Of the 66 people in a certain auditorium, at most 6 people have their birthdays in any one given month. Does at least one person in the auditorium have a birthday in January?

(1) More of the people in the auditorium have their birthday in February than in March.
(2) Five of the people in the auditorium have their birthday in March.
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Of the 66 people in a certain auditorium, at most 6 people have their birthdays in any one given month. Does at least one person in the auditorium have a birthday in January?

Pre-thinking: in the worst case scenario there are 11 months each with 6 people => 1 month out with 0.
We start from here because if we do NOT assume this, then all months have at least one person, and the question does not make sense.

(1) More of the people in the auditorium have their birthday in February than in March.
This could mean two things:
1) The month "out" is March (0 people), in this case January is one of the month with 6 people.
2) March and February cannot both have 6 people => in the worst case March has 5 people, and all months have at least one person now.
Both cases are sufficient

(2) Five of the people in the auditorium have their birthday in March.
So we take the 6 people of March, take one out and assign it to to a new month (because all the others have 6 already).
In this scenario all months have at least one person, sufficient

Hope this makes sense
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Apologies for flaring up an old topic.

Given both the question and the solution come from OG, there is certainly no conflict over the answer and its explanation. However, I find the answer incorrect. My reasoning for the same is as follows.

1) Agreed, this is sufficient
2) The statement reads as "Five of the people in the auditorium have birthdays in March". The statement to me reads as 5 people from the auditorium have birthdays in March, however it does not say these are the only 5 people whose birthday falls in March. Logically "only" is a required qualifier for the reader to conclude that there are only 5 such people. I don't think statement 2 will be incorrect if there are 6 people who have their birthdays in March. Therefore, the correct answer of this question should be (A).

I understand at the end of the day whatever OG mentions will be construed as correct answer, but I feel my logic is correct. I solicit your views on this.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
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Think about the sets questions you solve regularly.

55 of the 100 people drink tea. Do you take it as 55 drink tea and 45 do not or do you take it as 'at least 55 drink tea'?
When you are given that of the 100 people, 55 drink tea, it means only 55 drink tea.
Similarly, 5 of the people in the auditorium have their birthday in March means only 5 do.[/quote]


I do agree with VeritasPrepKarishma

Also considering the example put forward by Ankurdubey :

[/quote][/quote]

I am confused with what approach to follow, can any of the experts here suggest how to tackle such questions and come up with a concrete understanding of the question along with the right answer?

\(0\,\, \leqslant \,\,J,F,M, \ldots ,N,D\,\, \leqslant \,\,6\,\,\,\left( * \right)\,\,\,\,\,\,\left( {{\text{ints}}} \right)\)

\(J + F + M + \ldots + N + D = 66\,\,\,\,\,\,\, \Rightarrow \,\,\,{\mu _{{\text{all}}}} = 5.5\,\,\frac{{{\text{birthdays}}}}{{{\text{month}}}}\)

\(?\,\,\,:\,\,\,J\,\,\mathop \geqslant \limits^? \,\,1\)


\(\left( 1 \right)\,\,F > M\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( {**} \right)\,\,\,J = 0\,\,\,\, \Rightarrow \,\,\,{\mu _{{\text{all}}\, - \,\left\{ J \right\}}} = 6\,\,\frac{{{\text{birthdays}}}}{{{\text{month}}}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\underline {F = M} = \ldots = N = D = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\underline {{\text{impossible}}}\)


\(\left( 2 \right)\,\,\,M = 5\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( {***} \right)\,\,\,J = 0\,\,\,\, \Rightarrow \,\,\,{\mu _{{\text{all}}\, - \,\left\{ J \right\}}} = 6\,\,\frac{{{\text{birthdays}}}}{{{\text{month}}}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,F = \underline M = \ldots = N = D = \underline 6 \,\,\,\,\, \Rightarrow \,\,\,\,\,\,\underline {{\text{impossible}}}\)



We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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