IPMAT Indore 2024 · Quantitative Ability · Question 13

The structure of this set. The eight companies form a strictly increasing chain of age blocks: every employee of A is younger than every employee of B, who are all younger than every employee of C, and so on. So if you lined up all 245 employees by age, they would come out company by company, in order A, B, C, D, E, F, G, H — with no interleaving anywhere.

Total headcount.

32+28+43+39+35+29+23+16=245

The median rule you need throughout. For n values in sorted order:

n odd: the median is the \frac{n+1}{2}-th value.

n even: the median is the average of the \frac{n}{2}-th and \left( \frac{n}{2}+1 \right)-th values.

Step 1 — what the question really asks. A’s oldest employee must still be younger than every employee of B. So

\max \left( A \right)<\min \left( B \right)

To make \max \left( A \right) as large as possible, we must first make \min \left( B \right) as large as possible.

Step 2 — how large can B’s youngest be? B has 28 employees (even) with median 30, so

\frac{14\text{th}+15\text{th}}{2}=30 \Rightarrow 14\text{th}+15\text{th}=60

Since the list is sorted, 14\text{th}\le 15\text{th}, which forces 14\text{th}\le 30. Every employee from the 1st to the 14th is at most the 14th, so

\min \left( B \right)\le 30

This is attainable: let all 28 of B’s employees be aged exactly 30. Then the 14th and 15th are both 30 ✔ and \min \left( B \right)=30.

Step 3 — bound A. With \min \left( B \right)=30 and integer ages,

\max \left( A \right)\le 29

Step 4 — check 29 is actually achievable. A has 32 employees (even) with median 24, so the 16th and 17th must average 24, i.e. sum to 48. Take

employees 1–16 aged 24,

employee 17 aged 24,

employees 18–32 aged 29.

Then the 16th and 17th are both 24, median =24 ✔, all ages are integers ✔, and the maximum is 29<30 ✔

Answer: 29

Shortcut. Chain the two constraints — nothing else matters. \max \left( A \right)<\min \left( B \right), and \min \left( B \right) is capped by B’s lower median value: with 28 employees the 14th \le 30. So \min \left( B \right)\le 30 and \max \left( A \right)\le 29.

The reusable principle: in a median question, the median only constrains the middle positions. Everything above the median position is free to be as large as you like, and everything below as small as you like — which is exactly what lets A’s top 15 sit at 29 while its median stays at 24. Answer 29.