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16 Dec 2025, 10:14
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Total number of students is 90. Every student belong to Debate or Math club.
Let number of students in debate club and math club be D and M respectively. Let the number of students in both clubs be B.
We have to find if D>M
Statement A:
3/5 participants in only Math club.
This shows greater than 50% students belong to only math Club. [(3/5)*90 = 54]
Hence M-B = 54.
Whatever the number of D, D will have to be less than M as
M-B is over 50% of total. So D-B<M-B. Or D<M
This is sufficient to get an answer as No
Statement B:
B = (2/9)*90 = 20
This does not give values for D or M.
Insufficent to compare D and M
hence answer is A
Let number of students in debate club and math club be D and M respectively. Let the number of students in both clubs be B.
We have to find if D>M
Statement A:
3/5 participants in only Math club.
This shows greater than 50% students belong to only math Club. [(3/5)*90 = 54]
Hence M-B = 54.
Whatever the number of D, D will have to be less than M as
M-B is over 50% of total. So D-B<M-B. Or D<M
This is sufficient to get an answer as No
Statement B:
B = (2/9)*90 = 20
This does not give values for D or M.
Insufficent to compare D and M
hence answer is A
Bunuel
16 Dec 2025, 10:14
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Bunuel
We can plot in a 2 * 2 matrix
None = 0
Statement 1
Only math = 3/5 * 90 = 54
As 54 is more than half of the strength, we can conclude that number of students in the debate club cannot exceed the number of students in the math club.
This is just not possible.
S1 is sufficient.
Statement 2
2/9 = 20
Assume only math = only debate = 25
In that case the strength in both the clubs are equal. Also, we can adjust the numbers so that to arrive at different results. One in which the number of students in the math club club can be greater than the number of students in the debate club and vice versa.
Hence, this statement alone is not sufficient.
Option A
16 Dec 2025, 10:18
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Both statements are needed and sufficient
90 students, only Math 54 students, Both 20students => 16 students only Debating club, no, in Math club there are more students
90 students, only Math 54 students, Both 20students => 16 students only Debating club, no, in Math club there are more students
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16 Dec 2025, 10:26
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Okay important thing is here we have 0 in neither category
Neither=0
Question: D>M ?
Statement 1: Math only = 3/5*90= 54
90= 54+D+Neither-Both
90=54+D+0-Both two unknowns we can't solve not sufficient
Statement 2: Both = 2/9*90=20
90= M+D-20+0
Again one equation with two unknowns not sufficient
(1) and (2) combo works because we needed Both in the first statement. Answer is C
Neither=0
Question: D>M ?
Statement 1: Math only = 3/5*90= 54
90= 54+D+Neither-Both
90=54+D+0-Both two unknowns we can't solve not sufficient
Statement 2: Both = 2/9*90=20
90= M+D-20+0
Again one equation with two unknowns not sufficient
(1) and (2) combo works because we needed Both in the first statement. Answer is C
Bunuel
