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03 Jul 2025, 06:52
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Let students in H,S and A be 5x, 6x and 11x
Archery and Swimming both/11x = ?
S1
All students enrolled in Hiking are also enrolled in Archery.
No info on students enrolled in Swimming
Insufficient
S2
Every student enrolled in Archery is also enrolled in at least one other group.
No info on how many are enrolled in archery and swimming
Insufficient
Combined we have, all hiking students are in archery, so 5x students are in both Archery and Hiking. The remaining 6x students must be enrolled in Archery and Swimming both. This gives the requested ratio = 6x/11x
Sufficient
Archery and Swimming both/11x = ?
S1
All students enrolled in Hiking are also enrolled in Archery.
No info on students enrolled in Swimming
Insufficient
S2
Every student enrolled in Archery is also enrolled in at least one other group.
No info on how many are enrolled in archery and swimming
Insufficient
Combined we have, all hiking students are in archery, so 5x students are in both Archery and Hiking. The remaining 6x students must be enrolled in Archery and Swimming both. This gives the requested ratio = 6x/11x
Sufficient
Bunuel
03 Jul 2025, 07:01
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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?
Let the ratio of Hiking : Swimming: Archery students be = 5x:6x:11x
(1) All students enrolled in Hiking are also enrolled in Archery.
Hiking ⊆ Archery
So, of the 11x Archery students, at least 5x are also in Hiking
That leaves 6x Archery students who are not in Hiking. These 6x could be in swimming, swimming + archery or some in all 3 activity and remaining in others
Since we dont know how many of the 6x are in swimming, this statement alone is insufficient
(2) Every student enrolled in Archery is also enrolled in at least one other group.
Now, this states that every student enrolled in archery is also enrolled in one other activity too.
Out of the 11x in archery, 5x is in hiking
Now if from the 5x in hiking, some are also learning swimming too, then there will be some in archery who are NOT doing any activities as the ratios are divinded in H:S:A = 5:6:11
So, one can say that out of 11x the unique people who are going for archery, 5x of those unique ones have to be in hiking and remaining 6x unique people should be in swimming to make the ratio work
So statement 2 works out
Ans: B. Statement (2) alone is sufficient, but statement (1) is not.
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?
Let the ratio of Hiking : Swimming: Archery students be = 5x:6x:11x
(1) All students enrolled in Hiking are also enrolled in Archery.
Hiking ⊆ Archery
So, of the 11x Archery students, at least 5x are also in Hiking
That leaves 6x Archery students who are not in Hiking. These 6x could be in swimming, swimming + archery or some in all 3 activity and remaining in others
Since we dont know how many of the 6x are in swimming, this statement alone is insufficient
(2) Every student enrolled in Archery is also enrolled in at least one other group.
Now, this states that every student enrolled in archery is also enrolled in one other activity too.
Out of the 11x in archery, 5x is in hiking
Now if from the 5x in hiking, some are also learning swimming too, then there will be some in archery who are NOT doing any activities as the ratios are divinded in H:S:A = 5:6:11
So, one can say that out of 11x the unique people who are going for archery, 5x of those unique ones have to be in hiking and remaining 6x unique people should be in swimming to make the ratio work
So statement 2 works out
Ans: B. Statement (2) alone is sufficient, but statement (1) is not.
