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:

384

We know that , 

Distance = Speed × time

Let us consider that , Actual time taken = t hours

According to question,

60 × ( t + \(\frac{20}{60}\) ) = 75 × ( t - \(\frac{44}{60}\) ) 

4 × ( t + \(\frac{1}{3}\) ) = 5 × ( t - \(\frac{11}{15}\) )

4t + \(\frac{4}{3}\)  = 5t - \(\frac{11}{3}\) 

t = \(\frac{4}{3}\) + \(\frac{11}{3}\)

t = \(\frac{15}{3}\)

= 5 hours

Now,

Distance b/w H and O (D) = 60 × ( 5 + \(\frac{20}{60}\) ) 

= 60 × ( 5 + \(\frac{1}{3}\) )

= 60 × ( \(\frac{16}{3}\) )

= 320 km

And average speed = \(\frac{320}{5}\) = 64 km/h

Now,  Sum of D and S = 320 + 64

= 384