Last visit was: 20 Apr 2026, 23:08 It is currently 20 Apr 2026, 23:08
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   4   5   6   
805+ (Hard)|   Overlapping Sets|         
User avatar
vnar12
User avatar
Joined: 03 Jun 2024
Last visit: 31 Mar 2026
Posts: 51
Own Kudos:
32
 [1]
Given Kudos: 4
Posts: 51
Kudos: 32
 [1]
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:19
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The correct answer choice is (B). Statement 2 is sufficient but statement 1 is not.

I broke down what situations are defined by each statement.

Statement 1
Hiking - all 5 of these students are also in archery
Swimming - we don't learn anything specific, so we cannot answer the question about swimming
Archery - we know 5 of the 11 archery students are also in Hiking but we don't know for sure the other 6 are the same 6 as in the Swimming class.

Statement 2
Hiking - These 5 students are also in Archery
Swimming - These 6 students are also in Archery
Archery - The statement tells us that all 11 of these students are enrolled in Hiking or Swimming. Since we know the proportion of those in Swimming (6) we can calculate that 6 of the 11 archery students (fraction 6/11) are also in Swimming.
User avatar
Benedictng
User avatar
Joined: 21 Jan 2025
Last visit: 10 Jul 2025
Posts: 18
Own Kudos:
12
 [1]
Given Kudos: 1
Posts: 18
Kudos: 12
 [1]
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:20
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
(1) not enough information to tell, most archers could come from hiking, swimming or neither

(2) is enough, if all archers are in at least 1 other sport, for every 11 archers, 6 must be in swimming, and 5 in hiking.
User avatar
dthaman1201
User avatar
Joined: 08 Dec 2024
Last visit: 11 Mar 2026
Posts: 23
Own Kudos:
18
Given Kudos: 165
Products:
Forum Quiz
Posts: 23
Kudos: 18
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
5:6:11 is ratio

statement 1 tell us all hiking are in archery that means all 5 students are also in archery
this says nothing about swimming
statement 1 is not sufficient.

Statement 2 tell us all archery are in swimming or hiking or both
Above tell us people who are just doing archery and nothing else is 0
Again we dont know how many are in swimming
Statement 2 is not sufficent

Now together:
Archery = Archery & Hiking + Archery & Swimming + Archery & Hiking & Swimming
Archery & Hikking = Hikking
Archery & Hikking & Swimming = Hikking

Hence

Archery = Hikking + Archery & Swimming
Total archery =11
hikking=5
11-5=6
rATIO =6/11

SO BOTH TOGETHER R SUFFICIENT
User avatar
bart08241192
User avatar
Joined: 03 Dec 2024
Last visit: 28 Dec 2025
Posts: 75
Own Kudos:
64
Given Kudos: 13
Posts: 75
Kudos: 64
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
H:S:A ratio is: 5:6:11

Condition 1: We know H is inside A, but we're not sure if H intersects with S. Not enough info.

Condition 2: We know some of A is in both H and S, but we're not sure if H and S overlap.

Combining both conditions:
Now we know 5 out of 11 is in H.
11 - 5 = 6 is in S.
S/A = 6/11
User avatar
MinhChau789
User avatar
Joined: 18 Aug 2023
Last visit: 12 Apr 2026
Posts: 132
Own Kudos:
140
 [1]
Given Kudos: 2
Posts: 132
Kudos: 140
 [1]
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:33
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Assume the number of students of H:S:A = 5:6:11

(1) All students enrolled in Hiking are also enrolled in Archery.
Can't answer the question

(2) Every student enrolled in Archery is also enrolled in at least one other group.
Since there are only 3 groups, exactly all students of A must be enrolled in H & S with no remaining. So we can answer the question.

Answer: B
avatar
DachauerDon
avatar
Joined: 19 Apr 2025
Last visit: 27 Aug 2025
Posts: 29
Own Kudos:
19
Given Kudos: 26
Location: Germany
Schools: LBS
Schools: LBS
Posts: 29
Kudos: 19
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The prompt tells me that the ratio of participants for each activity is H5x:S6x:A11x, with x being the ratio multiplier.

Using statement (1):
I know that 11x/22x - 5x/22x = 6x/22x participate only in Archery OR Archery and Swimming. Theres no further way to figure out how many are in archery only and thus, statement (1) alone is not sufficient

Using statement (2):
I can't find out the amount of people that do archery only and the amount of people that do archery and swimming, statement (2) is not sufficient.

Both:
I know that all participants in hiking are also doing archery, but the second statements adds no helpful info to find what we're looking for. Thus, neither statement is sufficient.

Answer: E)
User avatar
kvaishvik24
User avatar
Joined: 31 Mar 2025
Last visit: 03 Mar 2026
Posts: 81
Own Kudos:
65
Given Kudos: 16
Posts: 81
Kudos: 65
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let students in Hiking, Swimming, and Archery be 5x, 6x, and 11x, respectively.

We want to find:
(students in both Swimming and Archery)/(students in Archery)=(S∩A)/11x

Using both statements:
  1. All hikers are archers, so 5x hikers are in Archery.
  2. Each archer is in at least one other group, so the remaining archers, 11x−5x=6x must be from Swimming.
Hence:
S∩A=6x
=6x/11x
=6/11

Answer: C) Both 1 and 2 are suffcient together.
User avatar
chasing725
User avatar
Joined: 22 Jun 2025
Last visit: 13 Jan 2026
Posts: 176
Own Kudos:
173
 [1]
Given Kudos: 5
Location: United States (OR)
Schools: Stanford
Schools: Stanford
Posts: 176
Kudos: 173
 [1]
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:46
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more

H : S : A = 5 : 6 : 11

1)

Out of 11 students enrolled in archery 1 student is enrolled in swimming

Ratio = 1 / 11

Out of 11 students enrolled in archery 6 student is enrolled in swimming

Ratio = 6 / 11

We can have multiple answers. Hence this statement alone is not sufficient to answer the question asked.

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

11 can only be split in 5 and 6. Because there is on other way we can group without violating the conditions. Therefore, six students in archery go to swimming.

We have an answer using this statement alone.

Option B
User avatar
Rakshit25
User avatar
Joined: 16 Jun 2020
Last visit: 23 Mar 2026
Posts: 120
Own Kudos:
40
Given Kudos: 25
Posts: 120
Kudos: 40
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 09:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Took some time but C

If the ratio is 5:6:11, then only option C helps you determine the unique value of the ratio of archery to swimming folks. Since all 5 are in archery, only 1 is left for swimming out of the 11, thus the ratio is 1/11
User avatar
AviNFC
User avatar
Joined: 31 May 2023
Last visit: 10 Apr 2026
Posts: 306
Own Kudos:
366
 [1]
Given Kudos: 5
Posts: 306
Kudos: 366
 [1]
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:00
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let Hiking=5x
Archery=11x
Swimming=6x

1. If all the 5x are part of 11x, we have no ifo of the 6x.. only archery is required....INSUFFICIENT
2. If only archery is 0, then the 11x of archery comes from hiking & swimming. These two max. themselves are 11x.. Thus archery+swim = 6x

proportion =6x/11x... SUFFICIENT


Ans B
User avatar
Raome
User avatar
Joined: 21 Apr 2025
Last visit: 16 Feb 2026
Posts: 114
Own Kudos:
31
Given Kudos: 84
Location: India
Posts: 114
Kudos: 31
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer (C)

H-5; S-6; A-11

a) H=A=5. Swimming info not given. Not sufficient
b) A=H or S. Not sufficient


c) H=A=5
A=S=6

Quote:


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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
Gmatismybestie
User avatar
Joined: 27 Jun 2025
Last visit: 01 Aug 2025
Posts: 29
Own Kudos:
24
Given Kudos: 2
Schools: UCR '27 (S)
Schools: UCR '27 (S)
Posts: 29
Kudos: 24
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
The Fraction we got: (5H:6S:11A) * X student.
(1) This statement inferred that all 5H will also enroll in Archery => in those 11A, 5 of them is enrolled Hiking. However, the question is about Swimming:Archery. => NO SUFFICIENT.
(2) This Statement inferred that each Archery enrollment, he/she also enroll one or more other groups. However, it could be enroll in Hiking, or in Swimming, or both. => NO SUFFICIENT.
Combine (1) and (2), 5H from Hiking also in 5A from who enroll Archery. So, the other 6A could be also enroll in Swimming or in both.

Answer: (E) Statements (1) and (2) TOGETHER are NOT sufficient.
User avatar
iCheetaah
User avatar
Joined: 13 Nov 2021
Last visit: 20 Apr 2026
Posts: 81
Own Kudos:
72
Given Kudos: 1
Location: India
Products:
GMAT Club Tests
Posts: 81
Kudos: 72
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let's put variables to the following:
  • Hiking (H)
  • Swimming (S)
  • Archery (A)

The prompt tells us that these three sports are in the following ratio:

H:S:A = 5:6:11

So we can say that,
  • H = 5x
  • S = 6x
  • A = 11x
We need to find the following: \(\frac{# of students in both A and S}{# of students in A}\)

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

This means that all of 5x students in hiking are a subset of the entire circle of students in archery (11x)



This still doesn't help us to get the shared yellow region, which is what we want in the numerator of our answer. So, insufficient.

Statement 2:

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

This tells us that all the 11x students in A, are split between H and S



This also doesn't help us in understanding the yellow shaded region, which is the numerator of our answer. So, insufficient


Statements 1 and 2 together:

This tells us that all of H is in A and all of A is within H and S, thus we get a situation, where both S and H are inside A:



This means, all of the 6x students in S are also taking A. So our answer becomes: \(\frac{6x}{11x}\)\(=\)\(\frac{6}{11}\)

Hence, sufficient.

Answer is C.
Show Spoiler
Attachment:
GMAT-Club-Forum-mzpr21qv.png
GMAT-Club-Forum-mzpr21qv.png [ 9.96 KiB | Viewed 217 times ]
Attachment:
GMAT-Club-Forum-8fs2uzvs.png
GMAT-Club-Forum-8fs2uzvs.png [ 9.53 KiB | Viewed 217 times ]
Attachment:
GMAT-Club-Forum-qnidz0om.png
GMAT-Club-Forum-qnidz0om.png [ 3.19 KiB | Viewed 217 times ]
User avatar
Duongduong273
User avatar
Joined: 08 Apr 2023
Last visit: 03 Feb 2026
Posts: 32
Own Kudos:
32
Given Kudos: 4
Posts: 32
Kudos: 32
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
For every 5 students in Hiking, there are 6 in Swimming and 11 in Archery:
H+HS+HA+HSA=5x
S+HS+AS+HSA=6x
A+AH+AS+HSA=11x
Among those enrolled in Archery, what fraction are also enrolled in Swimming: AS+HSA/A+AH+AS+HSA=AS+HSA/11x
(1) All students enrolled in Hiking are also enrolled in Archery => H=HS=0 => HA+HSA=5x and S+AS+HSA=6x => Can’t determine AS+HSA => Insufficient
(2) Every student enrolled in Archery is also enrolled in at least one other group => A=0 => AH+AS+HSA=11x => Can’t determine AS+HSA => Insufficient
(1)+(2) => AH+AS+HSA - (HA+HSA)= 11x-5x => AS= 6x =S+AS+HSA => S+HSA=0 => HSA=0 => AS+HSA=6x =>AS+HSA/11x = 6x/11x =6/11 => Sufficient
=> 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?

(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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
APram
User avatar
Joined: 23 Jun 2024
Last visit: 17 Mar 2026
Posts: 709
Own Kudos:
279
Given Kudos: 248
Location: India
Schools: ISB '26 HEC INSEAD Jan '26
GMAT Focus 1: 605 Q86 V78 DI76
GPA: 3.608
Schools: ISB '26 HEC INSEAD Jan '26
GMAT Focus 1: 605 Q86 V78 DI76
Posts: 709
Kudos: 279
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
H : S: A = 5:6:11
Question: (Number of students in both A and S) / Total number of students
lets assume H = 5x, S=6x and A=11x
Statement 1:
All 5x of Hiking are in Archery

We need more info to determine exact value of A&S together. We need to know how many enrolled in all A&H&S
Hence insufficient

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

This means A is subset of H or S or Both
=> A = |A&H| + |A&S| - |A&H&S|

We need more info on swimming to get unique value with this info given
hence insufficient

Combining both statements:
A = 11x
H = 5x and all 5x are in A
So remaining 6x must be in S or both H&S

11x will be such that some are in A&H and some are in A&S
We already know A&H =5x
So A&S = 6x

A&S = 6x
A = 11x

Thus A&S / A = 6/11 which is unique

Hence C is correct
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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
User avatar
asingh22
User avatar
Joined: 31 Jul 2024
Last visit: 31 Mar 2026
Posts: 68
Own Kudos:
57
Given Kudos: 8
Location: India
Schools: Foster '26 (II) Goizueta '26 (S) Scheller '26 Anderson '26 Darden '26 Kelley '26 Kellogg '26 Ross '26 Sloan '26 UNC '26 (S) Stern '26 Yale '26 Haas '26 Owen '26 (II)
GMAT Focus 1: 635 Q84 V78 DI82
GMAT Focus 2: 655 Q89 V80 DI78
GPA: 2.5
Schools: Foster '26 (II) Goizueta '26 (S) Scheller '26 Anderson '26 Darden '26 Kelley '26 Kellogg '26 Ross '26 Sloan '26 UNC '26 (S) Stern '26 Yale '26 Haas '26 Owen '26 (II)
GMAT Focus 2: 655 Q89 V80 DI78
Posts: 68
Kudos: 57
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hiking= 5, Swimming = 6, Archery = 11

1 Statement = all hiking is in archery,but we don't know anything about the swimming and archery ratio here
then not sufficient

2 Statement = every archery student is in at least 1 other group, but don't know anything about how many in hiking and how many in swimming
Not sufficient

Combined. =. Every archery student is in at least 1 other group then we know from 1st statement that all in hiking archery, then 5 archer are in hiking, and 6 are in swimming.
it will give us the ration of (swimming&archery )/Archery

Sufficient C
User avatar
DylanD
User avatar
Joined: 08 Jan 2025
Last visit: 20 Mar 2026
Posts: 39
Own Kudos:
20
Given Kudos: 178
Location: United States
Posts: 39
Kudos: 20
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Recognize this is a overlapping set question with 3 variables. Draw a 3-circle Venn diagram.
Variables- "H"; "S"; "A"; not = prefix "n"; number of overlapping variables = #/3
Given- 0/3 = 0; 5H:6S:11A -> Assume a total of 22 (5+6+11)
Rephrase the question- What is ((nHAS) + (HAS))/(11)?

(1)
1. Infer: HnAnS = 0, HnAS = 0
2. (HAnS) + (HAS) = 5 -> (HAS)? (nHAS)?
Eliminate A,D

(2)
1. Infer: nHAnS = 0
2. (HAnS) + (HAS) + (nHAS) = 11
(HAnS) + (HAS) = (11) - (nHAS) -> (nHAS)? (HAS)?
Eliminate B

(3) = (1)+(2)
1. Substitute equation from (1) into (2)
(5) = (11) - (nHAS)
(nHAS) = 6
2. Infer: H = 6 = (nHAS) + (HAS) -> (HAS) = 0
3. Can we answer the rephrased question? YES
(nHAS) + (HAS) / 11 = ((6) + (0))/ 11

Answer: C
User avatar
YashKakade
User avatar
Joined: 04 Jun 2025
Last visit: 16 Dec 2025
Posts: 12
Own Kudos:
11
 [1]
Given Kudos: 1
Posts: 12
Kudos: 11
 [1]
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:44
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given: There are 3 groups - Hiking, Swimming and Archery. All students are in at least one of these 3 groups. No. of students in Hiking: No. of students in Swimming : No. of students in Archery = 5:6:11

So, no. of students in Hiking = 5x ; no. of students in swimming = 6x ; no. of students in archery = 11x

Analyzing Statement 1:
All students enrolled in Hiking are also enrolled in Archery. So, the number of students enrolled in Archery but not Hiking = (11x - 5x) = 6x.
But this does not mean that this 6x and the number of students in swimming is the same. It might be possible that there are no students who are involved in both Hiking and Swimming. Hence, this statement alone is not sufficient.

Analyzing Statement 2:
Every student enrolled in Archery is also enrolled in at least one other group. This means that a student in archery is also involved in hiking or swimming or both.
Now, number of students in Hiking and archery = 5x. So number of students in archery but not in hiking = 11x - 5x = 6x. But as per the given information, number of students involved in swimming = 6x and students involved in archery but not hiking have to be involved in swimming. Hence, the students involved in archery and swimming = 6x. So fraction of archery students involved in swimming = 6x/11x = 6/11.

Hence, statement 2 alone is sufficient.
User avatar
PocusFocus
User avatar
IESE School Moderator
Joined: 27 Nov 2023
Last visit: 20 Apr 2026
Posts: 347
Own Kudos:
59
Given Kudos: 472
Location: Peru
Schools: IESE '26 (D) McCombs '26 (WL) Goizueta '26 (WL) Rotman '26 (A)
GMAT Focus 1: 575 Q86 V79 DI70
GMAT Focus 2: 525 Q82 V77 DI69
GMAT Focus 3: 575 Q83 V76 DI77
GMAT Focus 4: 625 Q84 V78 DI81
GMAT 1: 500 Q42 V18
WE:Corporate Finance (Manufacturing)
Schools: IESE '26 (D) McCombs '26 (WL) Goizueta '26 (WL) Rotman '26 (A)
GMAT Focus 4: 625 Q84 V78 DI81
GMAT 1: 500 Q42 V18
Posts: 347
Kudos: 59
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 10:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Following the draw (1) will not give further information about the quantity or rate of the intersection of Archery and Swimming, eliminate A and D

(2) Every student enrolled in Archery is also enrolled in at least one other group.
Following the draw, this info tell us that there are not students taking just Archery, with this we have not enough information to find a relation among the intersection in yellow and the total students taking archery. Then eliminate B

(1) + (2) if Hiking is included in Archery so we have "6K" students taking Archery and not takin any other course, but (2) force this student to be in another course, so this "6K" student must be in the Swimming curse, the 6k of 11k students is the relation we're looking for, so answer will be C
Show Spoiler

Attachments

20250702_174052942_iOS.png
20250702_174052942_iOS.png [ 351.09 KiB | Viewed 211 times ]

User avatar
HarshaBujji
User avatar
Joined: 29 Jun 2020
Last visit: 05 Apr 2026
Posts: 723
Own Kudos:
906
Given Kudos: 247
Location: India
Products:
My Reviews
Posts: 723
Kudos: 906
GMAT Club Olympics 2025 (Day 3): At a summer camp, there are exactly 02 Jul 2025, 11:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

Show more
Let's denote the #ppl in each astivity as h,s,a respectively any combination is hs, sa,ha & ahs. For combinations. We need to find as/s
We know that h,s,a are 5x, 6x, & 11x respectively.

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

So this means AH= H, AHS=SA,

We know that SA/S need to be found, We don't have any other ways to determine SA. Hence stmt 1 is insufficient.

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

A = AH+AS - AHS, So here also AS is not known insufficent.

Combining both 1 & 2.

Total = A + H + S -AH -AS -HS +AHS, => H + S - HS = Total. Total+HS =11x. Hence even after combining no way to find the ratio.

IMO E
   1   2   3   4   5   6   
Moderators:
Bunuel
Math Expert
109715 posts
kingbucky
498 posts
Gmat860sanskar
210 posts