The length of a rectangle is increased by 33. | vedclass

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(A) Let the original length be $l$ and the original width be $b$. The original area is $A = l 	imes b$.The length is increased by $33.33 \%$,which is

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Let the original length be $l$ and the original width be $b$. The original area is $A = l 	imes b$.The length is increased by $33.33 \%$,which is equivalent to an increase of $\frac{1}{3}$.New length $l' = l + \frac{1}{3}l = \frac{4}{3}l$.Let the new width be $b'$. To maintain the same area,$l 	imes b = l' 	imes b'$.$l 	imes b = \frac{4}{3}l 	imes b'$.$b' = \frac{3}{4}b = 0.75b$.The reduction in width is $b - b' = b - 0.75b = 0.25b$.Percentage decrease in width $= \frac{0.25b}{b} 	imes 100 \% = 25 \%$.

The length of a rectangle is increased by $33.33 \%$. By what percent should the width be decreased to maintain the same area? (in $\%$)

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