The length of a classroom floor exceeds its breadth by 25 m. The area remains unchanged when the length is decreased by 10 m and the breadth is increased by 8 m. Find the area of the floor in square metres (m^2).

Introduction / Context:
This is an algebra-based area invariance problem. When the area of a rectangle remains unchanged after changing length and breadth, you can equate the original area to the new area. The difference relationship between length and breadth (length exceeds breadth by 25 m) helps reduce the variables and makes the equation solvable.

Given Data / Assumptions:

• Let breadth = b m
• Then length = b + 25 m
• New length = (b + 25) - 10 = b + 15 m
• New breadth = b + 8 m
• Areas are equal: original area = new area

Concept / Approach:
Original area = b(b+25). New area = (b+15)(b+8). Set them equal and solve for b. Then compute area using the original dimensions.

Step-by-Step Solution:
Original area: A = b(b + 25) New area: A = (b + 15)(b + 8) Set equal: b(b + 25) = (b + 15)(b + 8) Expand LHS: b^2 + 25b Expand RHS: b^2 + 23b + 120 Equate and simplify: b^2 + 25b = b^2 + 23b + 120 25b - 23b = 120 => 2b = 120 => b = 60 Length = 60 + 25 = 85 Area = 60*85 = 5100 m^2

Verification / Alternative check:
New dimensions: length = 85 - 10 = 75, breadth = 60 + 8 = 68. New area = 75*68 = 5100 m^2, same as original, so the condition is satisfied exactly.

Why Other Options Are Wrong:
4870 m^2, 4987 m^2, 4442 m^2, 5200 m^2: these do not match the exact area obtained after satisfying both the difference condition and the unchanged-area condition.

Common Pitfalls:
Subtracting 10 from breadth instead of length, or adding 8 to length instead of breadth. Forgetting to equate areas because the area remains unchanged. Dropping the constant term (15*8 = 120) when expanding RHS.

Final Answer:
Area of the floor = 5100 m^2