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18 Jul 2026, 07:55
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1) We know the club counts are:
Drama = 5
Chess = 4
Robotics = 3
Art = 2
That's 14 students, leaving 3 students unaccounted for.
Those 3 could all belong to one club, or be split among multiple clubs. The number of valid teams would change. Not sufficient.
(2) We only know that exactly one club has fewer than 3 volunteers.
There are still many possible ways to distribute the 17 students among clubs. Not sufficient.
(1) + (2) Art has 2 volunteers, so Art must be the one and only club with fewer than 3 volunteers.
Therefore 3 students to belong to a single club.
So the club counts are fixed: 5, 4, 3, 2, 3
Option C is the correct ans.
Drama = 5
Chess = 4
Robotics = 3
Art = 2
That's 14 students, leaving 3 students unaccounted for.
Those 3 could all belong to one club, or be split among multiple clubs. The number of valid teams would change. Not sufficient.
(2) We only know that exactly one club has fewer than 3 volunteers.
There are still many possible ways to distribute the 17 students among clubs. Not sufficient.
(1) + (2) Art has 2 volunteers, so Art must be the one and only club with fewer than 3 volunteers.
Therefore 3 students to belong to a single club.
So the club counts are fixed: 5, 4, 3, 2, 3
Option C is the correct ans.
18 Jul 2026, 07:55
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We know that,
17 students/volunteers
Atleast 3 clubs
Drama club -> 5 volunteers
Chess club -> 4 volunteers
=> Remaining volunteers in other clubs = 17 - 9 = 8
Statement 1->
Robotics club -> 3 volunteers
Art club -> 2 volunteers
=> Remaining students for the other club = 8 - 3 - 2 = 3 volunteers
Now this can have multiple cases like [5,4,3,2,3] , [5,4,3,2,1,1,1], [5,4,3,2,1]
All of these will produce different number of teams
Statement 1 is insufficient
Statement 2->
Given that drama(5) and chess(4) club have more than 3 students, we can say that the remaining clubs need ot have exactly one club with less than 3 students/volunteers
Multiple distribution cases are possible for this statement such as
Case 1: [5,4,1,7]
Case 2: [5,4,2,3,3]
So we can say that,
Statement 2 is insufficient
Combining 1 and 2 ->
We know that
4 clubs are established with their volunteers which are ->
Drama [5] , Chess [4], Robotics [3] , Art [2]
We also know that only one club can have less than 3 members and art satisfies that criteria
=> The only possible remaining combination for the remaning 3 students is to be stacked as a 3 member team in another club as volunteers
=> New combination = [5,4,3,2,3]
Exact count of teams can be determined from this
C. Both statements together are sufficient but neither alone are enough
17 students/volunteers
Atleast 3 clubs
Drama club -> 5 volunteers
Chess club -> 4 volunteers
=> Remaining volunteers in other clubs = 17 - 9 = 8
Statement 1->
Robotics club -> 3 volunteers
Art club -> 2 volunteers
=> Remaining students for the other club = 8 - 3 - 2 = 3 volunteers
Now this can have multiple cases like [5,4,3,2,3] , [5,4,3,2,1,1,1], [5,4,3,2,1]
All of these will produce different number of teams
Statement 1 is insufficient
Statement 2->
Given that drama(5) and chess(4) club have more than 3 students, we can say that the remaining clubs need ot have exactly one club with less than 3 students/volunteers
Multiple distribution cases are possible for this statement such as
Case 1: [5,4,1,7]
Case 2: [5,4,2,3,3]
So we can say that,
Statement 2 is insufficient
Combining 1 and 2 ->
We know that
4 clubs are established with their volunteers which are ->
Drama [5] , Chess [4], Robotics [3] , Art [2]
We also know that only one club can have less than 3 members and art satisfies that criteria
=> The only possible remaining combination for the remaning 3 students is to be stacked as a 3 member team in another club as volunteers
=> New combination = [5,4,3,2,3]
Exact count of teams can be determined from this
C. Both statements together are sufficient but neither alone are enough
18 Jul 2026, 08:00
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A and B individually not sufficient as not enough info to count possibilities
But together they should be as we know total cases from each club - 5 , 4, 3, 2, 3 so together they will be sufficient
(C) is the answer
But together they should be as we know total cases from each club - 5 , 4, 3, 2, 3 so together they will be sufficient
(C) is the answer
