Question

Company BELIANCE hosted a party for 8 members of Company AXIAL. In the party no member of AXIAL had interacted with more than three members of BELIANCE. Out of all the members of BELIANCE, three members each interacted with four members of AXIAL and the remaining members each interacted with two members of AXIAL. The greatest possible number of company BELIANCE members in the party is:

• A. 9
• B. 10
• C. 11
• D. 12

Hint:

Count total interactions from both sides of the party; the AXIAL side gives a hard ceiling of 24, and you need the largest n for which the BELIANCE side stays within that ceiling.

Answer 1

The Correct Option is A

This problem is about counting the same set of interactions in two different ways, once from the BELIANCE side and once from the AXIAL side, and using the fact that they must agree.

1. Count from BELIANCE's side: if there are n BELIANCE members in total, 3 of them each spoke to 4 AXIAL members (12 interactions), and the rest, $(n-3)$ of them, each spoke to 2 AXIAL members ($2(n-3)$ interactions). Adding these gives a BELIANCE-side total of $12 + 2(n-3) = 2n + 6$ interactions.
2. Count from AXIAL's side: there are only 8 AXIAL members, and each spoke to at most 3 BELIANCE members. So the AXIAL-side total can never exceed $8 \times 3 = 24$ interactions.
3. Set up the inequality: since the two counts describe the same interactions, the BELIANCE-side total cannot exceed the AXIAL-side ceiling: $2n + 6 \leq 24$, which gives $n \leq 9$.
4. Check that this bound can be reached: at n = 9, the total is exactly 24, meaning every one of the 8 AXIAL members interacted with exactly 3 BELIANCE members. Nothing in the problem forbids this, so it is achievable, and n = 9 is a real maximum, not just a loose upper bound.

So the greatest possible number of BELIANCE members in the party is 9, which is option A. A value like 10, 11, or 12 would push the BELIANCE-side total above 24, which the AXIAL-side restriction rules out.

Let's summarize:

• Every interaction gets counted once from each side, so the two totals must be consistent with the stricter constraint.
• The AXIAL side, with only 8 members capped at 3 interactions each, sets the hard ceiling of 24.

Thus the answer is 9.