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(D) Let the original length be $L$ and the original breadth be $B.$ The original area is $A = L 	imes B.$According to the problem,the new length $L'

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) Let the original length be $L$ and the original breadth be $B.$ The original area is $A = L 	imes B.$According to the problem,the new length $L' = L + 0.20L = 1.2L$ and the new breadth $B' = B - 0.20B = 0.8B.$The new area is given as $192 \, m^2.$Therefore,$L' 	imes B' = 192.$Substituting the values,$(1.2L) 	imes (0.8B) = 192.$$0.96 	imes (L 	imes B) = 192.$Since $L 	imes B = A,$ we have $0.96 	imes A = 192.$$A = \frac{192}{0.96} = \frac{19200}{96} = 200 \, m^2.$Thus,the area of the original rectangle is $200 \, m^2.$

If the length of a rectangular field is increased by $20 \%$ and the breadth is reduced by $20 \%,$ the area of the rectangle will be $192 \, m^2.$ What is the area of the original rectangle? (in $m^2$)

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