The distance between the places H and O is D units. The average speed that gets a person from H to O in a stipulated time is S units. He takes 20 minutes more time than usual if he travels at 60 km/h, and reaches 44 minutes early if he travels at 75 km/h. The sum of the numerical values of D and S is:

Solution (By Examveda Team)

Let right time for the train to cover its journey (in minutes) be $$t$$ hours, then
As we know,
Distance = Speed × Time
According to the question,
$$\eqalign{ & 60 \times \left( {t + \frac{{20}}{{60}}} \right) = 75 \times \left( {t - \frac{{44}}{{60}}} \right) \cr & \Rightarrow 4\left( {t + \frac{1}{3}} \right) = 5\left( {t - \frac{{11}}{{15}}} \right) \cr & \Rightarrow 4t + \frac{4}{3} = 5t - \frac{{11}}{3} \cr & \Rightarrow 5t - 4t = \frac{{11}}{3} + \frac{4}{3} \cr & \Rightarrow t = \frac{{11 + 4}}{3} \cr & \Rightarrow t = 5{\text{ hours}} \cr & {\text{Distance }}\left( {\text{D}} \right) = 60 \times \left( {5 + \frac{{20}}{{60}}} \right) \cr & \Rightarrow {\text{D}} = 60 \times \left( {5 + \frac{1}{3}} \right) \cr & \Rightarrow {\text{D}} = 60 \times \frac{{16}}{3} \cr & \Rightarrow {\text{D}} = 320{\text{ km}} \cr} $$
Distance between H and O is 320 km
Average speed (S) = $$\frac{{320}}{5}$$ = 64 km/hr
∴ Sum of numerical value of D and S = 320 + 64 = 384