CAT 2020 Exam Shift 1 Solved Paper

Question : 63 of 76

Marks: +1, -0

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}

Solution:

Given

x_{0}=(x_{1}, x_{2}, . . . . . . . . . x_{12})

x_{1}=x_{2}=x_{3}=x_{4}=9 ; x_{5}=x_{6}=. . . . . . x_{12}=8

∴ x_{0}=\max (9,9,9,9,8,8 . . . .8)

The minimum value if

x_{0}

is 9 .

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