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Question 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

  1. 8
  2. 10
  3. 12
  4. 9

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Explanatory Answer

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
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

The answer is, "9"

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