The length of a rectangle increases by 20 % a | vedclass

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Question

(D) Let the initial length be $l$ and breadth be $b$. The initial perimeter is $P_1 = 2(l + b)$.After the increase,the new length $l' = l + 0.20l = 1.

Vedclass · Accepted Answer

$	ext{Data inadequate}$

Vedclass · Answer

(D) Let the initial length be $l$ and breadth be $b$. The initial perimeter is $P_1 = 2(l + b)$.After the increase,the new length $l' = l + 0.20l = 1.2l$ and the new breadth $b' = b + 0.30b = 1.3b$.The new perimeter is $P_2 = 2(1.2l + 1.3b) = 2.4l + 2.6b$.The percentage increase in perimeter is given by $\frac{P_2 - P_1}{P_1} 	imes 100$.Substituting the values: $\frac{(2.4l + 2.6b) - 2(l + b)}{2(l + b)} 	imes 100 = \frac{0.4l + 0.6b}{2(l + b)} 	imes 100 = \frac{0.2l + 0.3b}{l + b} 	imes 100$.Since the ratio of $l$ to $b$ is not provided,the percentage increase depends on the values of $l$ and $b$. Therefore,the data is inadequate.

The length of a rectangle increases by $20 \%$ and the breadth by $30 \%$. Then the perimeter of the rectangle will increase by:

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