Two men start together walking to a certain destination, one at the speed of 3 km/h and another at the speed of 3.75 km/h. The latter arrives half an hour before the former, the distance is:

Correct option is C

Given:
Speed of the first man $$( S_1)$$​ = 3 km/h.
Speed of the second man $$(S_2) =$$​3.75 km/h.
The second man arrives half an hour earlier than the first man.
Concept Used:
Time taken to cover a distance =$$\frac{\text{Distance}}{\text{Speed}}$$​​
Solution:
Let the distance to the destination be D km.
$$T_1 = \frac{D}{S_1} = \frac{D}{3} \, \text{hours}$$​​
$$T_2 = \frac{D}{S_2} = \frac{D}{3.75} \, \text{hours}$$​
The second man arrives half an hour earlier, so:
$$T_1 - T_2 = 0.5 \, \text{hours}$$​​
$$\frac{D}{3} - \frac{D}{3.75} = 0.5$$​

$$\frac{5D}{15} - \frac{4D}{15} = 0.5$$​

$$\frac{D}{15} = 0.5$$​
D = $$0.5 \times 15 = 7.5 \, \text{km}$$​