## When tickets to a popular concert went on sale, a group of five friends collectively purchased a total of 30 tickets. If

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### When tickets to a popular concert went on sale, a group of five friends collectively purchased a total of 30 tickets. If

by AAPL » Mon Sep 20, 2021 10:08 am

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When tickets to a popular concert went on sale, a group of five friends collectively purchased a total of 30 tickets. If no two friends purchased the same number of tickets, did anyone friend purchase at least nine tickets?

1) Among the five friends, the median number of tickets purchased exceeded the mean number of tickets purchased

2) One of the five friends purchased exactly three tickets

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### Re: When tickets to a popular concert went on sale, a group of five friends collectively purchased a total of 30 tickets

by Manan073 » Mon Sep 20, 2021 12:23 pm

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## Global Stats

Let the number of tickets purchased be a, b, c, d, and e, where a<b<c<d<e
a<b<c<d<e and a+b+c+d+e=30
a+b+c+d+e=30. The question asks whether e≥9
e≥9.
(1) Among the five friends, the median number of tickets purchased exceeded the mean number of tickets purchased. The mean number of tickets is 30/5=6 and the median number of tickets is the middle term, c. Hence we are told that c>6
c>6. This implies that the least value of c is 7, the least value of d is 8 and the least value of e is 9: e≥9e≥9. Sufficient.
(2) One of the five friends purchased exactly three tickets. If a=3a=3 and e<9
e<9, then the maximum possible sum is 3+5+6+7+8=29<30. So, this case is not possible. If b or c is 3 and e<9e<9, the maximum is even less than that. Therefore e must be more than or equal to 9. Sufficient.