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

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Intern
Joined: 24 May 2013
Posts: 28

Kudos [?]: 23 [1], given: 21

Concentration: Operations, General Management
WE: Engineering (Energy and Utilities)
When tickets to a popular concert went on sale, a group of f [#permalink]

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04 Jun 2014, 04:26
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Difficulty:

95% (hard)

Question Stats:

44% (02:15) correct 56% (02:11) wrong based on 97 sessions

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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, 04:27, edited 1 time in total.
RENAMED THE TOPIC.

Kudos [?]: 23 [1], given: 21

Intern
Joined: 24 May 2013
Posts: 28

Kudos [?]: 23 [0], given: 21

Concentration: Operations, General Management
WE: Engineering (Energy and Utilities)

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04 Jun 2014, 04:29
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.

Kudos [?]: 23 [0], given: 21

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139706 [1], given: 12794

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

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04 Jun 2014, 04:52
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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.

Hope it's clear.
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Kudos [?]: 139706 [1], given: 12794

Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 139706 [0], given: 12794

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04 Jun 2014, 04:54
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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Kudos [?]: 139706 [0], given: 12794

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Posts: 14201

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

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11 Aug 2017, 22:16
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: When tickets to a popular concert went on sale, a group of f   [#permalink] 11 Aug 2017, 22:16
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