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:

Answer: Option B

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,
$$\begin{array}{rl}& 60\times \left(t+\frac{20}{60}\right)=75\times \left(t-\frac{44}{60}\right)\\ & \Rightarrow 4\left(t+\frac{1}{3}\right)=5\left(t-\frac{11}{15}\right)\\ & \Rightarrow 4t+\frac{4}{3}=5t-\frac{11}{3}\\ & \Rightarrow 5t-4t=\frac{11}{3}+\frac{4}{3}\\ & \Rightarrow t=\frac{11+4}{3}\\ & \Rightarrow t=5\text{ hours}\\ & \text{Distance }\left(\text{D}\right)=60\times \left(5+\frac{20}{60}\right)\\ & \Rightarrow \text{D}=60\times \left(5+\frac{1}{3}\right)\\ & \Rightarrow \text{D}=60\times \frac{16}{3}\\ & \Rightarrow \text{D}=320\text{ km}\end{array}$$
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