Jharkhand PSC Prelim Exam 2006 Solved Paper

Let length of the rectangle = $\mathit{x}$
Breadth = $\mathit{y}$
∴ Area = $\mathit{x}\mathit{y}$
According to question,
Length increase by 20% then new length = $\mathit{x}+\mathit{x}\times \frac{20}{100}=\frac{6\mathit{x}}{5}$
Breadth decrease by 20%New Breadth = $\mathit{y}-\mathit{y}\times \frac{20}{100}=\frac{4\mathit{y}}{5}$
∴ New area = $\frac{6\mathit{x}}{5}\times \frac{4\mathit{y}}{5}=\frac{24\mathit{x}\mathit{y}}{25}$
∴ Difference between % of both area
$\frac{\mathit{x}\mathit{y}}{25}\times \frac{100}{\mathit{x}\mathit{y}}=4\%$