A train when moves at an average speed of 40 km/h, reaches its destination on time. When its average speed becomes 35 km/h, then it reaches its destination 15 min late. Find the distance.

Correct option is D

Given:
Average speed of the train when it reaches on time, $$v_1 = 40 \, \text{km/h} .$$​
Average speed of the train when it is late,  $$v_2 = 35 \, \text{km/h} .$$​
Time delay when the train is late, $$t = 15 \, \text{minutes} = \frac{15}{60} = 0.25 \, \text{hours}.$$​
Formula Used:
Time taken to reach the destination = $$\frac{\text{Distance}}{\text{Speed}} .$$​
Solution:
Let the distance to the destination be  D  km.
Time taken to reach the destination at speed  $$v_1 :$$​​
$$t_1 = \frac{D}{v_1} = \frac{D}{40} \, \text{hours}$$​
Time taken to reach the destination at speed $$v_2:$$​​
$$t_2 = \frac{D}{v_2} = \frac{D}{35} \, \text{hours}$$​​
The difference in time taken is equal to the delay:
$$t_2 - t_1 = t$$​​
Substitute the expressions for $$t_1$$ and  $$t_2 :$$​
$$\frac{D}{35} - \frac{D}{40} = 0.25$$​​

$$\frac{8D}{280} - \frac{7D}{280} = 0.25$$​​

$$\frac{D}{280} = 0.25$$​​
$$D=0.25 \times 280 = 70 \, \text{km}$$​ 
The distance to the destination is 70 km.