# The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

A. 40%

B. 42%

C. 44%

D. 46%

### Solution(By Examveda Team)

Let original length = x metres and original breadth = y metres
\eqalign{ & {\text{Original}}\,{\text{are}} = \left( {xy} \right){m^2} \cr & {\text{New}}\,{\text{length}} = \left( {\frac{{120}}{{100}}x} \right)m = \left( {\frac{6}{5}x} \right)m \cr & {\text{New}}\,{\text{breadth}} = \left( {\frac{{120}}{{100}}y} \right)m = \left( {\frac{6}{5}y} \right)m \cr & {\text{New}}\,{\text{area}} = \left( {\frac{6}{5}x \times \frac{6}{5}y} \right){m^2} = \left( {\frac{{36}}{{25}}xy} \right){m^2} \cr & {\text{The}}\,{\text{difference}}\,{\text{between}}\,{\text{the}}\,{\text{original}}\,{\text{area = }}xy \cr & {\text{and}}\,{\text{new}}\,{\text{area}}\,\frac{{36}}{{25}}xy\,{\text{is}} \cr & = \left( {\frac{{36}}{{25}}} \right)xy - xy \cr & = xy\left( {\frac{{36}}{{25}} - 1} \right) \cr & = xy\left( {\frac{{11}}{{25}}} \right)\,or\,\left( {\frac{{11}}{{25}}} \right)xy \cr & \therefore {\text{Increase}}\,\% \cr & = \left( {\frac{{11}}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 44\% \cr}

1. 