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

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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, 09:58
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?

(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) ?


adkikani

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 the membe  [#permalink]

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New post 27 Feb 2018, 10:29
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?

(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) ?


Hi adkikani

As you rightly mentioned that total number of members have to be in the range 11 to 39.

In the first scenario we have 3 member at One table and 4 members at other let's say \(x\) tables

So number of members will be \(= 4x+3\)

In the second scenario we have 3 member in One table and 5 members at other let's say \(y\) tables

So from this we have number of members \(= 5y+3\)

but number of members in both scenarios has to be equal so \(4x+3=5y+3=>4x=5y\)

Now in the given range there is only one possibility for the above equation to hold true \(4x=5y=20\).

Hence number of members \(= 20+3=23\)

In your method you will have to form two sets as per the given relationship and then find the common value, which will be 23
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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 05 Dec 2018, 23:24
x = number of members

x-3 is divisible by 4
x-3 is divisible by 5

If x-3 is divisible by 4 and 5 it must also be divisible by the LCM(4,5), which is 20.

Thus (x-3)/20 = (some Integer)
x-3 = 20(*some integer*)
X = 20(*some integer*) + 3
lets test values
x = 20(1) + 3 = 23
x = 20 (2) + 3 = 43 wait... this is outside the range of 10<x<40, so x must be 23

Thus 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 22 Jun 2019, 02:59
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, am writing on the forum first time i think, must say am a huge huge fan of your work...

Umm, could you please explain your method once more for my slow brain to get..
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Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe   [#permalink] 22 Jun 2019, 02:59

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