IPMAT Indore 2024 · Quantitative Ability · Question 15

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 F must satisfy. F has 29 employees (odd), so its median is the

\frac{29+1}{2}=15\text{th}

employee, whose age is therefore exactly 54. Sorted, F looks like

\underset{14\text{below}}{\underset{⏟}{{f}_{1}\le \cdots \le{f}_{14}}}\le \underset{15\text{th}}{\underset{⏟}{54}}\le \underset{14\text{above}}{\underset{⏟}{{f}_{16}\le \cdots \le{f}_{29}}}

Step 2 — minimise the top 14. Each must be at least the median, so the cheapest choice is 54 each:

14\times 54=756

Step 3 — minimise the bottom 14. These may be below 54, but every F employee must be older than every E employee. So we need \min \left( F \right)>\max \left( E \right), and to push F’s floor down we must push E’s ceiling down.

E has 35 employees (odd) with median 49, i.e. its 18th employee is 49. Employees 18 through 35 are all \ge 49, so \max \left( E \right)\ge 49. The smallest possible \max \left( E \right) is therefore 49 — achieved by making all of E’s employees exactly 49.

With \max \left( E \right)=49 and integer ages,

\min \left( F \right)\ge 50

so the cheapest bottom 14 are all aged 50:

14\times 50=700

Step 4 — add the three pieces.

700+54+756=1510

Step 5 — confirm the chain still holds. F’s ages run 50 (×14), 54, 54 (×14). All exceed E’s 49 ✔, and all are below G’s employees, whose median is 59 and which can all be set to 59 or above ✔

Answer: 1510

Shortcut. Three blocks, each pushed to its floor.

\underset{\text{below median}}{\underset{⏟}{14\times 50}}+\underset{\text{median}}{\underset{⏟}{54}}+\underset{\text{above median}}{\underset{⏟}{14\times 54}}=700+54+756=1510

The only subtle number is the 50. It comes from E, not F: E’s median of 49 means E’s oldest cannot be pushed below 49, so F’s youngest cannot drop below 50.

General rule for these chain-of-medians questions: to minimise a block, set everything at or above its median equal to the median, and everything below it equal to one more than the previous block’s median.