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:

Distance between places H and O = D units

Average speed of the person from H to O = S

He takes 20 minutes more time than usual if he travels at 60 km/h

$$=$$>  $$\frac{D}{60}-\frac{D}{S}=\frac{20}{60}$$ .............(1)

He reches 44 minutes early than usual if he travels at 75 km/h

$$=$$>  $$\frac{D}{S}-\frac{D}{75}=\frac{44}{60}$$ ..............(2)

Solving (1)+(2)

$$\frac{D}{60}-\frac{D}{75}=\frac{20}{60}+\frac{44}{60}$$

$$=$$>  $$\frac{15D}{60\times75}=\frac{64}{60}$$

$$=$$>   D = 320

From (1), $$\frac{320}{60}-\frac{320}{S}=\frac{20}{60}$$

$$=$$>  $$\frac{16}{60}-\frac{16}{S}=\frac{1}{60}$$

$$=$$>  $$\frac{16}{S}=\frac{16}{60}-\frac{1}{60}$$

$$=$$>  $$\frac{16}{S}=\frac{15}{60}$$

$$=$$>   S = 64

$$\therefore\ $$Sum of D and S = 320 + 64 = 384

Hence, the correct answer is Option D