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Club X has more than 10 but fewer than 40 members. Sometimes
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Updated on: 13 Sep 2017, 17:17

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

let m=number of members assume (m-3)/4-(m-3)/5=1 m=23 23/6 gives a remainder of 5 E

Originally posted by gracie on 15 May 2016, 13:10.
Last edited by gracie on 13 Sep 2017, 17:17, edited 1 time in total.

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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01 Dec 2016, 03:34

1

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

Question boils down to the following:

\(10 < X < 40\), and

\(x = 3 (mod_4)\) ---> dividing X into groups of 4 leaves remainder 3

\(x = 3 (mod_5)\) ---> dividing X into groups of 5 leaves same remainder 3

Then:

\(x = LCM (4, 5) + 3 = 23\)\(\)

Now we need to divide X into groups of 6 and find the remainder for the last table.

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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04 Dec 2016, 00:09

This One is a Great Official Question. Here is my solution => Let the number of members be N As per Question N=> (10,23) N=4k+3 N=5k'+3 Combing the above two Equations => N=20k''+3 N=> 3,23,43,63,83... But N=> (10,23) So N must be 23 Hence N=23 Now 23=> 6k+5 So 5 members will be left out when 6 members are arranged on each table. Hence E _________________

Club X has more than 10 but fewer than 40 members. Sometimes
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21 Jan 2017, 13:30

Hello All,

1. Members sit as 4m + 3 ( 4 members on all table and 3 on last table) 2. Members sit as 5n + 3 ( 5 members on all table and 3 on last table)

Since 10 < X < 40 - Range of Members.

So only possible number which is divisible by 4 & 5 is 20 and adding 3. Total members are 23. So now sitting with 6 members per table we have 3. 6m + 5 = 23. So remainder on last table are 5

Answer is E.
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Re: Club X has more than 10 but fewer than 40 members. Sometimes
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25 Mar 2017, 09:33

from the beginning we know the the number of people sitting by several tables is devisible by 4 and 5, hence it is multiply of 20. There is only one multiply of 20 in the range. 20+3 (since 3 people always set separately) 23/6=3tables +5 left 5 is the answer (E)

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30 Apr 2017, 07:20

1

Top Contributor

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

This is a nice remainder question in disguise. For this question, we'll use a nice rule that that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Let N = the TOTAL number of members.

Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables... With 4 members at each table, then N is multiple of 4 However, we still have one more table to consider. Since the last table has 3 members, we know that N is 3 greater than a multiple of 4 In other words, when we divide N by 4, the remainder is 3 By the above rule, some possible values of N are: 11, 15, 19, 23, 27, etc NOTE: I started at 11, since we're told that 10 < N < 49

Sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables Using the same logic as above, this question tells us that, when we divide N by 5, the remainder is 3 By the above rule, some possible values of N are: 13, 18, 23, 28, 33, 38

Let's check the two results. First we learned that N can equal 11, 15, 19, 23, 27, 31, 35, 38 Next we learned that N can equal 13, 18, 23, 28, 33, 38 Once we check the OVERLAP, we can see that N equals 23

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? If N = 23, then we'll have 3 tables with 6 members and the remaining 5 members will sit at the other table.

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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10 Jun 2017, 09:51

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

I just did the LCM of 5&4 +3 . So basically 20+3/6 . Not sure if this method will hold if the range of members was to change, but it worked on this instance.

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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09 Sep 2017, 04:34

Bunuel wrote:

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.

Hi Bunuel,

I approached the question like this:

Total members is 40>M>10, so 29 members. Now if each table has 6 members, that is 29/6 then the last table would have 5 (29-24) members on the table. Is this approach right? I did the problem in less than a minute with this approach. Thanks in advance.

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09 Sep 2017, 04:42

SinhaS wrote:

Bunuel wrote:

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.

Hi Bunuel,

I approached the question like this:

Total members is 40>M>10, so 29 members. Now if each table has 6 members, that is 29/6 then the last table would have 5 (29-24) members on the table. Is this approach right? I did the problem in less than a minute with this approach. Thanks in advance.

How did you get that there must be 29 members from 40>M>10? Does 29 members satisfy ANY of the conditions (3 members at one table and 4 members at each of the other tables and 3 members at one table and 5 members at each of the other tables)?
_________________

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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09 Sep 2017, 04:48

Bunuel wrote:

SinhaS wrote:

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.

Hi Bunuel,

I approached the question like this:

Total members is 40>M>10, so 29 members. Now if each table has 6 members, that is 29/6 then the last table would have 5 (29-24) members on the table. Is this approach right? I did the problem in less than a minute with this approach. Thanks in advance.

How did you get that there must be 29 members from 40>M>10? Does 29 members satisfy ANY of the conditions (3 members at one table and 4 members at each of the other tables and 3 members at one table and 5 members at each of the other tables)?[/quote]

I think my approach was not the right way to solve this problem, I took there can be a max 29 members within the range. Although i came to the same answer, this solution is not right i believe.

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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27 Feb 2018, 08:14

Quote:

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?

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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27 Feb 2018, 08:58

adkikani wrote:

Quote:

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?

Without having a common number of the members of the club, it is not possible to solve this problem. So, yes it is necessary for us to have a unique number of the members of the club.

However, we don't need to perform trial and error to arrive at such a number. In the range 10 < x < 40 Case 1: 11,15,19,23,27,31,35,39 are possible of form 3x + 4 Case 2: 13,18,23,28,33,38 are possible of form 5x + 3

The only overlap happens at 23 and can be assumed to be the number of people at the club.
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Re: Club X has more than 10 but fewer than 40 members. Sometimes
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27 Feb 2018, 09:29

adkikani wrote:

Quote:

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?

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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27 Feb 2018, 10:31

adkikani wrote:

Quote:

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?

Re: Club X has more than 10 but fewer than 40 members. Sometimes
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15 Oct 2018, 00:53

victoraditya wrote:

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 .

yes.. I used the same approach.But Bunnel's approach makes a lot of sense.