Students of a class are preparing for a drill and are made to stand in rows. If 4 students are extra in a row, then there would be 2 rows less. But there would be 4 more rows if 4 students are less in a row. The number of students in the class is

### Concept & Logic This problem involves setting up simultaneous equations based on the area model formula: $\text{Total} = \text{Rows} \times \text{Students per Row}$. ### Step-by-Step Solution * **Given:** Let the number of students per row be $s$ and the number of rows be $r$. $$\text{Total Students (N)} = s \times r$$ * **Condition 1:** If 4 extra students are in a row, there are 2 less rows. $$(s + 4)(r - 2) = sr$$ $$sr - 2s + 4r - 8 = sr$$ $$-2s + 4r = 8 \Rightarrow 2r - s = 4 \Rightarrow s = 2r - 4$$ * **Condition 2:** If 4 less students are in a row, there are 4 more rows. $$(s - 4)(r + 4) = sr$$ $$sr + 4s - 4r - 16 = sr$$ $$4s - 4r = 16 \Rightarrow s - r = 4$$ * **Solve the system of equations:** Substitute $s = 2r - 4$ into the simplified second equation ($s - r = 4$): $$(2r - 4) - r = 4$$ $$r - 4 = 4 \Rightarrow r = 8 \text{ rows}$$ * **Find students per row:** $$s = 8 + 4 = 12 \text{ students/row}$$ * **Calculate total students:** $$N = 12 \times 8 = 96$$ ### Exam Strategy & Shortcut For equations of the form $(s + a)(r - b) = sr$ and $(s - c)(r + d) = sr$, simplifying them always yields two linear equations: $ar - bs = ab$ and $cs - dr = cd$. Memorizing this expansion saves time. Here, $4r - 2s = 8$ and $4s - 4r = 16$. Adding them immediately gives $2s = 24 \Rightarrow s = 12$. Substitute back: $r = 8$. Total = $12 \times 8 = 96$. ### Common Pitfall Failing to expand the binomials properly (e.g., forgetting the $ab$ constant term) or mixing up the signs when moving variables across the equals sign. ### Final Answer Therefore, the correct answer is **96**.