The length of a rectangle is increased by 60%. By what percent would the width have to be decreased so as to maintain the same area ?

Answer: Option A

Let original length = x and original breadth = y
Then, original area = xy
New length :
$$\begin{array}{rl}& =\frac{160x}{100}\\ & =\frac{8x}{5}\end{array}$$
Let the new breadth = z
Then,
$$\begin{array}{rl}& \Rightarrow \frac{8x}{5}\times z=xy\\ & \Rightarrow z=\frac{5y}{8}\end{array}$$
∴ Decrease in breadth :
$$\begin{array}{rl}& =\left(\frac{3y}{8}\times \frac{1}{y}\times 100\right)\%\\ & =37\frac{1}{2}\%\end{array}$$