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

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Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133072 [0], given: 12403

Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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16 Jul 2012, 04:49
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Difficulty:

75% (hard)

Question Stats:

49% (01:18) correct 51% (01:21) wrong based on 842 sessions

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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.
(2) Five of the people in the auditorium have their birthday in March.

Diagnostic Test
Question: 45
Page: 26
Difficulty: 650
[Reveal] Spoiler: OA

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Kudos [?]: 133072 [0], given: 12403

Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133072 [3], given: 12403

Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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16 Jul 2012, 13:08
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SOLUTION

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?

Basically the question is whether we can distribute 66 birthdays between 12 moths so that January to get 0.

(1) More of the people in the auditorium have their birthday in February than in March. Let 10 months (except March and January) have 6 birthdays each (maximum possible) --> 6*10=60. As in March there was less birthdays than in February than maximum possible for March is 5 --> total 60+5=65, so even for the worst case scenario (maximum for other months) still 1 birthday (66-65=1) is left for January. Sufficient.

(2) Five of the people in the auditorium have their birthday in March. Again: let 10 months have 6 birthdays each (maximum possible) --> 6*10=60 + 5 birthdays in March = 65. The same here: even for the worst case scenario (maximum for other months) still 1 birthday (66-65=1) is left for January. Sufficient.

Hope it helps.
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Kudos [?]: 133072 [3], given: 12403

Intern
Joined: 05 May 2012
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Kudos [?]: 3 [2], given: 12

Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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16 Jul 2012, 19:43
2
KUDOS
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.

Kudos [?]: 3 [2], given: 12

Intern
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Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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16 Jul 2012, 19:46
1
KUDOS
Q is does any one have their b'day in jan

given: max people(with their b'day) in any given month is 6. no of people is 66 .

stmt 1 : no of people in feb > no of people in march

lets consider the number of b'day people in feb as 6 (max)
so no of people in march cud be 5 or 4 or 3 or 2 or 1 or 0 . now max total is (6 in feb) + (5 in march) = 11 (66-11=55)
other remaining months from april to december can hold max 6 people each which means 6 * 9 = 54 people covered.
and only one person remaining out of 55 who fits into jan. therby atleast 1 person's b'day is in jan.

sufficient

stmt 2 : no of people having bday in march is 5

so 66-5 = 61.
no of months other than march and jan is 10 . so considering each months holds max of 6 . (10 * 6 = 60)
remaining 1 left who cud be in jan.

sufficient

IMO D

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Intern
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Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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18 Jul 2012, 00:41
1
KUDOS
Any month can have a max of 6 people celebrating their birthdays. No information is given about the min people or a lower cap of people.

(1)Feb>Mar implies that it can be a combination of any of the following.
Feb,Mar : 1,0 ; 2,0 ; 2,1 ; 3,0 ; 3,1 ; 3,2 and so on.
No information is given about the other 10 months. We cannot simply populate the values for the other 10 months with just the cap for the maximum number of birthdays.(Although a minimum number would have been useful in this regard)

NOT SUFFICIENT

(2)5 people have their birthdays in March. This does not give any information about the other months. Combining with the information given in the question, we can say that the condition has not been violated.

NOT SUFFICIENT

(1&2) Imply that Feb has 6 birthdays. Still we do not have information about the other months. 66-(6+5) = 55 member can have their birthdays distributed in 10 months in any number of ways as long as any of the remaining months do not exceed 6 birthdays. No information on the number of birthdays in Jan.

NOT SUFFICIENT

Hence 'E' must be the solution.

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Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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17 Jul 2012, 02:52
lateapp wrote:
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.

Agreed. In order for for there to be no January birthdays, Feb-Dec must each have 6. Statements to the contrary answer the question.

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Manager
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Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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09 Jan 2014, 11:14
The question really asks us if there is a month - except january - where less than 6 people have their birthday.

Both statements alone tell us the answer to this is "yes", and both more or less tell us the same thing, actually.

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Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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14 Jun 2016, 22:20
After reading the prompt,I realized that in order to satisfy the condition that the 'NO' answer to the DS question requires that each month we have 6 people having their birthdays. Why? Because again the prompt says that we have 66 people and in a month we cant have more than 6 people's birthdays.
So if Jan =0
then 0,6,6,6,6,6,6,6,6,6,6,6
If we violate the possibility of 0,6,6,6,6,6,6,... then we get a firm answer of YES to the DS.

Statement #1 -- says that Feb and March are unequal. So obviously they cant be equal. and hence the answer is YES.sufficient
statement #2--5 have birthdays in March. Again we violate the possibility of 0,6,6,6,6,...... So again we get a firm answer YES. sufficient

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Re: Of the 66 people in a certain auditorium, at most 6 people [#permalink]

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21 Jun 2017, 02:34
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Re: Of the 66 people in a certain auditorium, at most 6 people   [#permalink] 21 Jun 2017, 02:34
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