Walking at 4 mph, a person covers a certain distance in a given time. If he increases his speed to 9 mph, he covers 7.5 miles more in the same time. How much distance does he cover originally?

Correct option is D

Given:
Speed 1 (S₁): 4 mph
Speed 2 (S₂): 9 mph
Extra distance covered at S₂: 7.5 miles
Formula Used:
Distance = Speed × Time
Time is constant for both cases, so:
Time = $$\frac {\text {Distance at S₁}}{ S₁} = \frac{{ \text {Distance at S₁ + Extra Distance}}}{ S₂}$$​​
Solution:
Let the original distance covered be D miles.
​$$\frac{D}{4} = \frac{D + 7.5}{9}$$​​
9D = 4(D + 7.5)
9D = 4D + 30
5D = 30
D = 6
Thus, the original distance covered is 6 miles.