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(I) The first number in the series is odd.
X th number = a + x -1
Ave of X consecutive numbers = [a + (a+X-1)]/2 = a + (x -1)/2

Ave depends on X.
If X is odd, x-1 is even, x-1/2 is whole => Ave is either odd or even.
If X is even, x-1 is odd, x-1/2 always ends in .5 => Ave is not odd/even

(II) The sum of the numbers is odd.
Ave = Sum/X = Odd/X

Ave depends on X.
If X is odd, ave = odd/odd => Ave is odd.
If X is even, ave = odd/even => Ave is a decimal, not odd/even

Any other easy way? Even plug-in seems too clumsy.

(1) is insufficient.
(2) The sum of the numbers is odd ---> the number of odd integers is odd. There're 2 possibilities
- The number of odd and even numbers is odd ---> the average is odd
- The number of odd and even numbers is odd ---> the average is always .5 , not odd or even.
---> insufficient.

(1)&(2) ---> insufficient. The ans is E _________________

"Life is like a box of chocolates, you never know what you'r gonna get"

(1) is insufficient. (2) The sum of the numbers is odd ---> the number of odd integers is odd. There're 2 possibilities - The number of odd and even numbers is odd ---> the average is odd - The number of odd and even numbers is odd ---> the average is always .5 , not odd or even. ---> insufficient.

(1)&(2) ---> insufficient. The ans is E

No the anwer is B. Out of the 2 possibilites in 2nd condition, you are given the first one only where they add up to odd.
S

(1) is insufficient. (2) The sum of the numbers is odd ---> the number of odd integers is odd. There're 2 possibilities - The number of odd and even numbers is odd ---> the average is odd - The number of odd and even numbers is odd ---> the average is always .5 , not odd or even. ---> insufficient.

(1)&(2) ---> insufficient. The ans is E

No the anwer is B. Out of the 2 possibilites in 2nd condition, you are given the first one only where they add up to odd. S

Saurya_s. Please consider:
- 1,2,3,4 and 5. They add up to 15, odd and the average is 3, odd

- 1,2,3,4,5 and 6. They add up to 21, odd but the average here is 3.5, not odd or even.
So B is insufficient _________________

"Life is like a box of chocolates, you never know what you'r gonna get"

(1) is insufficient. (2) The sum of the numbers is odd ---> the number of odd integers is odd. There're 2 possibilities - The number of odd and even numbers is odd ---> the average is odd - The number of odd and even numbers is odd ---> the average is always .5 , not odd or even. ---> insufficient.

(1)&(2) ---> insufficient. The ans is E

Ok, you are right and thanks for the exxplaanation. U need to correct this line
- The number of odd and even numbers is odd to even.
Thanks a lot
S

(1) is insufficient. (2) The sum of the numbers is odd ---> the number of odd integers is odd. There're 2 possibilities - The number of odd and even numbers is odd ---> the average is odd - The number of odd and even numbers is odd ---> the average is always .5 , not odd or even. ---> insufficient.

(1)&(2) ---> insufficient. The ans is E

No the anwer is B. Out of the 2 possibilites in 2nd condition, you are given the first one only where they add up to odd. S

Saurya_s. Please consider: - 1,2,3,4 and 5. They add up to 15, odd and the average is 3, odd

- 1,2,3,4,5 and 6. They add up to 21, odd but the average here is 3.5, not odd or even. So B is insufficient

bigtooth:

in above example, you get ODD on the first one but not an integer on the second one. Since second option is not even an ineteger. Wouldn't B enough to answer the question?