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When tickets to a popular concert went on sale, a group of f

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When tickets to a popular concert went on sale, a group of f [#permalink]

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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 any one 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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jun 2014, 05:27, edited 1 time in total.
RENAMED THE TOPIC.

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Re: Concert and tickets [#permalink]

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New post 04 Jun 2014, 05:29
Why is statement 1 sufficient:
Case 1:

1+5+7+8+9 = 30
Median > Mean here and one guy has 9 tickets

2+4+6+8+10 = 30
Median = Mean (violates statement 1) but still answers question wiith YES as question asks for Atleast 9..So please clarify.

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Re: When tickets to a popular concert went on sale, a group of f [#permalink]

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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 any one friend purchase at least nine tickets?

Let the number of tickets purchased be a, b, c, d, and e, where \(a<b<c<d<e\) and \(a+b+c+d+e=30\).
The question asks whether \(e\geq{9}\).

(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 \(c>6\). 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: \(e\geq{9}\). Sufficient.

(2) One of the five friends purchased exactly three tickets. If \(a=3\) and \(e<9\), 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 \(e<9\), the maximum is even less than that. Therefore e must be more than or equal to 9. Sufficient.

Answer: D.

Hope it's clear.
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Re: Concert and tickets [#permalink]

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New post 04 Jun 2014, 05:54
nidhi12 wrote:
Why is statement 1 sufficient:
Case 1:

1+5+7+8+9 = 30
Median > Mean here and one guy has 9 tickets

2+4+6+8+10 = 30
Median = Mean (violates statement 1) but still answers question wiith YES as question asks for Atleast 9..So please clarify.


When you plug the numbers you must choose them so that they satisfy a statement.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: When tickets to a popular concert went on sale, a group of f [#permalink]

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Re: When tickets to a popular concert went on sale, a group of f   [#permalink] 11 Aug 2017, 23:16
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