In a school, 600 students are to be seated on benches, with an equal number of…

Concept: A product-constraint word problem (number of benches × students per bench = fixed total) gives two equations for the same total under two different bench/per-bench splits. Equating the two expressions for the total eliminates one variable, reducing the system to a single quadratic equation.

1. Let x = original number of benches and y = original number of students seated per bench.

2. Since all 600 students are seated with y students per bench across x benches: x·y = 600 — Equation (1).

3. Reducing the benches by 10 (x − 10 benches) requires 2 more students per bench (y + 2 per bench) to still seat all 600 students: (x − 10)(y + 2) = 600 — Equation (2).

4. Expanding Equation (2): xy + 2x − 10y − 20 = 600.

5. Substituting xy = 600 from Equation (1): 600 + 2x − 10y − 20 = 600, which simplifies to 2x − 10y = 20, i.e. x − 5y = 10, so x = 5y + 10.

6. Substituting x = 5y + 10 into xy = 600: (5y + 10)y = 600, i.e. 5y2 + 10y − 600 = 0, i.e. y2 + 2y − 120 = 0.

7. Solving this quadratic: y = [−2 ± √(4 + 480)] / 2 = (−2 ± 22) / 2, giving y = 10 or y = −12. Since y is a count of students, y = 10.

8. Then x = 5(10) + 10 = 60.

Cross-check: with x = 60 and y = 10, x·y = 600 ✓. With 10 fewer benches (50) and 2 more students per bench (12): 50 × 12 = 600 ✓. Both conditions hold, confirming the original number of benches is 60.