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

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Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 12 Dec 2012, 06:31
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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 the membe  [#permalink]

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New post 12 Dec 2012, 06:39
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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 the membe  [#permalink]

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New post 18 Dec 2012, 13:20
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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 the membe  [#permalink]

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New post 14 Dec 2012, 03:16
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Ans:

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 the membe  [#permalink]

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New post 30 Apr 2014, 06:26
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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 the membe  [#permalink]

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New post 28 May 2014, 10:09
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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 the membe  [#permalink]

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New post 27 Aug 2014, 02:58
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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 the membe  [#permalink]

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New post 30 Oct 2014, 22:54
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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 the membe  [#permalink]

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New post 29 Mar 2015, 02:32
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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 the membe  [#permalink]

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New post 15 Jun 2015, 18:21
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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 the membe  [#permalink]

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New post 15 Jun 2015, 23:17
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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 the membe  [#permalink]

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New post 11 May 2016, 00:35
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10<x<40
in both cases 3 members sit on one table
remaining members = x-3
so 7<x-3<37
x-3 should be multiple of 4 and 5 to fit all members perfectly on tables.
between 7 and 37 only 20 is such number.
x-3 = 20
x= 23
If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table
=> 23/6
remainder ---> 5
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 11 May 2016, 07:05
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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


Although this problem appears to be a general word problem it is actually testing us on our understanding of remainders when dividing integers. We are first told that the total number of members, which we can denote as “T”, is between 10 and 40. Next, we are told two important pieces of information:

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

2) “Sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables.”

Let’s now translate these into two mathematical expressions.

1) T/4 = Quotient + Remainder 3

2) T/5 = Quotient + Remainder 3

Because T is being divided by 4 and 5, we are really looking for the following:

T/20 = Quotient + remainder 3.

Since T must be between 10 and 40, there is only one value in that range which, when divided by 20, produces a remainder of 3. That value is 23. We can now use this value to complete the question. We are finally asked:

“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?”

This is same as asking the following: what is the remainder when 23 is divided by 6? We can see that 6 divides into 23 3 times with a remainder of 5.

Answer: E.
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 14 May 2016, 22:36
If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table

What does the statement above means? specially the "Except one". I got tricked by it thinking that 22 (23 except 1 or 22-1) members were seated with 6 members at each table :(
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 14 May 2016, 23:09
powellmittra wrote:
If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table

What does the statement above means? specially the "Except one". I got tricked by it thinking that 22 (23 except 1 or 22-1) members were seated with 6 members at each table :(


hi

Quote:
If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table

this tells us that if there were x tables, x-1 tables had 6 members on those table and <6, say y, on the ONE remaining table..
MEANS - y is the remainder when you divide TOTAL by 6
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 01 Dec 2016, 04:34
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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


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.

\(\frac{23}{6}\) -----> \(rem = 5\)

Answer E.
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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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


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.

Answer:

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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 09 Sep 2017, 05:34
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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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 09 Sep 2017, 05:42
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)?
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe  [#permalink]

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New post 27 Feb 2018, 09:14
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?

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


pushpitkc niks18 Hatakekakashi

What is wrong with below thinking?

I know my total members in X have to be between 11 and 39.
Quote:
Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables,

9*4 = 36
1*3 = 3
Total = 39 . .. . .(1)

Quote:
and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables

7*5 = 35
1*3 = 3
Total =38 .. ... .(2)

Do i need to perform trial and error to make (1) = (2) ?
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe   [#permalink] 27 Feb 2018, 09:14

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