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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
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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
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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] New post 27 Aug 2014, 01:58
I did this the long way, wrote out the seating arrangements possible under each scenario and found that 23 people is the only situation which applies to both seat configurations. Then as the others have pointed out 23 / 6 = 3 remainder 5
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 11 Sep 2014, 11:27
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



10<x<40

x= k4+3 ...11,15,19,23
x=a5+3...13,18,23
x=b6+?

x= 23 then 23/6 = remainder = 5
Re: Club X has more than 10 but fewer than 40 members. Sometimes   [#permalink] 11 Sep 2014, 11:27
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