TheCATExam
Sign in

CAT 2020 Slot 1 — QA Question 15

Basics (Functions)Easy

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

Answer & solution

Correct answer: 9

  • A

    10

  • B

    8

  • C

    12

  • 9

Solution

Easy

x0x_0 is the largest of the twelve monthly counts. To make the largest as small as possible, spread the 100100 students as evenly as possible across 1212 months. The most even split decides the minimum.

1

Pigeonhole bound. If every month had at most 88, the total would be at most 12×8=96<10012\times 8 = 96 < 100, which is impossible.

12×8=96<100⇒ some month must hold≥9⇒ x0≥9\begin{aligned} &12\times 8 = 96 < 100\\ &\Rightarrow\ \text{some month must hold} \ge 9 \quad\Rightarrow\ x_0 \ge 9 \end{aligned}
2

Show 99 is achievable. Use eight months of 88 and four months of 99.

8×8+4×9=64+36=100(valid distribution)⇒ x0=max⁡=9\begin{aligned} &8\times 8 + 4\times 9 = 64 + 36 = 100 \quad\text{(valid distribution)}\\ &\Rightarrow\ x_0 = \max = 9 \end{aligned}
x0min⁡=9— option (d)x_0^{\min} = 9\quad\text{— option (d)}

Related Basics (Functions) questions

See all Functions & Graphs questions →
CAT 2020 Slot 1 QA Q15 — Basics (Functions) solution | TheCATExam