IPMAT Indore 2023 · Quantitative Ability · Question 1

Step 1 — count from one player’s point of view. Let each group have n players, so there are 4n players in all. Fix a player A in group {G}_{1}.

\text{Games played by}A=\left( n-1 \right)+2n+3n+4n=10n-1

Step 2 — total over all players, then halve. Every one of the 4n players plays 10n-1 games, but each game involves two players and is therefore counted twice:

\text{Total matches}=\frac{4n\left( 10n-1 \right)}{2}=2n\left( 10n-1 \right)

Step 3 — apply the condition.

2n\left( 10n-1 \right)>1000 \Rightarrow n\left( 10n-1 \right)>500

Step 4 — test integers.

n=7: 7\times 69=483 \left( \text{too small}\right) n=8: 8\times 79=632>500 ✓

With n=8 there are 2\left( 8 \right)\left( 79 \right)=1264 matches.

Answer: 8

Shortcut. The whole question is “count once per player, then divide by 2”. Everything else is decoration: the four clauses just add up to \left( n-1 \right)+2n+3n+4n=10n-1 for every player, regardless of which group they are in — the structure is symmetric.

So total =2n\left( 10n-1 \right)\approx 20{n}^{2}. Setting 20{n}^{2}>1000 gives {n}^{2}>50, so n\ge 8 immediately, and one check confirms it. Approximating first and verifying second turns a two-minute problem into a twenty-second one.