A local club has between 24 and 57 members. The members of the club ca

We have to find the number of members in the final group when the groups are formed by 2 members max(pair).
So final group will either have 2 members or 1 member (leaving us with choice A and choice B).

For the first condition, assume that there is x number of the members in each group except final one. Total members= x*4+3
For the second condition, assume that there is y number of the members in each group except final one. Total members= y*5+3
x*4+3=y*5+3
4x=5y

the possibe integer values that can satisfy above equation are x=5, y=4 or x=10, y=8 or x=15, y=12

We are given that the number of the members should be between 24 and 57

the only value that satisfy above condition is x=10, y=8

total number of the members 10*4+3=43=> odd number, the number of members in the last group will certainly be 1

The correct answer is A.
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The members of the club can be separated into groups of which all but the final group, which will have 3 members, will have 4 members -

This means the number will be a multiple of 4+3 so the number could be 24+3, 28+3, 32+3, 36+3,40+3 etc. Right here is enough to prove that when this number is divided by 2 (pairs) we will get a remaining group of 1 but just to be sure we go to second statement. Here we can say that the number of members is a multiple of 5 and then a+3 so 25+3, 30+3, 35+3 or 40+3

43 is common, 43/2 will give a remainder of 1.

This could be a great DS question too.

lowest possible membership=4*5+3=23
23/5=Q4+R3
23/4=Q5+R3
23/11=Q2+R1
next lowest membership=23+4*5=43
43/21=Q2+R1
1
A

all but the final group, which will have 3 members, will have 4 members.
all group=4 members (divisor)
final group=3 members (remainder)
N(total member)=4a+3

all groups but the final group, which will have 3 members, will have 5 members
all group=5 members (divisor)
final group=3 members (remainder)
N(total member)=5b+3

N=N
4a+3=5b+3
4a=5b
so, 4a=5b=20 [lcm of 4,5]
N=20+3=23

so, first (total number of member) satisfying 2 conditions= 23
but [A local club has between 24 and 57 members]. so, we need to find a general number that will satisfy all condition.

Finding next pool
lcm of 4,5=20
The general number=20c+23 [will satisfy all condition] [where c>0]

Question asked: If the members are separated into many groups of 6 (dividend) then how many members will be in the final group (remainder)?


(20c+23)/6
[(18c+18)+(2c+5)]/6

(18c+18)/6 = remainder 0
(2c+5)]/6= remainder 1 [c=1]

(rem 0 + rem 1)/6
= remainder 1

Answer: E

(n-3)/4-1=(n-3)/5
n=23
23+4*5=43
43/6 gives a remainder of 1
E

Let’s let x = the number of initial groups. The number of members in these x groups will be 4x. The final group has 3 members. Thus, the total number of members is 4x + 3.

Similarly, the number of initial 5-member groups can be represented as y, and the final group has 3 members. Thus, the total number of members is equal to 5y + 3.

From the information above, we deduce that three less than the number of members must be a multiple of both 4 and 5; therefore, the number of members must be 3 more than a multiple of 20. The total membership can’t be 23 members because 23 is below the lower bound provided for the members. Similarly, the number of members can’t be 63, which is higher than the upper bound provided for the number of members. The only possible number left is 43, and it is consistent with the provided information about the number of members.

Now, since 43 divided by 6 produces a remainder of 1, when the 43 members are separated into groups of 6, there will be one person left for the final group.

Answer: E

There are 2 givens:
1) We can separate the members into number of groups, all groups are 4 members, only 1 group is 3.
2) We can separate the members into number of groups, all groups are 5 members, only 1 group is 3.

Now subtract 3 (last group members) from the possible members,
Therefore the remaining members are from 21 to 54.

The number has to be divisible by both 5 & 4 (Note the above givens),
The only number in the range (21 to 54) that satisfies this condition is 40.
Therefore the total number of members are 43.

Now, we need to separate 43 members into groups, where the number of groups that contains 6 members are maximized.
Then we can divide 42 members into 7 groups leaving only 1 member in the last group.

So the answer is 1 (E)

Since 2 leaves a remainder of 0 or 1 , ans is A

Came to say this. Easiest 2sec approach.

I am not sure whether this is the correct way to look at the problem but it does get the answer:

It says maximum groups possible: minimum number of people required in a group i.e 2
Therefore any odd number will leave 1 as a remainder. No need to look at the pairing mentioned in the question.

Kindly if anyone can verify about this approach, it would be really helpful.
Thanks.

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