12 Days of Christmas GMAT Competition - Day 3: Of all the students in

Member of chess club = 22
Members of swim team = 10
Member of chess and swim = 2

Total members = Members of chess + Members of swim - Members of chess and swim
= 22 + 10 - 2
= 30 --- (1)

Female members = Members of swim team = 10
Male members = 2 * Female members = 20
Total students = male members + female members = 10 + 20 = 30 --- (2)

From (1) and (2), as there are no extra students who are not members of swim team or chess club, answer should be 0.

Answer: A

using a venn diagram:

we see, that chess has total 22 members, swim team has 10 members who are all female.
its given 2 members are part of both the teams.
also given male = 2*female

therefore all 2 common members and only swim are all female equaling 10.
only chess members are all male: 22-2=20 which is twice of females.

hence there are no remaining members which arent part of any group. answer is 0

number of students in chess club is 22
number of students in swim team is 10
2 students are common in both

and given that number of male students are double of female

And all the female are in swim team

To find the students in neither team, we have to first know the number of students, boys and girls

Total number of students in swim team is 10
If we consider the number of girls as less than 10 in swim team, the double of it(the boys count) will be less than 20, which contradicts the total number of students in both the teams

So we take the number of girls as 10 (the max possible number) and the number of boys become 2(10) = 20

Hence the total number of students in the school are 30, and all of them are in either swim team or chess club.

So, there are 0 students who are members of neither the chess or swim club.

[*]Let F be the number of females.
[*]Let M be the number of males. Given M=2F
[*]Total students = M+F=3F.
[*]Chess club members: 22. (C)
[*]Swim team members: 10. (S)
[*]Students in both the chess club and the swim team: 2.

We know that C U S = 32-2 => 30.

So Total - none = 30.
We also know that al female are in S. So max f=10, => f<=10.

This implies total <=30.
Hence none has to be 0.


So #students are members of neither the chess club nor the swim team is 0 (IMO A)

Of all the students in a certain school, 22 are members of a chess club, 10 are members of a swim team, and 2 are members of both. If all females are members of the swim team and the number of males is twice the number of females, how many students are members of neither the chess club nor the swim team?

A. 0
B. 3
C. 6
D. 9
E. 12

Total unique members in either club = 22+10−2 = 30......................(1)

All females are member of the swim team, and males is twice the number of females => M = 2F

Total no. of students in school = M + F = 3F

For (1) to satisfy, F has to be 10

Total no. of students in school = 30

Now, Total no. of students in school - Students in neither club = Students in Chess club + Students in Swimming club - Students in both clubs
=> 30 - Students in neither club = 22 + 10 - 2
=> 30 - Students in neither club = 22 + 10 - 2
=> Students in neither club = 0

Answer A.

Of all the students in a certain school, 22 are members of a chess club, 10 are members of a swim team, and 2 are members of both.

If all females are members of the swim team and the number of males is twice the number of females, how many students are members of neither the chess club nor the swim team?

Let the number of students who are neither members of chess club nor the swim team be x

	Chess	Not Chess	Total
Swim	2	8	10
Not Swim	20	x	20+x
Total	22	8+x	30+x

Total students = 30 + x

Let the number of females be f
Number of males = 2f
Total students = f + 2f = 3f

3f = 30 + x
Since all females are members of the swim team, maximum value of f = 10

30 = 30 + x
x = 0

The number of students who are members of neither the chess club nor the swim team = 0

IMO A

Problem Analysis:
- Total chess club members = 22
- Total swim team members = 10
- Members of both clubs = 2

Calculating Exclusive Memberships:
- Chess only members = Total chess - Both clubs = \(22 - 2 = 20\)
- Swim only members = Total swim - Both clubs = \(10 - 2 = 8\)

Total Members in Clubs:
- Total club members = Chess only + Swim only + Both = \(20 + 8 + 2 = 30\)

Gender-based Team Membership and Ratios:
- All females (F) are on the swim team.
- Number of males (M) is twice the number of females (M = 2F).

Total Population Breakdown:
- Let x be the number of students neither in chess nor swim.
- Total students = Club members + Neither = \(30 + x\)

Relationship of Total Males and Females to Club Membership:
- Total swimmers (including both clubs) = 10, hence \(F \leq 10\)
- Using the ratio of males to females, \(M + F = 30 + x\) and \(M = 2F\), we substitute to find:
- \(2F + F = 30 + x \Rightarrow 3F = 30 + x\)

Determine Maximum Females on Swim Team:
- Since \(3F = 30 + x\) and \(F \leq 10\), this implies \(3F \leq 30\), setting \(x + 30 \leq 30 \Rightarrow x \leq 0\).

Concluding the Value of x:
- Given x must be a non-negative integer and x \leq 0, the only possibility is \(x = 0\).

Answer:
- The number of students who are members of neither the chess club nor the swim team is 0.

Correct Answer: A. 0

	Chess	Swim	Both	Neither	Total
Male					2x
Female	0	x	0	0	x
Total	22	10	2	N

22+10+2+N = 3x
x = (N + 34)/3

For x to be an integer N+34 must be a multiple of 3

Putting all the values of N from options
When N =
0 x=34/3 not divisible
3 x=37/3 not divisible
6 x=40/3 not divisible
9 x=43/3 not divisible
12 x=46/3 DIVISIBLE

Option E

With given info we can fill the numbers in the above table. From another piece of info which says no of males(M)=2F, its already given all females are members are swimmers so

F=10 and therefore M=20, total should be 30. With this we can fill the rest of info and get neither swimmer or chess club members as '0'.

Since in this case we have 22 people in chess club and 10 people in swim club and 2 in both of them.
Also, total number of males is twice the number of females.

In order to satisfy 2nd point we need atleast 10 females and therefore males will be 20.

Now chess club has 20 males and 2 females and swim club will have all 10 females.

Hence, there are no student who is neither a part of swim club and chess club.

All the females are members of swim team => # of females = 10
Since # men =2*(# of female)
# men = 20
i.e. there are total 30 students in the school.

# of students playing Chess OR Swimming:
(C OR S) = C + S - ( C AND S)
(C OR S) = 22 + 10 - 2 =30

Students playing neither chess nor swimming = total - # of students playing Chess OR Swimming=30 - 30 =0

Hence answer is option A) 0

The total number of students is 3f (where f is the number of females and 2f is the number of males).

All females are on the swim team, and there are 10 members in the swim team, so f ≤ 10.

The number of students in at least one club is 30 (22 in the chess club, 10 in the swim team, minus 2 in both). The total number of students is 30 + x (where x is the number of students in neither club).

The equation becomes 3f = 30 + x. Since f must be ≤ 10, the only valid value for f is 10, because any value less than 10 would make x negative, which isn’t possible.

With f = 10, we find x = 0. Thus, the number of students in neither club is 0, and the answer is A. 0

22 in the chess club, 10 in the swim team, 2 in both
All females are in the swim team
male are twice the females.
Students in either club
22+10−2=30 2
Females : Since all females are in the swim team:
F=10F = 10F=10, highest possible value is 10
Total students:
3F=3×10=30
Students in neither club:
IMO A

Total number of member = chess + swim - both = 22 + 10 - 2 = 30

Given all female are swim team, so 10 (include 2 as chess club) will the female number

Male = twice of female = 10 *2 = 20

So Neither students are members of chess club nor the swim team is 30-10-20 = 0
The answer is A

Let me break this down to make it easy


We are tasked with determining the number of students who are not members of either the chess club or the swim team, given the following data:

1. Chess Club Members: 22
2. Swim Team Members: 10
3. Members of Both: 2
4. All females are on the swim team.
5. The number of males is twice the number of females.

Step 1: Total number of students in the chess club or swim team
Using the inclusion-exclusion principle
Total in chess club or swim team =(Chess Club Members)+(Swim Team Members)−(Members of Both). {Total in chess club or swim team} = ({Chess Club Members}) + ({Swim Team Members}) - ({Members of Both}). Total in chess club or swim team=22+10−2=30.{Total in chess club or swim team} = 22 + 10 - 2 = 30.

Step 2: Total number of students in the school
Let the number of females be ff. Since the number of males is twice the number of females, the total number of students is:
Total students=f+2f=3f.{Total students} = f + 2f = 3f.
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Step 3: Total students not in the chess club or swim team
The total number of students who are members of neither is:
Students in neither=Total students−Students in chess club or swim team. {Students in neither} = {Total students} - {Students in chess club or swim team}. Students in neither=3f−30.{Students in neither} = 3f - 30.
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Step 4: Determine ff (number of females)
All females are swim team members, and the total number of swim team members is 10. Therefore, the number of females is:
f=10.
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Step 5: Calculate the total number of students
Total students=3f=3×10=30.{Total students} = 3f = 3 times 10 = 30.
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Step 6: Calculate the number of students in neither
Students in neither=3f−30=30−30=0. {Students in neither} = 3f - 30 = 30 - 30 = 0.
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Final Answer:
The number of students who are members of neither the chess club nor the swim team is 0.

The answer is A 0.

Chess - C Swim -S Female = F and Male = M , F=2M

Total students = F+M = F+2F = 3F

students either in C or S = 22+10-2 = 30

total students = 3F = 30 ; F=10 ; M=20 --> (3F = T = 30)

Students in neither chess or swim = total students - students in either chess or swim

= 30-30 =0