manolas89 wrote:
as you said (2) can cleary answer the question but i think you are wrong for (1) .(1) is suff too because we know it;s about 6 teams and the rule says even*even=even , even*odd=even .
Yes, even*even=even and even*odd=even but this is not applicable to this problem.
A line of people waiting to enter a theater consists of 7 separate and successive groups. The first person in each group purchases one ticket for each person in the group and for no one else. If n is the total number of tickets sold for the first 6 groups, is n an even group? (1) There are no more than 4 people in each group --> n is the sum of 6 positive integers each of which is at most 4. From this we cannot say whether n is even or odd. Consider {1, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 2}. Not sufficient.
(2) The 19th person in line purchases the tickets for the 7th group. This implies that there are 18 people in the first 6 groups, thus n=18=even. Sufficient.
Answer: B.
Hope it's clear.
I didn't think that the question was clear ' If n is the total number of tickets sold for the first 6 groups, is n an even group? '
n is the number of tickets not a group, so I think the questions should be is 'n' even just like that.