CAT 2020 Quant Question [Slot 1] with Solution 13

Question

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

  1. 9
  2. 10
  3. 8
  4. 12
Option: 1
Solution:

Given \({x_0} = \left( {{x_1},{x_2}, \ldots  \ldots  \ldots {x_{12}}} \right)\)

If \({x_1} = {x_2} = {x_3} = {x_4} = 9;{x_5} = {x_6} =  \ldots  \ldots { \times _{12}} = 8\)

\(\therefore {x_0} = \max (9,9,9,9,8,8 \ldots .8)\)

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

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CAT 2020 Quant questions with Solutions

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