Question

Among 100 students, \(x_1\) have birthdays in January, \(x_2\) have birthdays in February, and so on. If \(x_0\) = \(max(x_1,x_2,....,x_{12})\), then the smallest possible value of \(x_0\) is

• A. 9
• B. 10
• C. 8
• D. 12

Answer 1

The Correct Option is A

The correct answer is (A) : \(9\)

Given \(x_0=(x_1,x_2,....,x_{12})\)

If \(x_1=x_2=x_3=x_4=9; x_5=x_6=......×12 = 8\)

∴ \(x_0 = max(9,9,9,9,8,8...8)\)

The minimum value if \(x_0\) is \(9\).

Answer 2

\(x_0= \) max\((x_1, x_2,...., X_{12})\)
So, \(x_o \) will be lesser  when the numbers are as close as possible.
\(\frac {100}{12}=8.33\)
Since, max\((x_1, x_2,...., X_{12})\) will be minimum if \((x_1,x_2...x_{12})\)\(=(9,9,9,9,8,8,8,8,8,8,8,8,)\)

So, the correct option is \((A): 9\)