CAT 2020 Slot 1 QA Question & Solution
AlgebraMedium
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
Options
8
9
10
12
Solution
$x_0 = max (x_1,x_2,....,x_{12})$
$x_0$ will be minimum if x1, x2, ...., x12 are close to each other
100/12=8.33
$\Rightarrow$ 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)
$\Rightarrow$ Option B is correct.