The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800 km. The speed of the train in the onward journey is :

Solution(By Examveda Team)

Let the speed in return journey be x km/hr
Then, speed in onward journey :
$$\eqalign{ & = \left( {\frac{{125}}{{100}}x} \right){\text{ km/hr}} \cr & = \left( {\frac{5}{4}x} \right){\text{ km/hr}} \cr} $$
Average speed :
$$\eqalign{ & {\text{ = }}\left( {\frac{{2 \times \frac{5}{4}x \times x}}{{\frac{5}{4}x + x}}} \right){\text{ km/hr}} \cr & {\text{ = }}\left( {\frac{{10x}}{9}} \right){\text{ km/hr}} \cr & \therefore \left( {800 \times \frac{9}{{10x}}} \right) = 16 \cr & \Leftrightarrow x = \left( {\frac{{800 \times 9}}{{16 \times 10}}} \right) \cr & \Leftrightarrow x = 45 \cr} $$
So, speed in onward journey :
$$\eqalign{ & = \left( {\frac{5}{4} \times 45} \right){\text{ km/hr}} \cr & = 56.25{\text{ km/hr}} \cr} $$