(B) The original area of the rectangle is $A_1 = a \times b = ab$.The new width is $a' = a - 0.20a = 0.80a$.The new length is $b' = b + 0.10b = 1.10b$

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(B) The original area of the rectangle is $A_1 = a \times b = ab$.The new width is $a' = a - 0.20a = 0.80a$.The new length is $b' = b + 0.10b = 1.10b$.The new area is $A_2 = a' \times b' = (0.80a) \times (1.10b) = 0.88ab$.To express the new area as a percentage of the original area,we calculate $(A_2 / A_1) \times 100 = (0.88ab / ab) \times 100 = 88 \%$.Therefore,the area of the new rectangle is $88 \%$ of the original area $ab$.

Competitive Quantitative Aptitude Mensuration Measurement of Area

Easy

$A$ rectangle has width $a$ and length $b$. If the width is decreased by $20 \%$ and the length is increased by $10 \%$,what is the area of the new rectangle as a percentage of the original area $ab$?

A
$80$
B
$88$
C
$110$
D
$120$

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Measurement of Area

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$A$ rectangle has width $a$ and length $b$. If the width is decreased by $20 \%$ and the length is increased by $10 \%$,what is the area of the new rectangle as a percentage of the original area $ab$?

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The length of a rectangular plot of land is three times as much as its breadth. $A$ playground measuring $1200 \ ft^2$ occupies $\frac{1}{4}$ of the total area of the plot. What is the length of the plot in feet?

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