A train increases its normal speed by 12.5% and reaches its destination 20 min earlier. What is the actual time taken by the train in the journey ?

Answer: Option C

Let the normal speed of the train be x km/hr
Then, new speed
$$\begin{array}{rl}& =\left(112\frac{1}{2}\%\text{ of }x\right)\text{ km/hr}\\ & =\left(\frac{225}{2}\times \frac{1}{100}\times x\right)\text{ km/hr}\\ & =\left(\frac{9}{8}x\right)\text{ km/hr}\end{array}$$
Let the distance covered be d km
Then,
$$\begin{array}{rl}& \Rightarrow \frac{d}{x}-\frac{d}{\left(\frac{9x}{8}\right)}=\frac{20}{60}\\ & \Rightarrow \frac{d}{x}-\frac{d}{\left(\frac{9x}{8}\right)}=\frac{1}{3}\\ & \Rightarrow \frac{d}{x}-\frac{8d}{9x}=\frac{1}{3}\\ & \Rightarrow \frac{d}{9x}=\frac{1}{3}\\ & \Rightarrow d=3x\end{array}$$
∴ Actual time taken:
$$\frac{d}{x}=\frac{3x}{x}=3\text{ hours = 180 minutes}$$