Of the 66 people in a certain auditorium, at most 6 people

Assume that January has 0 birthdays => all other months have 6 birthdays to total 66.

Statement 1: "More of the people in the auditorium have their birthday in February than in March" => March needs to be less than February, say March has 5 so that March (5) < February (6). The one birthday which has been 'plucked' from March, can only be accommodated in January since all other months are maxed out. => Jan has atleast one. Sufficient.

Statement 2: "Five of the people in the auditorium have their birthday in March" - same logic as for statement 1. Sufficient.

Answer: (D)

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.

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.

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

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

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


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

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


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

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



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

Regards,
Fabio.

This question needs some logic + inference based thinking. If you get that part right, the sum becomes very easy

Soln - Lets assume every month had 6 birthdays, so 12 * 6 = 72 (Note that we have 66 people and we are getting 72 i.e. 6 extra people)
.
So we can have a situation in which Feb-Dec have 6 people and Jan has none. (11 months have 6 people = 11*6 = 66 people)
.
Inference - Even if 1 month has less than 6 then Jan WILL have at least 1
.
.
Each statement alone conveys that 1 month has less than 6 people i.e Jan WILL have at least 1
.
.
(D)

Only scenario when there can be no birthday in January is that in all other months there should be 6 birhtdays ( which is maximum for any given month as per question).
1 - This means Feb and March have different figure and not both as 6. Which means there has to be atleast 1 birthday in Jan.
Best case scenario - Let birthday in Feb be 6 and in March it cannot be 6 so let it be 5. To get the least no. of birthday in Jan, let in all other months the no. of birthday be 6.
So, total no. of birthdays excluding Jan would be = 6x10 (for 10 months excluding Jan and Mar) + 5 (for March) = 65. So the 66th person's bday has to fall in Jan.
This is sufficient.

2 - If 5 have birthday in March which makes it less than 6 which is required for 0 bday in Jan. Now if we take the maximum for all the other months, it will be 6x10 = 60. So, in all it will be 60+5 (in march) = 65. So, the 66th bday has to fall in Jan.
This is also sufficient.

Each statement alone is sufficient to answer the question. Hence D.

Here is the rephrased question:
Does every other month have exactly 6 birthdays?

(2) Five of the people in the auditorium have their birthday in March
^^ March=5.
The answer is NO.
-->Sufficient (cross out choices-->A,C and E)

(1) More of the people in the auditorium have their birthday in February than in March
February (e.g., 6)>March (e.g., 5)
The answer is NO.
-->Sufficient (cross out choice-->B).
So, the correct choice is D
Hope it helps..

Solution:

We need to determine whether at least one person in the auditorium has a birthday in January. The only way this can’t happen (i.e., no one has a birthday in January) isif exactly 6 people have a birthday in each of the other 11 months. In other words, if we know there is a month (besides January) that has fewer than 6 people whose birthdays are in that month, then at least one person in the auditorium will have a birthday in January.

Statement One Alone:

More of the people in the auditorium have their birthday in February than in March.

Since more people in the auditorium have their birthday in February than in March, there must be fewer than 6 people whose birthdays are in March (since the most February could have is 6). Therefore, from our stem analysis, there is at least one person whose birthday is in January. Statement one alone is sufficient.

Statement Two Alone:

Five of the people in the auditorium have their birthday in March.

From our stem analysis, we see that there must be at least one person whose birthday is in January. Thus, statement two is also sufficient.

Answer: D

We can fidget with the numbers a bit and get insufficiency for both statements.

Of the 66 people in a certain auditorium, at most 6 people have their birthdays in anyone 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.
(6 x 9) + 7 + 5 <--- February is 7 and March is 5
(5 x 9) + 9 + 7 + 5 <--- January is 9, February is 7, March is 5
Insufficient.
(2) Five of the people in the auditorium have their birthday in March.
Use the exact same numbers as above.
(6 x 9) + 7 + 5 <--- February is 7 and March is 5
(5 x 9) + 9 + 7 + 5 <--- January is 9, February is 7, March is 5
Insufficient.

C: Still insufficient.

Could you say which book this question is from, please?

This question is presented in several official guides: OG 2015, OG 2016, OG 2017, OG 2018, OG 2019, and OG 2020.

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