IPMAT Indore 2024 · Quantitative Ability · Set 1: Eight companies – headcount and median age

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.

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 — find the median position. There are 245 employees — an odd number — so the median is the

\frac{245+1}{2}=123\text{rd}

value in the sorted list.

Step 2 — use the chain to locate the 123rd person. Because the companies form a strictly increasing chain, the sorted list is simply A, then B, then C, and so on. Build the cumulative count:

Position 123 falls inside company D.

Step 3 — which employee of D is it?

123-103=20

so it is D’s 20th employee by age.

Step 4 — recognise that this is D’s own median. D has 39 employees — odd — so D’s median is its

\frac{39+1}{2}=20\text{th}

employee, which the table gives as 45.

The overall median is D’s median, exactly.

Answer: 45

Shortcut. Cumulate, then land. 245 employees \Rightarrow the 123rd value. Running totals 32,60,103,142 show 123 sits in D, at D’s 123-103=20th place — and 20 is precisely D’s median position since D has 39 (odd) employees.

So the answer is just D’s tabulated median, 45 — no construction of individual ages is needed. Whenever the overall median position coincides with a company’s own median position, the table hands you the answer directly.

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.