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(D) Let the original length be $L$ and breadth be $B$. The original area is $A = L 	imes B$.New length $L' = L + 0.20L = 1.20L$.New breadth $B' = B -

Vedclass · Accepted Answer

$8\% 	ext{ increase}$

Vedclass · Answer

(D) Let the original length be $L$ and breadth be $B$. The original area is $A = L 	imes B$.New length $L' = L + 0.20L = 1.20L$.New breadth $B' = B - 0.10B = 0.90B$.New area $A' = L' 	imes B' = (1.20L) 	imes (0.90B) = 1.08 	imes (L 	imes B) = 1.08A$.Percentage change in area = $\frac{A' - A}{A} 	imes 100 = \frac{1.08A - A}{A} 	imes 100 = 0.08 	imes 100 = 8\%$.Alternatively,using the net percentage change formula $m + n + \frac{m 	imes n}{100}$:Impact on area = $20 + (-10) + \frac{20 	imes (-10)}{100} = 10 - 2 = 8\% 	ext{ increase}$.

If the length of a rectangle is increased by $20\%$ and its breadth is decreased by $10\%$,what will be the impact on its area?

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