A student rides on bicycle at 8 km/h and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/h and reaches school 5 minutes early. How far is the school from his house?

Correct option is C

Given:
Speed on the first day = 8 km/h, and the student is 2.5 minutes late.
Speed on the second day = 10 km/h, and the student is 5 minutes early.

Formula used:
Distance = Speed $$\times$$​ Time

Solution:
Let the actual time required to reach the school be T hours and Distance = d km.
2.5 min in hours = $$\frac{2.5}{60} = \frac{1}{24}$$​​
5 min in hours = $$\frac{5}{60} = \frac{1}{12}$$

​​For the first day:
​$$\frac{d}{8} = T + \frac{1}{24}$$​​
For the second day:
​$$\frac{d}{10} = T - \frac{1}{12}$$

Now,
​$$\frac{d}{8} - \frac{d}{10} = T + \frac{1}{24} - T + \frac{1}{12}$$​​

​$$\frac{d}{8} - \frac{d}{10} = \frac{1}{24} + \frac{1}{12}$$

$$\frac{5d - 4d}{40} = \frac{1+2}{24}$$​​​

​$$\frac{d}{40} = \frac{3}{24}$$

$$\frac{d}{40} = \frac{1}{8}$$

d = $$\frac{40}{8}$$

d = 5km

so, 5km  is the school from his house.

Thus, the correct answer is (c).