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 ?

Solution(By Examveda Team)

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