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Club X has more than 10 but fewer than 40 members. Sometimes

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Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 12 Dec 2012, 05:31
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Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
[Reveal] Spoiler: OA
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 12 Dec 2012, 05:39
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Walkabout wrote:
Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


3 members at one table and 4 members at each of the other tables, means that the total number of members is 3 more than a multiple of 4: x=4m+3.
3 members at one table and 5 members at each of the other tables, means that the total number of members is 3 more than a multiple of 5: x=5n+3.

Thus x-3 is a multiple of both 4 and 5, so a multiple of 20. Therefore x is 3 more than a multiple of 20. Since 10<x<40, then x=23.

The remainder when 23 is divided by 6 is 5.

Answer: E.
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 14 Dec 2012, 02:16
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let the number of people be n , now 10<n<40. Also n=(3+ multiple of 4) and n=(3+ multiple of 5). Therefore n-3 is a multiple of both 4 and 5, one such number is 20. N=23, when 6 members sit at tables then people left are 5, therefore the answer is (E).
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 18 Dec 2012, 12:20
It took me 2 minutes of understanding and stupid calculation:
1) 3+4n=x 7 11 15 19 23 27 31 35 39
2) 3+5m=x 8 13 18 23 28 33 38

X=23 only.
23 (mod 6)=5
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 14 Aug 2013, 00:26
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 30 Apr 2014, 05:26
I did it this way and also got to the correct answer... is my reasoning correct here?

3+4+4 = 11 members
3+5+5 = 13 members
6+x = needs to be more than 10
so x is minimum of 5

x=5
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 01 May 2014, 01:15
Bunuel wrote:
Bumping for review and further discussion.


How much time should a question like this take in exam ?
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 01 May 2014, 08:43
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 28 May 2014, 09:09
My solution:

10<Members of Club X<40

X=4q+3
X=5p+3

Therefore, general formula based on both statements is X= 20k+23
Thus according to this particular statement X could ONLY take as values 23

23/6 gives a remainder of 5

Answer: E
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 01 Jun 2014, 20:42
GDR29 wrote:
Therefore, general formula based on both statements is X= 20k+23

Answer: E


Is there a typo? Did u mean X= 20k+3?
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 02 Jun 2014, 00:10
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Re: Club X has more than 10 but fewer than 40 members. Sometimes   [#permalink] 02 Jun 2014, 00:10
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