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Expert replies
by ikaplan » Mon Jul 23, 2012 10:20 am
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

OA: E

Source: OG13
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Source: — Problem Solving |

by Anurag@Gurome » Mon Jul 23, 2012 10:28 am
ikaplan 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?
Say, the number of members of the club = n
Hence, 10 < n < 40

n = (Multiple of 4 + 3)
n = (Multiple of 5 + 3)

Hence, (n - 3) is multiple of both 4 and 5, i.e. multiple of 20.
Hence, n is 3 more than a multiple of 20.

Only possible value of n is 23.

Now, the remainder when 23 is divided by 6 is 5.

The correct answer is E.
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by ikaplan » Mon Jul 23, 2012 10:49 am
I think I had difficulty with understanding the questions stem :(

Can you please clarify what does "the members sit at tables with 3 members at one table and 4 members" really mean?

I understood that the sitting arrangement is as follows: /3/ /3/ /3/ /3/... /1/ /1/ /1/ /1/
where / / represents a table
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by Anurag@Gurome » Mon Jul 23, 2012 10:55 am
ikaplan wrote:I think I had difficulty with understanding the questions stem :(
"3 members at one table and 4 members at each of the other tables"

This means in only one table there are 3 members and all the other tables have 4 members each.
Hence, if there are x tables, then in one table there are 3 members and in the rest (x - 1) tables have 4 members each.
Therefore, total number of members = 3*1 + 4*(x - 1) = (4x - 1) = (Multiple of 4 - 1) = (Multiple of 4 + 3)

Hope that helps.
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by ikaplan » Mon Jul 23, 2012 12:40 pm
Thanks, I get it now.
"Commitment is more than just wishing for the right conditions. Commitment is working with what you have."
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by zeza » Sat Dec 29, 2012 6:11 am
I am confused with the stem. How is the information about number of members (10<n<40) important? What if there is a club with max number of members (40) or min number of members (10)? Is it possible to start from this information?
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by ToxicAvenger » Mon Mar 07, 2016 1:13 pm
This assumes that each time, the number of members that attend the club is constant, while the number of tables available to them varies.

This is not very realistic: it is much more likely that the number of attendees varies every time as opposed to removing some of the tables available to them.
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by Matt@VeritasPrep » Thu Mar 17, 2016 9:07 pm
That's definitely a realistic objection, but not one that the GMAT would heed: the scenario is the scenario, unfortunately.
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