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.
OA E
Source: MGMAT
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Mean Even
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logitech
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Yes it does. Let's have a discussion first. I can always PM you the solution, if you want.cramya wrote:Picked numbers for this. Dont know if there's an easier solution.
Does MGMAT provide one??
LGTCH
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cramya
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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.
Stmt I)
0 2 4 6 mean is odd
2 4 6 12 mean even
INSUFF
Stmt II
0 2 4 mean is odd
2 4 6 mean even
StmT I and II together(using same example from I)
0 2 4 6 mean is odd
2 4 6 12 mean even
Done E)
(1) n is even.
(2) All of the integers in set X are even.
Stmt I)
0 2 4 6 mean is odd
2 4 6 12 mean even
INSUFF
Stmt II
0 2 4 mean is odd
2 4 6 mean even
StmT I and II together(using same example from I)
0 2 4 6 mean is odd
2 4 6 12 mean even
Done E)
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mals24
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Well you can use logic for this question.
St 1 says n is even.
That means your denominator is even.
Now if the sum of the integers is even - mean might be even or odd or not even an integer. Not all even integers are evenly divisible by an even integer.
For instance 16 is not divisible by 6. So the mean is neither even nor odd.
If the sum of the integers is odd - mean is not an integer, and hence its neither even nor odd.
So st 1 INSUFF
St 2 says all integers in set X are even
Again same logic
If n is odd the mean might be even or might not be an integer.
Example: 10/5 or 16/7.
If n is even the mean might not be an integer. (16/6 or 24/4 or 28/4).
So st 2 INSUFF
Combining 1 and 2, we know n is even and sum is even. But still we don't know if the mean is an integer or not (16/6 or 28/4 or 24/4).
So INSUFF
Answer is E
I hope you get the logic.
Whats MGMATs explanation btw.
St 1 says n is even.
That means your denominator is even.
Now if the sum of the integers is even - mean might be even or odd or not even an integer. Not all even integers are evenly divisible by an even integer.
For instance 16 is not divisible by 6. So the mean is neither even nor odd.
If the sum of the integers is odd - mean is not an integer, and hence its neither even nor odd.
So st 1 INSUFF
St 2 says all integers in set X are even
Again same logic
If n is odd the mean might be even or might not be an integer.
Example: 10/5 or 16/7.
If n is even the mean might not be an integer. (16/6 or 24/4 or 28/4).
So st 2 INSUFF
Combining 1 and 2, we know n is even and sum is even. But still we don't know if the mean is an integer or not (16/6 or 28/4 or 24/4).
So INSUFF
Answer is E
I hope you get the logic.
Whats MGMATs explanation btw.
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EricLien9122
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It's hard to predict the out come of divisibility, because the outcome can be a fraction regardless the even or odd rules.
Q: set x has n number of int. is mean even?
Statement 1: n is even
mean can be fraction, odd or even
insufficient
Statement 2: all int. in x are even
mean can be fraction or even.
1+2, still not sufficient.
Q: set x has n number of int. is mean even?
Statement 1: n is even
mean can be fraction, odd or even
insufficient
Statement 2: all int. in x are even
mean can be fraction or even.
1+2, still not sufficient.
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logitech
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Manhattan GMAT wrote:
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. .
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
I solved it using the logic that division of Even/Even can be even or odd.logitech wrote:Manhattan GMAT wrote:
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. .
as in case of 12/6 = 6 (even) or 12/4 = 3 (Odd).
So, even combining both stmt, we cant say Mean = Even ( Sum of all even numbers n) / n (even) will be definitely even.
