If the circumference of a circle is decreased by 50% then the percentage of decrease in its area is :

Answer: Option D

Let the original circumference be x
Then, new circumference = 50% of x = $$\frac{x}{2}$$
Let original radius = r and new radius = R
$$\begin{array}{rl}& 2\pi r=x\\ & \Rightarrow r=\frac{x\times 7}{2\times 22}\\ & \Rightarrow r=\frac{7x}{44}\end{array}$$
$$\begin{array}{rl}& 2\pi R=\frac{x}{2}\\ & \Rightarrow R=\frac{x}{2}\times \frac{7}{2\times 22}\\ & \Rightarrow R=\frac{7x}{88}\end{array}$$
Original area :
$$\begin{array}{rl}& =\pi r^2\\ & =\left(\frac{22}{7}\times \frac{7x}{44}\times \frac{7x}{44}\right)\\ & =\frac{7x^2}{88}\end{array}$$
New area :
$$\begin{array}{rl}& =\pi R^2\\ & =\left(\frac{22}{7}\times \frac{7x}{88}\times \frac{7x}{88}\right)\\ & =\frac{7x^2}{352}\end{array}$$
Decrease in area :
$$\begin{array}{rl}& =\left(\frac{7x^2}{88}-\frac{7x^2}{352}\right)\\ & =\frac{21x^2}{352}\end{array}$$
∴ Decrease % :
$$\begin{array}{rl}& =\left(\frac{21x^2}{352}\times \frac{88}{7x^2}\times 100\right)\%\\ & =75\%\end{array}$$