Question

The length and breadth of a square are increased by $60\%$ and $40\%$ respectively. The area of the resulting rectangle exceeds the area of the square by

• A. $224\%$
• B. $24\%$
• C. $124\%$
• D. $100\%$

Answer 1

The correct option is A $224\%$

Let $x$ be the length of each of side of a square.

Then, area of square $=x^2$

New length $=x+60\%ofx$

$=x+\frac{60}{100}\times x$

$=x+\frac{3x}{5}$

$=\frac{8x}{5}unit$

New breadth $=x+40\%ofx$

$=x+\frac{40}{100}\times x$

$=x+\frac{2x}{5}$

$=\frac{7x}{5}unit$

New area $=\frac{8x}{5}\times \frac{7x}{5}=\frac{56x^2}{25}sq.unit$

Percentage increased $=\frac{\frac{56x^2}{25}}{x^2}\times 100$

$=56\times 4$

$=224\%$