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

Question

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?

Let No.of members be N
N=3 + 4p=3+5m
=>4p=5m
p=5,m=4
p=10,m=8 (these values and above these values are not possible because N will then exceed 40)
therefore N = 3 +4x5=23

As per required condition,
23=6x3+5=18+5

Therefore E

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

Bunuel · Accepted Answer

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.

fukirua · Answer

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

ScottTargetTestPrep · Answer

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.

priyamne · Answer

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).

Tmmmmy · Answer

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

GDR29 · Answer

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

imperfectdark · Answer

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

Attachments

Tables.png [ 4.85 KiB | Viewed 132459 times ]

victoraditya · Answer

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 .

TudorM · Answer

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?

KarishmaB · Answer

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".

vitaliyGMAT · Answer

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.

BrentGMATPrepNow · Answer

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:  E

RELATED VIDEO
 https://www.youtube.com/watch?v=XCxgIR042tE

GMATinsight · Answer

Please find solution attahed

avigutman · Answer

Video solution from Quant Reasoning starts at 21:33: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=_uwR9AZas9M

EMPOWERgmatRichC · Answer

Hi All,

This prompt is wordy, but it gives us information in a logical order (which we can take notes on AS we read):

1) Club X has MORE than 10 members but FEWER than 40.
2) The members can sit with 3 at ONE table and 4 at EACH of the other tables.
3) The members can sit with 3 at ONE table and 5 at EACH of the other tables.

We’re asked if 6 members sit at EACH table except for ONE additional table (which will have FEWER than 6 members sitting at it), how many members will be at that last table.

This question has a great ‘brute force’ aspect to it. The TOTAL number of members MUST be ‘3 more’ than a multiple of 4 AND ‘3 more’ than a multiple of 5. Since we know that there are FEWER than 40 members in total, there can’t be that many integers that fit BOTH of those restrictions, so we can just list out the possibilities for each and find the one that matches…

‘3 more’ than a multiple of 4: 11, 15, 19, 23, 27, 31, 35, 39
‘3 more’ than a multiple of 5: 13, 18, 23, 28, 33, 38

The only number that appears in both lists is 23, so that must be the total number of members. That would create three tables of 6 and one table of 5.

Final Answer:  E

GMAT Assassins aren’t born, they’re made,
Rich

Abolnasr · Answer

My approach:
Let number of patrons be X

This X need 1 more to be divided evenly on 4
And needs 2 more to be divided evenly on 5

The only number with this criteria from 10 to 40 is 23

23+1=24, 23+2=25

23/6=3R5

Paras96 · Answer

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?

Let No.of members be N
N=3 + 4p=3+5m
=>4p=5m
p=5,m=4
p=10,m=8 (these values and above these values are not possible because N will then exceed 40)
therefore N = 3 +4x5=23

As per required condition,
23=6x3+5=18+5

Therefore E

Let No.of members be N 
N=3 + 4p=3+5m
=>4p=5m
p=5,m=4
p=10,m=8 (these values and above these values are not possible because N will then exceed 40)
therefore N = 3 +4x5=23

As per required condition,
23=6x3+5=18+ 5

Therefore E

Abolnasr · Answer

Wow I tried to solve it systemically and it worked!
X/4=K+3/4
X/5=Y+3/5
=====
X = 4K + 3
X = 5 Y + 3
K = 5 
Y = 4

X = 23
X/6=3R5

Ishaan24 · Answer

From the 1st combination: 
- 3 ppl at only one table and 4 each at the rest => the total number of people is ODD. 
- So options B & D are eliminated.

Taking the 1st 2 combinations: Let x be the total number of ppl (which is ODD)
=> x= 4m+3 & x= 5n+3 (1 table of 3 ppl separate in each combination)

=> x-3 is divisible by 4m & 5n 
=> x-3 = 20 
=> x = 23 => 6*3 + 5 - Final table combination => 5 at the individual table.

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

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