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805+ (Hard)|   Overlapping Sets|         
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H:S:A=5:6:11
Statement-1:
This tells all the Hiking i.e 5x students are in in archery but it doesn’t tell us how many Archery students are in Swimming, so we cannot find the fraction of Archery students who are in Swimming.
Statement-2:
It tells all the archery Students are enrolled in any one other course. But it also doesn't tell us how many Students are there in Swimming.

Both statements combined:
In 11k Students from archery 5x students enrolled to hiking so the rest will be enrolled in swimming i.e, 11x-5x=6x

Ratio= 6x/11x=6/11
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Based on the statement we got the percentage of student enrolled in three groups. Now coming to the question, among those enrolled in archery, what fraction are also enrolled in swimming, we will look at both statements.
From statement 1, if 5 students from hiking are enrolled in archery also there are two possiblities- one is the 6 from swimming are enrolled in archery which makes the total 11 or they are not enrolled which makes only archery 6.
If we look at second statement every student enrolled in archery is enrolled in atleast one other so there is only one possiblity that from 11 students, 6 are in swimming and 5 are on hiking. There is no other possiblity as overlap will occur and all conditions will not be satisfied. So it can be answered by 2nd but not 1st.
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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For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery

We have to find for students in Archery How many are enrolled in Swimming so that ration can be obtained

So lets say 5x students are enrolled in Hiking

Then 6x students are in swimming & 11x students in Archery are enrolled

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

Hiking = 5x students

so 11x students also have 5x students. So remaining students are 6x but we don't know whether they are enrolled in swimming or not.

So statement is insufficient.

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

Students enrolled in Archery = 11x

all 11x are enrolled in either swimming or hiking. But we don't know how many are in hiking & how many in swimming.

So statement is again insufficient.


Lets combine both statements

From paragraph & statement 1 we know 11x students are in Archery & 5x ae in Hiking

So 6X are remaining

from statement 2 we know that complete 11x is enrolled in Hiking & swimming

Hiking has 5x students so automatically remaining students i.e 6x are enrolled in swimming.

So ration is 6x/11x of students enrolled in swimming of total students in Archery.

Answer is C. Both statements combined are sufficient.
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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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We are asked: Among those in Archery (11x), what fraction are also in Swimming?
So we need to find the number of students in both Archery and Swimming, divided by 11x.

Statement 1: All students enrolled in Hiking are also enrolled in Archery
This tells us that the 5x in Hiking are included in the 11x Archery group. But it tells us nothing about who is in Swimming or about overlap between Swimming and Archery. Not sufficient.

>statement 2 says all archery students are also in hiking or swimming or both
>so archery is a subset of hiking union swimming
>hiking has 5x people, swimming has 6x
>their union can have at most 11x people
>archery also has 11x
>so the union of hiking and swimming must exactly match archery, that means all 5x in hiking are also in archery
>all 6x in swimming are also in archery
>therefore 6x students are in both archery and swimming
total archery students = 11x
required fraction = 6x divided by 11x = 6 over 11
statement 2 is sufficient
final answer is b
Classic case of C trap, approaching step by step is key in DS.
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---
(1) if every H in A
then
5 hikers be in 5 Archers and remaining 6 A be either only Archer or be swimmers as well;

(2) if every A be part of other group as well then for base condition; 5A be Hikers and 6A be swimmers; numbers can't be adjusted in this case; hence we know
6/11 of A be swimers
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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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Let Hiking students = H
Let Swimming students = S
Let Archery students = A

We are given
H : S : A = 5 : 6 : 11

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

We have to find (A&S)/A

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

This means H is included in A
=> 5x of H is part of 11x
Total = 5x + 6x + 11x - {Part of two groups} + {Part of three groups}
=> Total = 5x + 6x + 11x - {A&H} - {A&S} - {S&H} + {A&H&S}
=> Total = 5x + 6x + 11x - 5x - {A&S} - {S&H} + {A&H&S}
=> Total = 6x + 11x - {A&S} - {S&H} + {A&H&S}
=> Total = 17x - {A&S} - {S&H} + {A&H&S}
With this there isn't enough information to figure out the value of (A∩S)/A

Not Sufficient


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

H and S are also enrolled in A as 5x + 6x = 11x
=> Everyone is enrolled in Archery
=> Total = 11x
Total = 5x + 6x + 11x - {Part of two groups} + {Part of three groups}
=> 11x = 5x + 6x + 11x - {A&H} - {A&S} - {S&H} + {A&H&S}
=> 11x = 5x + 6x + 11x - {A&H} - {A&S} - 0 + {A&H&S}
=> 11x = 5x + 6x + 11x - 5x - 6x - 0 + {A&H&S}
=> 11x = 11x + {A&H&S}
=> {A&H&S} = 0
Hence, none of the students were part of all three groups
=> {A&S} = 6x

Now (A∩S)/A = {A&S}/A
=> (A∩S)/A = 6x/11x = 6/11

Sufficient

IMHO Option B
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From the given information,
Hiking : Swimming : Archery = h : s : a = 5 : 6 : 11
Let h = 5x
s = 6x
a = 11x
To find out s/a
Considering statement 1,
The statement translates to, H is a subset of A
Let students in hiking only, hiking and swimming only, hiking and archery only, and hiking, swimming & archery be represented by ho, hso, hao and hsa respectively. Similiarly in all terms "o" represents only.
H = ho + hso + hao + hsa
S = so + hso + sao + hsa
A = ao + aso + hao + hsa
From H is a subset of A, ho = 0, but this does not directly lead to an answer to the question
∴Not Sufficient

Considering statement 2,
The statement translates to, A is a subset of H ∪ S
This means no student is in one activity groups only; ho = 0, so = 0, & ao = 0
Also, A = H + S
aso + hao + hsa = hso + hao + hsa+ hso + sao + hsa
2hso + hsa =0; The only way this an be true is if, hso and hsa = 0
So students in hiking and swimming only is 0 and students in all of hiking, swimming and archery is also 0

∴ s/a = sao/a = 6x/11x = 6/11
The correct option is Option B
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3 groups: Hiking (H), Swimming (S), and Archery (A)
[*]Every student is in at least one group
[*]For every 5 Hikers, there are 6 Swimmers and 11 Archers
=> H:S:A=5:6:11

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

Need to know how many Archers are also in Swimming

(1) All students in Hiking are also in Archery
=> 5x people who are in both H and A. Don't know how many Archers are also in Swimming.
=> Insufficient.

(2) Every student in Archery is also in at least one other group
=> All 11x Archers are also in Hiking or Swimming or both. Don't know exactly how many Archers are in Swimming.
=> Insufficient

(1) + (2):
  • 5x people who are in both H and A
  • 11x Archers are also in Hiking or Swimming or both
=> 11x-5x=6x Archers must be in Swimming or both
=> Fraction = 6x/11x = 6/11
=> Sufficient.

Answer: C
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There are three groups with sizes in ratio 5:6:11 for Hiking (H), Swimming (S), and Archery (A).

Statement (1) says all hikers are also in archery, but gives no info about swimming overlap, so it's not enough to find what fraction of archers swim.

Statement (2) says every archer is also in either hiking or swimming, meaning no archery-only students; since |H| + |S| = |A|, hiking and swimming must be disjoint and cover all archers.

Therefore, the fraction of archers who swim is |S|/|A| = 6/11. So, only statement (2) alone suffices to answer the question.

Option (B) Statement 2 alone is sufficient.
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we don't need individual numbers as we need the ans in fraction.
We have Hiking = 5, Swimming = 6, Archery = 11 (the total numbers of these will be multiples of 5, 6 and 11 only)

A) All students enrolled in Hiking are also enrolled in Archery.
doesn't give us any info of swimming, not sufficient.

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

Hiking + Swimming = 11 and Archery = 11, since every student of archery to involved with atleast one other thing, they are either in hiking or swimming.

Required fraction = 6/11 sufficient B
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H:S = 5: 6
H:A = 5: 11

Statement 1: This can be interpreted as H is a subset of A. This does not give us info on relation between A and S. NOT SUFFICIENT.
Statement 2: This can be interpreted as A is a superset which has H and S as a subset. Thus S is a subset of A hence required fraction is 6/11. B is SUFFICIENT.

Bunuel
Quote:
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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We need to Find the ratio of A:S
consider there are A=11 S=6 and H=5

1) Doesn't tell us anything about Summing students. not sufficient

2) Tells us all 11 students from A are participating in other activities as well.

5 participate in Hiking. We still have 6 left so all students Must also participate in Swimming. therefore 11:6 is the ratio
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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We are given that for every 5 students in hiking we have 6 students in swimming
And for every 5 students in hikinh we have 11 students in Archery.
So activity wise students can be 5x, 6x, 11x

Statement 1 says all students who are enrolled in hiking are also enrolled in archery
which means 5x are a part of 11x. which leaves 6x students in archery, but we don't know whether these 6x are part of swimming too or not.
So, Statement 1 is insufficent.

Sattement 2 says every student enrolled in archery is also enrolled in at least one other group.
so all 11x are also enrolled in hiking and swimming but we don't kno how many of those are enrolled in which sport.
Thus, Statement 2 is also insufficient.

Both statement combined. Now, because of Statement 2 we know that those left 6x students from Statment 1 are also enrolled in Swimming.
Hence, fraction of people in archery also enrolled in swimmming is \(\frac{6x}{11x}\) = \(\frac{6}{11}\)

So, Answer would be (c), both together are sufficient.
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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H:S:A=5:6:11
A) All students enrolled in Hiking are also enrolled in Archery.
This doesn't give us much about the relation between A and S , all we know is all hikers are in archery (not sufficient)
B) Every student enrolled in Archery is also enrolled in at least one other group.
-Again this doesn't give us much about the relation between A and S , we get to know students in A are also in S or H(not sufficient)
Combining both : we can find out H+S=A as 5x+6x=11x
so A/S =6/11
IMO:C
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IMO Answer is C

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?

So Hiking has say 5x students, Swimming has 6x students and Archery has 11x students.

Statement(1)- All students enrolled in Hiking are also enrolled in Archery- so Archery has 5x students who have both hiking and archery, and rest 6x students ( can be only archery or both archery and swimming)- Insufficient

(2) Every student enrolled in Archery is also enrolled in at least one other group.
So 11x students are part of at least one other group, can be hiking or swimming- Hence Insufficient

Statements 1 and 2 together- 5x students are in archery and hiking. Rest 6x students are in archery and swimming.
Hence both statements we can find what fraction of those in archery are also enrolled in Swimming.
Option 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?

HikingHiking+SwimmingSwimmingNoneTotal
Archery11K
No Archery
Total5K-yy6K-y

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

HikingHiking+SwimmingSwimmingNoneTotal
Archery5K-yyx6K-x11K
No Archery006K-x-y06K-x-y
Total5K-yy6K-y6K-x17K-x-y

Among those enrolled in Archery, the fraction also enrolled in Swimming = x/11K
Since x and K are unknown
NOT SUFFICIENT

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

HikingHiking+SwimmingSwimmingNoneTotal
Archeryyx011K
No Archery6K-x-y
Total5K-yy6K-y

Among those enrolled in Archery, the fraction also enrolled in Swimming = x/11K
Since x and K are unknown
NOT SUFFICIENT


(1) + (2)
(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.


HikingHiking+SwimmingSwimmingNoneTotal
Archery5K-yyx6K-x=011K
No Archery006K-x-y06K-x-y
Total5K-yy6K-y6K-x17K-x-y

6K - x = 0
x = 6K

Among those enrolled in Archery, the fraction also enrolled in Swimming = x/11K = 6K/11K = 6/11
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?

H:S:A = 5:6:11

"(1) All students enrolled in Hiking are also enrolled in Archery."
No information on swimming -> not sufficient.

"(2) Every student enrolled in Archery is also enrolled in at least one other group."
Because "enrolled in at least one" means it can be in one, on both groups. Not sufficient.

(1) + (2) = We know among students in Archery, the proportion is 5 in Hiking.
Therefore, the remaining is 6 as per (2), so answer choice C, both together are sufficient, in my opinion.
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GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 03 Jul 2025, 02:07
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1 cannot guarantee whether those in Hiking and archery are also in Swimming or not, hence AD are not the Answer.
2. This is interesting, if one student has to be enrolled from Archery, the ratio also must be maintained at all times. Hence no matter there are 11, 22, 33 folks. the ratio will remain constant, hence fraction of those in Swimming from Archery will be constant, hence B.
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GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 03 Jul 2025, 02:20
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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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Let's consider number of Students in Hiking, Swimming and Archery 5x,6x and 11x respectively.

out of 11x of Archery how much are enrolled in Swimming ??

AD -> nothing is given about swimming.
BCE -> All students that is 11x are also enrolled in at least one group->

Case 1 - No student is enrolled in all 3 groups, then -> Maximum number of archery students that can enroll in swimming will be 6 x and rest 5x will enroll in Hiking.
Case 2 - Some students were enrolled in all 3 groups -> we won't be able to determine the answer.

Combining both -> we still are not sure how many students of hiking are enrolled in swimming. hence, answer is E
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