 and 4444 for both statements

and will come out to be E

hope it helps

This is very tricky because 0 is actually an even number and there not many restriction on the type of integers that are permitted in set X.

0 4 4 4

This conforms to the criteria in statements 1 and 2; however, the result is an odd mean. But if we have

2 2 2 2

Then the mean is even

E

For a set X containing n integers, the mean equals the sum of integers s divided by the number of integers n. If s is odd, the mean cannot be even. If s is even, there are no guaranteed outcomes for establishing that the mean of set X will be even. Therefore, there is no simple way to rephrase the question, but note that if we determine that s is odd we have achieved sufficiency.

(1) INSUFFICIENT: Statement (1) tells us nothing about whether s is odd or even. As noted above, if s were odd, we could determine that the mean of set X is not even. If s were even, observe that an even number by another even number could produce either an odd result (e.g. 12/4 = 3), an even result (e.g. 12/2 = 6), or a non-integer (e.g. 14/6).

(2) INSUFFICIENT: Statement (2) tells us that the sum s must be even. As described above, an even sum s does not provide a guaranteed outcome for the mean of set X regardless of whether n is odd or even.

(1) AND (2) INSUFFICIENT: Given that both s and n are even, there are no guaranteed outcomes for the mean of set X.

The correct answer is E.