Change in total area of four walls: Room dimensions are in ratio L:B:H = 7:3:1. If breadth and height are doubled while length is halved, by what percent does the total area of the four walls change?

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Introduction / Context:The total area of four walls (lateral surface area) of a rectangular room equals perimeter of base times height: 2(L + B) * H. We are to compare original and modified wall areas using given dimension changes. Given Data / Assumptions: Original L:B:H = 7k : 3k : 1k. Modified: L' = (1/2)*7k = 3.5k, B' = 2*3k = 6k, H' = 2*1k = 2k. Concept / Approach:Compute original area A1 = 2(L + B)H and new area A2 = 2(L' + B')H'. Then find percentage increase = ((A2 − A1)/A1)*100. Step-by-Step Solution: A1 = 2(7k + 3k)*1k = 2(10k)*k = 20k^2.A2 = 2(3.5k + 6k)*2k = 2(9.5k)*2k = 38k^2.Increase = 38k^2 − 20k^2 = 18k^2.Percent increase = (18 / 20)*100% = 90%. Verification / Alternative check: Use k = 1 to numerically verify: Original 20, New 38 ⇒ increase 18 ⇒ 90%. Why Other Options Are Wrong: 88%, 85%, 84% result from miscomputing either the halving or doubling steps in the formula. Common Pitfalls: Using surface area of all six faces instead of only the four walls.Forgetting the factor 2(L + B) rather than L + B for the perimeter. Final Answer: 90%

Change in total area of four walls: Room dimensions are in ratio L:B:H = 7:3:1. If breadth and height are doubled while length is halved, by what percent does the total area of the four walls change?

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