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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
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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
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 30 Oct 2014, 21:54
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Simple solution quickly would be -
E
We know that the remainder is 3 in both cases when 4 or 5 people sit --> such one number is 23 (which also is between 20 and 40).
And hence, 23/6 gives remainder =5 .
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 29 Mar 2015, 01:32
I got tricked by "tables" in the problem stem and I considered more than 1 table of 3 person eacg.

Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables...

Isn't the problem poorly worded?
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 15 Jun 2015, 17:21
I got 5 as my answer.

For people that are more visual (like me)..

I put a 3 down and another 3 down representing the first two tables, then figured out how many "4s" were needed for the first set and how many "5s" were needed for the second set to add up and equal to the same number for both sets. Since 3 was constant between the two sets, it meant that 4 and 5 needed to have the same number of people, or the LCM, which is 20. Therefore 20 plus 3 is a total of 23 people.

Now you know that the table with a set of 6s needs to add up to 23, but the last table needs to still be less than 6. So the only combination for 6s is 3 tables of 6 people to equal 18 and then a table of 5 people.
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Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] New post 15 Jun 2015, 22:17
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TudorM wrote:
I got tricked by "tables" in the problem stem and I considered more than 1 table of 3 person eacg.

Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables...

Isn't the problem poorly worded?


No, it isn't. This is GMAT language - especially considering that the question is official - and hence you will be required to successfully comprehend such questions.

"Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables,"

The statement explains how the members sit at tables: 3 at ONE table and 4 at EACH of the other tables. Practice questions from the official guide to get comfortable with "GMAT language".
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Re: Club X has more than 10 but fewer than 40 members. Sometimes   [#permalink] 15 Jun 2015, 22:17
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