DS: For a set X containing n integers ("Even Mean")

[aaliquo]

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.

For this question, I understand why the first and second statements are insufficient on their own, but I don't understand why they're not sufficient together.

When I put the statements together I know that the Sum (numerator) is even, and n (the denominator) is also even. If an even number is divided by another even number wouldn't the result also be even?

Thanks!

[snhassan1970]

Not really, lets take a simple example fo 4 even integers which satisfies both (1) and (2).

(2,4,6,8)--> Sum=20; --> Average(=20/4) = 5.--> An odd integer.

Thus, even we devide an even number by another even number, the result is not necessarily an even number. It could be odd and even it could be a fraction.

Hope this helps!

[aaliquo]

Thanks for the response, now I see. Next time I'll try using real numbers to verify the answer.

[RonPurewal]

If an even number is divided by another even number wouldn't the result also be even?

Thanks!

nope.

(even) / (even) can be ANYTHING:
it can be even (24/6 = 4)
it can be odd (24/8 = 3)
it can be a fraction with two odds (10/6 = 5/3)
it can be a fraction that is odd/even (6/8 = 3/4)
it can be a fraction that is even/odd (8/6 = 4/3)

the only thing it can't be is even/even, because that fraction could be reduced.

[yousuf_azim]

What is the ans?

BR

[shrive_]

E is answer

[StaceyKoprince]

thanks!

[sajaysuresh]

How is the answer E?

If we have a set say {2,4} mean is 3 which is odd.
If we have a set say {2,4,6,8} mean is 5 which is odd.

So mean is always odd. Shouldn't C be the answer?

Or can you give me some examples where mean is even?

[RonPurewal]

...ok, but if you just keep poking around, before long you'll discover sets like {2, 6}, whose mean is 4, or {2, 4, 8, 10}, whose mean is 6.

together with your examples, these prove that the answer is E.

[RonPurewal]

...oh, are you thinking that these have to be CONSECUTIVE even integers? because, yeah, don't think that. (: there's no reason why that would have to be true.