If the circumference of a circle is increased by 20%, what will be the effect on the circle ?

Answer: Option B

Let the original circumference be x units.
Then,
New circumference = 120% of x = $$\left(\frac{6x}{5}\right)$$
Let original radius = r and new radius = R
$$\begin{array}{rl}& 2\pi r=x\\ & \Rightarrow r=\frac{7x}{2\times 22}\\ & \Rightarrow r=\frac{7x}{44}\end{array}$$
And,
$$\begin{array}{rl}& 2\pi R=\frac{6x}{5}\\ & \Rightarrow R=\frac{6x}{5}\times \frac{7}{2\times 22}\\ & \Rightarrow R=\frac{21x}{110}\end{array}$$
Original are :
$$\begin{array}{rl}& =\pi r^2\\ & =\left(\frac{22}{7}\times \frac{7x}{44}\times \frac{7x}{44}\right)\\ & =\frac{7x^2}{88}\end{array}$$
New area :
$$\begin{array}{rl}& =\pi R^2\\ & =\left(\frac{22}{7}\times \frac{21x}{110}\times \frac{21x}{110}\right)\\ & =\frac{63x^2}{550}\end{array}$$
Increase area :
$$\begin{array}{rl}& =\left(\frac{63x^2}{550}-\frac{7x^2}{88}\right)\\ & =\frac{77x^2}{2200}\end{array}$$
∴ Increase % :
$$\begin{array}{rl}& =\left(\frac{77x^2}{2200}\times \frac{88}{7x^2}\times 100\right)\%\\ & =44\%\end{array}$$