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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]

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

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23 May 2016, 09:02
Was it coincidence that I got this answer right?

I thought X= 29 since 10<X<40. ( 39-11+1)
So, if each table has 6 members and 1 table has less than that. Then,
29/6 will leave a remainder 5.

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

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01 Dec 2016, 03:34
1
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.

$$\frac{23}{6}$$ -----> $$rem = 5$$

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

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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

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

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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

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

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14 Feb 2017, 02:43
Let,the x represents the members.

Now, the members are more than 10 but fewer than 40.

so, 10
By following the question, we get

(1) Sometimes the members sit at tables with 3 members at one table & 4 members at each of the others.

so, x = 3+ 4 m

or x-3 = 4 m. -----(A)

(2) Again, sometimes they sit at tables with 3 members at one table & 5 members at each of the other tables.

so, x = 3+ 5 n

or, x-3 = 5n. -------(B)

here, ( x-3) is the multiple of 4 & 5. or 4*5= 20.

But there are also many multiple of 4 & 5

i. e 40,60,80. ....

but if we take another multiple of 4 & 5 then it will break the condition 10
so, here x= 20 satisfies the above condition.

Now, x-3 = 20

or, x = 23.

Following the question, dividing 23 by 6

23/6

here, quotient = 3
Remainder = 5.

Posted from my mobile device
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Re: Club X has more than 10 but fewer than 40 members. Sometimes  [#permalink]

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14 Feb 2017, 02:47
Let,the x represents the members.

Now, the members are more than 10 but fewer than 40.

so, 10
By following the question, we get

(1) Sometimes the members sit at tables with 3 members at one table & 4 members at each of the others.

so, x = 3+ 4 m

or x-3 = 4 m. -----(A)

(2) Again, sometimes they sit at tables with 3 members at one table & 5 members at each of the other tables.

so, x = 3+ 5 n

or, x-3 = 5n. -------(B)

here, ( x-3) is the multiple of 4 & 5. or 4*5= 20.

But there are also many multiple of 4 & 5

i. e 40,60,80. ....

but if we take another multiple of 4 & 5 then it will break the condition 10
so, here x= 20 satisfies the above condition.

Now, x-3 = 20

or, x = 23.

Following the question, dividing 23 by 6

23/6

here, quotient = 3
Remainder = 5.

Posted from my mobile device
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Posts: 156
Re: Club X has more than 10 but fewer than 40 members. Sometimes  [#permalink]

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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

Be careful with adding 3 at the beginning!!!
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Re: Club X has more than 10 but fewer than 40 members. Sometimes  [#permalink]

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30 Apr 2017, 07:20
1
Top Contributor
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.

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

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03 May 2017, 13:38
lcm of 4 & 5=20
the general number=20a+(20+3)=20a+23
(20a+23)/6
23/6 [a=0]
remainder=5
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Re: Club X has more than 10 but fewer than 40 members. Sometimes  [#permalink]

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

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09 Sep 2017, 04:34
Bunuel 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.

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  [#permalink]

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09 Sep 2017, 04:42
SinhaS wrote:
Bunuel 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.

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  [#permalink]

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09 Sep 2017, 04:48
Bunuel wrote:
SinhaS 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.

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

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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?

(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  [#permalink]

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27 Feb 2018, 08:58
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) ?

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  [#permalink]

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27 Feb 2018, 09:29
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) ?

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  [#permalink]

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27 Feb 2018, 10:31
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) ?

pushpitkc and niks18 have already posted what i wanted to say.. sorry for the delayed response

always prepare cases and 3K=4 = 3(k+1) +1
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Re: Club X has more than 10 but fewer than 40 members. Sometimes  [#permalink]

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

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02 Nov 2018, 04:31
i let the members be and considered the # of tables: meaning if 1 table is occupied with 3 mems, then x-1 remain:

range(10 to 40)--> 4(x-1) + 3(1)
range(10 to 40)--> 5(x-1) + 3(1)

then i found out the common number 23 and got the equation

range(10-40)--> 20(x-1) + 23. The only value that satisfies the range is 1, thus 23 is the number of members.
and 23/6 leaves a rem of 5
Re: Club X has more than 10 but fewer than 40 members. Sometimes &nbs [#permalink] 02 Nov 2018, 04:31

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