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 = * (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.
Correct Answer: 303
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