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805+ (Hard)|   Overlapping Sets|         
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Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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The ratio of the three activity is given to us

Hiking : Swimming : Archery = 5 : 6 : 11

(1) All students enrolled in Hiking are also enrolled in Archery.

Let's assume there are just 5 students in hiking, 6 students in swimming and 11 students in archery.

Among these 11 students who are enrolled in archery 4 students are enrolled in swimming as well.

Ratio = 4/11

Another possibility could be among these 11 students who are enrolled in archery 2 students are enrolled in swimming as well.

Ratio = 2/11

We can have multiple possible combinations.

Statement 1 is not sufficient.

(2) Every student enrolled in Archery is also enrolled in at least one other group.

Let's assume there are just 5 students in hiking, 6 students in swimming and 11 students in archery.

We have to divide these 11 students in a manner that they are assigned to either Hiking or Swimming. As the max capacity of Hiking is 5, the remaining must be in swmming. Hence there is no possible way other than to put 5 in hiking and 6 in swimming.

Ratio is always 6 /11

The statement alone is sufficient.

Option B
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GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 11:42
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Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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There are three activity groups : Hiking H, Swimming S, and Archery A.

Every student is enrolled in at least one of these groups H, S, and A.

H: S: A = 5x : 6x : 11x

To find : ( swimming + archery) / archery ?

Statement 1:

(1) All students enrolled in Hiking are also enrolled in Archery.

if 5 students in H enroll for A, out of 11 in A, we see 5 is also enrolled for H.

Remaining 6 can be either A only or ( A+S).

Hence Insufficient

Statement 2:

(2) Every student enrolled in Archery is also enrolled in at least one other group.

if every student in A is enrolled in at least one other group.

out of 11x = 5x is present in Hiking. And 6x out of 11x is present in Swimming and archery.

Fraction = 6x/11x = 6/11.

Hence, Sufficient

Option B
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Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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asked: ratio of students in Archery and swimming to total number of students in archery

Statement 1:
No mention about number of students enrolled in swimming. Hence B,C,E

Statement 2:
Although it it mentioned about other group but no mention about only swimming. So C, E

Combined 1 and 2:
No mention of students that are in all 3 activities so difficult to find the asked ratio Hence Option E
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Based on the question stem we can determine the ratio of Hiking:Swimming:Archery to be 5:6:11. Statement 1 is insufficient because it doesn't provide us with information needed to determine the overlap of swimming and archery. Statement 2 alone is sufficient because we now know that each student in archery will take on at least one other group. Because we have our ratio to be Hiking and swimming being exactly the same amount of students as the archery group we know that there is only one possible outcome, being that for each 11 students in the archery group: 5 will also be in group hiking and 6 will be in the swimming group. Answer B.

Regards,
Lucas
Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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(1) if all students are enrolled in hiking are also in archery than for the ratios 5 out of the 11 archery students are also hikers. This itself is insufficient. (2) The archery students could all be in swimming or hiking we don't know, this is insufficient. Together, we know that 5 are in hiking so the other 6 must be in swimming. So C. sufficient together.
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The stem says the ratio between the sports is as follows = H:S:A = 5:6:11

We need to find kids enrolled in (S + A)/A

1) Basically hiking is a subset of archery, but what about the swimming kids? Maybe there's no overlap at all, maybe just 1 of the 6 are enrolled in archery.. we don't know. So this doesn't seem too useful to find the ratio

2) This means all the 11x kids are also enrolled in swimming and hiking, notice 5+6 = 11 only... so 6x kids are those who are enrolled in swimming + archery
we know total archery is 11x. We can find the fraction now = 6x/11x = 6/11 - sufficient. B
Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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per stem: 5x:6x:11x is the ratio of the three groups = 22 total parts * X --> need to solve for X

statement 1: 5X=11X ; doesn't seem sufficient to me. No count info provided to determine x, and there for # of each category

statement 2: similar story; no count info provided.

perhaps together, it can be deduced, but my math skills are not that good yet, so although guessing E isn't best practice when you're stuck, I will guess E in this instance.
Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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IMO, both statements together are sufficient. Option C is the correct answer. Let's take a look how,

Lets assume there are 5 hikers, 6 swimmers and 11 archers

Statement I alone: " All students enrolled in Hiking are also enrolled in Archery."
It means all hikers are archers. But we do not have any information regarding swimmers in this option. Therefore, insufficient.

Statement II alone: " Every student enrolled in Archery is also enrolled in at least one other group.
"
This means all archers are also either swimmers or hikers or both. It doesn't give us any information about the fraction enrolled in swimming.

Statement I and II together:
Out of 11 archers, 5 are hikers and 6 are swimmers.

Therefore option C, is the correct answer.

Some Archers

Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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As we don't know the total number of students, the given statement seem to be insufficient (we have only ratio information).
Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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Given:

3 activities: Hiking (h), Swimming (s) and Archery (a)
And every student enrolled in at least one class.
H:S=5:6
H:A=5:11

Question:

Fraction?: students enrolled in archery and swimming / Students in archery

Statement 1: All students enrolled in hiking also in Archery.

So lets assume Hiking has 5 students, who are all part of archery. and we know for 5 students there are 11 archery students. and 6 in swimming.

5 students out of 11 have swimming, remaining 6 might have swimming or none of them could have swimming we cant determine from information given.

This statement is not sufficient.

Statement 2: Every student enrolled in Archery is also enrolled in at least one other group.

We know for every 5 Hiking students we have 11 Archery students and 6 Swimming students, so ratio h:a:s=5:11:6 needs to be maintained at all time

So if we assume 11 archery students, they can be divided into hiking and swimming. Lets try putting maximum number of archery students in hiking. but we know ratio of 5:11 is there so for 11 students in archery we have 5 students in hiking and 6 students are left.
Now, 6 students has to be in some other group so they all will be part of swimming and swimming has 6 member capacity as well.

So fraction will be 6/11

Statement 2 alone is sufficient.

Answer:B

Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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Thats a cool one!

So, what we are looking for is basically any hints about the relations that include Archery and/or swimming.

(1) - Alone insufficient, as we now know, that 5 of the 11 of the Archer-Students does Hiking as well. But we dont know, if there are any "Only-Swimming Students"
(2) - Alone insufficient, as this does not help us at all. We dont know, how this translate to the distribution to the two options given.

Together we know, that all Students Hike, do swimming as well.
So we have two options for the Swimming ones.
Swimming -> Hiking -> Must swim
Swimming -> Hiking.

Therefore, we have a 1:1 relation of people that swim and do archery.
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None of the statements is relatable to swimming by itself.

Together, we know:
5 are in hiking and archery (statement 1)
6 remaining from archery can't be in hiking, hence, must be in swimming too (statement 2)

Would mean 6/11 are in both, BUT the 5 in Hiking/Archery could be in Swimmig as well, making both statements together not enough.

E. Correct answer
Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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Hiking : swimming : Archery = 5:6:11

S1 - If all students enrolled in hiking are also enrolled in Archery out of every 11 students in Archery 5 are in hiking as well. But we don't have any info about 6 other students out of 11 students, whether they are into swimming or not, So insufficient

S2 - If Every student enrolled in Archery is also enrolled in at least one other group and there are Exactly 3 activities, this means all of the 11 students who are in archery are also enrolled in either swimming or hiking. But since we don't have more information, we can't say among those enrolled in Archery, what fraction are also enrolled in Swimming.
Insufficient

S2+S1 = of the 11 students who are in archery 5 are in hiking this means 6 are enrolled in swimming. so
students enrolled in swimming/students enrolled in archery = 6/11
Sufficient
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Hiking = H = 5*a
Swimming = S = 6*a
Archery = A = 11*a

being a some positive integer.

We can reason assuming a=1, so H=5, S=6 and A=11

The question is:
(Archery intersect Swimming)/Archery?

(1)
The 5 students in H are also in A.
We don't know how many of the remaining 11-5=6 in Archery are also in Swimming.

INSUFFICIENT

(2)
No student is only in Archery. No enough data to answer the question.

INSUFFICIENT

(1)+(2)
As no student is only in Archery, the remaining 6 in Archery (taking into account (1)) must be in Swimming. It is consistent with the fact that S=6.

6/11 is the answer.

SUFFICIENT

IMO C
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At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.

# of students' ratio of Hiking : Swimming : Archery = 5 : 6 : 11
Assuming 5 / 6 / 11 students in each,
(1) Just Archery or Archery and Swimming = 11 - 5 =6 >> but doesn't tell the exact enrolled in Swimming out of 6. Insufficient
(2) Every studnet in Archery: Archery and Hiking, or Archery and Swimming, or Archery and Hiking and Swimming >> but doesn't tell the exact enrolled in Swimming. Insufficient
(1+2) All Hiking = 5 enrolled in Archery, but we won't know how many of them are included in Archery and Hiking or Archery and Hiking and Swimming, each. >> Insufficient

Answer: (1) and (2) together are not sufficient
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Let's use a simpler representation with the relationships.
Let region a be students only in H.
Let region b be students only in S.
Let region c be students only in A.
Let region d be students in H and S only.
Let region e be students in H and A only.
Let region f be students in S and A only.
Let region g be students in H, S, and A.

From the problem statement:
n(H)=a+d+e+g=5x
n(S)=b+d+f+g=6x
n(A)=c+e+f+g=11x
Every student is in at least one group: a+b+c+d+e+f+g=N total

From Statement (1): All students enrolled in Hiking are also enrolled in Archery (H⊆A).
This means:

a=0 (no one is only in H).

d=0 (no one is in H and S but not A). This is because if d>0, then there are students in H and S but not A. If H⊆A, then any student in H must be in A. So, if they are in H and S, they must also be in A. So d must be 0.
So, n(H)=e+g=5x.

From Statement (2): Every student enrolled in Archery is also enrolled in at least one other group.
This means no student is only in Archery.
So, c=0.

Now combine (1) and (2):
a=0
d=0
c=0

Let's rewrite the initial equations with these zeroes:
n(H)=e+g=5x
n(S)=b+f+g=6x
n(A)=e+f+g=11x

We need to find
n(S∩A)/ n(A) = f+g/ (e+f+g)
​
From n(A)=e+f+g=11x.
We have e+g=5x.
Substitute this into the equation for n(A):
n(A)=(e+g)+f=11x
5x+f=11x
f=6x

Now we have f=6x.
We need to find
f+g / 11x
​
We still need g.

Let's use the total number of students. "Every student is enrolled in at least one of these groups."
N total =a+b+c+d+e+f+g.

With a=0,c=0,d=0:
N total =b+e+f+g.
We know e+g=5x and f=6x.
So, N total=b+5x+6x=b+11x.

We also know n(S)=b+f+g=6x.
Substitute f=6x:
b+6x+g=6x
This implies b+g=0. Since b and g represent counts of students, they must be non-negative.
Therefore, b=0 and g=0.

Now we have:
a=0
b=0
c=0
d=0
g=0

Let's check consistency:
n(H)=e+g=e+0=e=5x.
n(S)=b+f+g=0+f+0=f=6x.
n(A)=c+e+f+g=0+e+f+0=e+f=11x.
Is e+f=5x+6x=11x? Yes, it is consistent.

So, the only non-zero regions are e (Hiking and Archery only) and f (Swimming and Archery only).
This means:

Students in Hiking are only in Hiking and Archery (e).

Students in Swimming are only in Swimming and Archery (f).

No student is in all three groups (g=0).

No student is only in one group (a=0,b=0,c=0).

No student is in Hiking and Swimming only (d=0).

The question asks for the fraction of students enrolled in Archery who are also enrolled in Swimming.
This is

n(S∩A)/n(A)
n(S∩A)=f+g. Since g=0, n(S∩A)=f.
n(A)=e+f+g. Since g=0, n(A)=e+f.

So we need to find f/ (e+f)

We found f=6x and e=5x.
Therefore, the fraction is
6x / (5x+6x) = 6x/11x = 6/11

This means that combining both statements is sufficient to answer the question.

Final answer is C
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GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 19:26
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Correct Answer: Statement 1 and Statement 2 are not sufficient
We need to find, "among those enrolled in Archery, what fraction are also enrolled in Swimming"?
Given we have
Let x be number of students
Hiking= 5x
Swimming= 6x
Archery= 11x

(1) All students enrolled in Hiking are also enrolled in Archery.
Here in Archery we have 11x and from that 5x are there in Hiking so remaining 6x can be in Swimming or Archery.
But here we don't know exactly how many students opted for both.
So this statement is not sufficient.

(2) Every student enrolled in Archery is also enrolled in at least one other group.
Here in Archery we have 11x and this 11x can have Archery (A), Hiking(H) or Archery(A), Swimming(S) or in all three
Let’s assume
Archery (A), Hiking(H)= a
Archery(A), Swimming(S)= b
all three= c
we need to find b+c/11x
But we still don't know values of a, b and c.
So this statement is not sufficient too.

If we combine both the statements.
than we have a= 5x, b=6x
But we still don't know value of c so this also becomes insufficient.
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GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 19:47
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Lets break it down
Consider H=Hiking, A=Archery and S=Swimming

(1) All students enrolled in H are also enrolled in A.
Gives that for all students in H also belongs to A, but no relation about A&S is given.
Insufficient

(2) Every student enrolled in Archery is also enrolled in at least one other group.
Given that all of A also also part of either H or S. But there is no specific number given.
Insufficient.

Combining 1 & 2
Using 1 All H belong to A, making
A= 11 - H = 11 - 5 = 6 students in A but not in H.
Using 2 Since all of A are part of either H or S.
Unique A = 0
A & H = 6
Leaving A & S = 11 - (A & H) = 11 - 6 = 5

Therefore ratio would be (A&H)/A = 5/11
Sufficient!
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GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 20:05
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Statement 1:
This statement does not give enough information; there could be either of the options below, or something in between them.


Statement 2:
Since the ratio is 5:6:11, if everyone in archery is also in another group, the only option to keep the ratio right is for everyone to be in exactly 2 groups.




Bunuel
At a summer camp, there are exactly three activity groups: Hiking, Swimming, and Archery, and every student is enrolled in at least one of these groups. For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery. Among those enrolled in Archery, what fraction are also enrolled in Swimming?

(1) All students enrolled in Hiking are also enrolled in Archery.
(2) Every student enrolled in Archery is also enrolled in at least one other group.


 


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Statement (1) All students enrolled in Hiking are also enrolled in Archery.

This statement does not tell us anything about how many students are enrolled in Swimming and Archery. At most it says 5 of the students in Hiking are also in Archery, so those in Archery only + swimming & archery are 6, but we don't know how these 6 are distributed between archery alone and swiming+archery.
This statement is not sufficient.

Statement (2) Every student enrolled in Archery is also enrolled in at least one other group.
This means there are 0 students enrolled in Archery alone.

(Archery + Hiking) + (Archery + Swimming) + (Archery + Hiking + Swimming) = 11
Now if we keep
(Archery + Hiking) = 4 (< max of Hiking = 5)
then (Archery + Hiking + Swimming) will have to 1 or 0 since max of Hiking is 5
But even if we take (Archery + Hiking + Swimming) =1, we can have max (Archery + Swimming) = 5
(Archery + Hiking) = 4, (Archery + Swimming) = 5, (Archery + Hiking + Swimming) =1 > all added come to 10, not 11.

From above observation, there is no way other than having:
(Archery + Hiking) = 5, (Archery + Swimming) = 6, (Archery + Hiking + Swimming) = 0 > all add up to 11
This is the only case which meets the constraints of Hiking total = 5, Swimming total =6, Archery total = 11

So, Archery + Swimming = 6
Fraction of Archery also doing Swimming = 6/11.
Statement 2 is sufficient.

Answer is B
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