Q.13: Among 100 students, x1 have birthdays in January, x2 have birthdays in February, and so on.
If x0 = max(x1, x2, …, x12), then the smallest possible value of x0 is
Correct Answer: 9
x0 = 100 (If all 100 had their birthdays on January)
If x0 = 3 (Not possible, because all x1 ,…..x12 together will be only 36)
So x0 will be lesser when the numbers are as close as possible
100/12=8.5
So if we add 8 (12 times) = 96
So, need to have 8,8,8,8,8,8,8,8,9,9,9,9 (Adding these 12 we will get 100)
x0 = 9
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