A rectangle has length L and breadth B with L = 2B. If the length is decreased by 5 cm and the breadth is increased by 5 cm, the area increases by “5 sq. cm more than 70 sq. cm,” i.e., by 75 sq. cm. Find the original length L.

Introduction / Context:This problem embeds a slightly awkward phrase describing the net area increase as “5 sq. cm more than 70 sq. cm,” meaning the increase is 75 sq. cm. With L = 2B initially, we form the new area from (L − 5)(B + 5), subtract the old area LB, set that difference to 75, and solve for B (hence L).

Given Data / Assumptions:

• L = 2B (original dimensions).
• New dimensions: L′ = L − 5, B′ = B + 5.
• Area increase ΔA = 75 sq. cm.

Concept / Approach:Compute ΔA = L′B′ − LB = (L − 5)(B + 5) − LB and substitute L = 2B. This reduces to a single quadratic in B that solves cleanly to an integer value. Then recover L.

Step-by-Step Solution:ΔA = (L − 5)(B + 5) − LB = LB + 5L − 5B − 25 − LB = 5L − 5B − 25With L = 2B ⇒ ΔA = 5(2B) − 5B − 25 = 5B − 25Set ΔA = 75 ⇒ 5B − 25 = 75 ⇒ 5B = 100 ⇒ B = 20 cmL = 2B = 40 cm

Verification / Alternative check:Original area = 40 * 20 = 800. New area = 35 * 25 = 875. Increase = 75, as required.

Why Other Options Are Wrong:30, 21, and 45 cm do not satisfy the derived relation ΔA = 5B − 25 = 75 (they imply non-integer B or wrong increases).

Common Pitfalls:Misreading the phrase about area increase; distributing terms incorrectly when expanding (L − 5)(B + 5); or forgetting to apply L = 2B before solving.

Final Answer:40 cm