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07 Aug 2025, 03:56
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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.
07 Aug 2025, 03:56
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Official Solution:
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
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
01 Apr 2026, 01:57
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Really clever framing! Option C looked almost certain until one notices that everybody in swimming technically does the other two.
Bunuel
