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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.
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(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.
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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
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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
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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
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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)
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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.
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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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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
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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
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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
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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.


 


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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 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.
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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.
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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.


 


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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.


 


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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
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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
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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.
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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.
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
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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 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
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