The length of a rectangle is increased by 60 | vedclass

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(A) Let the original length be $L$ and the original width be $W$. The original area is $A = L 	imes W$.The new length $L' = L + 0.60L = 1.6L$.Let the

Vedclass · Accepted Answer

$37 \frac{1}{2} \%$

Vedclass · Answer

(A) Let the original length be $L$ and the original width be $W$. The original area is $A = L 	imes W$.The new length $L' = L + 0.60L = 1.6L$.Let the new width be $W'$. To maintain the same area,$L' 	imes W' = L 	imes W$.$1.6L 	imes W' = L 	imes W \implies W' = \frac{W}{1.6} = 0.625W$.The decrease in width is $W - 0.625W = 0.375W$.The percentage decrease is $\frac{0.375W}{W} 	imes 100 = 37.5 \%$.Since $37.5 = 37 \frac{1}{2}$,the width must be decreased by $37 \frac{1}{2} \%$.

The length of a rectangle is increased by $60 \%$. By what percent would the width have to be decreased so as to maintain the same area?

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