IB DP Math MAA HL : IB Style Mock Exams – Set 5 Paper 2

Method 1 — Using combinations
Let \( n \) be the original number of students.

If everyone played each other twice, the total games would be: \( 2 \times \binom{n}{2} = n(n-1) \).

Stephen should have played \( 2(n-1) \) games (one against each other student, twice).
He actually played 7, so he missed \( 2(n-1) – 7 \) games.

Therefore, the total games played is: \( n(n-1) – [2(n-1) – 7] \).

This is given as 513: \( n(n-1) – 2n + 2 + 7 = 513 \)
\( n^2 – n – 2n + 9 = 513 \)
\( n^2 – 3n + 9 = 513 \)
\( n^2 – 3n – 504 = 0 \)
\( (n – 24)(n + 21) = 0 \)
\( n = 24 \) (since \( n > 0 \)).

Method 2 — Separating Stephen
Let there be \( N \) students excluding Stephen.

Games among these \( N \) students: each pair plays twice, so \( 2 \times \binom{N}{2} = N(N-1) \).

Stephen played 7 games against some of these \( N \) students.
Total games = \( N(N-1) + 7 = 513 \)
\( N(N-1) = 506 \)
\( N^2 – N – 506 = 0 \)
\( (N – 23)(N + 22) = 0 \)
\( N = 23 \).

Thus total original students = \( N + 1 = 24 \).

\( \boxed{24} \)