## Answer & Solution

Answer:** Option C**

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