# Set Theory, Calendars, Clocks and Binomial Theorem

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## Sets and Unions

In its annual fest, a college is organizing three events: B-quiz, Finance & Marketing. The college has a strength of 510 students.The students were allowed to participate in any no. of events they liked. While viewing the statistics of the performance, the general secretary noticed:-
1. The number of students who participated in atleast two events were 52% more than those who participated in exactly one game.
2. The no. of students participating in 1,2 or 3 events respectively was atleast equal to 1.
3. The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.
What can be the maximum no. of students who participated in exactly 3 games?
1. 200
2. 300
3. 303
4. 304

Choice C. 303

## Detailed Solution

Let the no. of students who participated in 0,1,2,3 games be A, B, C, D respectively.
Then from the information we have we can conclude that
C + D = 152% 0f B = 1.52 of B -----------------------(1)
Since the total no. of students the college has is 420,
A + B + C + D = 510 ----------------------------------(2)
From (1) and (2), we can conclude that
A + 2.52B = 510
B = $\frac{25}{63}$ * (510 – A)
For A to be minimum, 510 – A should give us the largest multiple of 63. Since, 63 x 8 = 504 we will get A = 6.
So B = 200 and C + D = 1.52 B = 304
Thus for the students participating in exactly 3 games to be maximum, the number of students participating in exactly 2 games has to be minimized and made equal to 1.
Thus no of students who participated in exactly 3 games = 304 – 1 = 303.

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