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02 Jul 2025, 08:00
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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.
(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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02 Jul 2025, 08:30
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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?
We are given that H : S : A = 5 : 6 : 11.
The question asks for the value of (A and S) to A.
(1) All students enrolled in Hiking are also enrolled in Archery.
This implies that the Hiking group is entirely contained in the Archery group. However, this statement gives no information about the Swimming group. For example, we can have a case where all 5 students in Hiking are also in Archery, and only 1 of the 6 students in Swimming is in Archery, making the fraction equal to 1/11. Or we can have a case where all 6 students in Swimming are in Archery, making the fraction equal to 6/11. Not sufficient.
(2) Every student enrolled in Archery is also enrolled in at least one other group.
This means that no student is enrolled in Archery only. Now, notice that H : S : A = 5 : 6 : 11 implies that the total number of students in Hiking and Swimming equals the number in Archery: 5x + 6x = 11x. These two pieces together imply that both the Hiking and Swimming groups are entirely contained in the Archery group.
For example, suppose there are 5 students in Hiking, 6 in Swimming, and 11 in Archery. Since each of the 11 students in Archery is also in at least one other group, there must be 11 additional group assignments among them. But there are only 5 + 6 = 11 group assignments available outside Archery. That’s exactly enough. So, 5 out of 11 must be in Hiking, and 6 out of 11 must be in Swimming. This makes the value of (A and S) to A equal to 6:11. Sufficient.
Answer: B.
GMAT Club Official Explanation:
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?
We are given that H : S : A = 5 : 6 : 11.
The question asks for the value of (A and S) to A.
(1) All students enrolled in Hiking are also enrolled in Archery.
This implies that the Hiking group is entirely contained in the Archery group. However, this statement gives no information about the Swimming group. For example, we can have a case where all 5 students in Hiking are also in Archery, and only 1 of the 6 students in Swimming is in Archery, making the fraction equal to 1/11. Or we can have a case where all 6 students in Swimming are in Archery, making the fraction equal to 6/11. Not sufficient.
(2) Every student enrolled in Archery is also enrolled in at least one other group.
This means that no student is enrolled in Archery only. Now, notice that H : S : A = 5 : 6 : 11 implies that the total number of students in Hiking and Swimming equals the number in Archery: 5x + 6x = 11x. These two pieces together imply that both the Hiking and Swimming groups are entirely contained in the Archery group.
For example, suppose there are 5 students in Hiking, 6 in Swimming, and 11 in Archery. Since each of the 11 students in Archery is also in at least one other group, there must be 11 additional group assignments among them. But there are only 5 + 6 = 11 group assignments available outside Archery. That’s exactly enough. So, 5 out of 11 must be in Hiking, and 6 out of 11 must be in Swimming. This makes the value of (A and S) to A equal to 6:11. Sufficient.
Answer: B.
General Discussion
MaxFabianKirchner
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Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26
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02 Jul 2025, 08:21
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The goal is to find the value of the fraction: (Number in Archery and Swimming) / (Total in Archery). Using the given ratio, this is (A intersection S) / 11k.
Statement (1) alone is insufficient. It tells us that all 5k Hiking students are also in Archery, but gives no information about the overlap with the 6k Swimming students.
Statement (2) alone is insufficient. It tells us there are no 'Archery only' students, but we don't know how the overlap is split between Hiking and Swimming.
Combining both statements:
From (1), we know the 5k students in Hiking are a subset of the 11k students in Archery.
From (2), we know the entire Archery group must overlap with other groups. The 5k overlap with Hiking is accounted for. This leaves the other 6k Archery students (11k - 5k) who must overlap with Swimming.
The total number of students in Swimming is also 6k. If the overlap with Archery must account for 6k students, and there are only 6k swimmers in total, it means every swimmer must also be in Archery.
Therefore, the number of students in both Archery and Swimming is exactly 6k.
The fraction is (6k) / (11k) = 6/11. Since we found a single, definite value, the statements together are sufficient.
The correct answer is (C).
Statement (1) alone is insufficient. It tells us that all 5k Hiking students are also in Archery, but gives no information about the overlap with the 6k Swimming students.
Statement (2) alone is insufficient. It tells us there are no 'Archery only' students, but we don't know how the overlap is split between Hiking and Swimming.
Combining both statements:
From (1), we know the 5k students in Hiking are a subset of the 11k students in Archery.
From (2), we know the entire Archery group must overlap with other groups. The 5k overlap with Hiking is accounted for. This leaves the other 6k Archery students (11k - 5k) who must overlap with Swimming.
The total number of students in Swimming is also 6k. If the overlap with Archery must account for 6k students, and there are only 6k swimmers in total, it means every swimmer must also be in Archery.
Therefore, the number of students in both Archery and Swimming is exactly 6k.
The fraction is (6k) / (11k) = 6/11. Since we found a single, definite value, the statements together are sufficient.
The correct answer is (C).
