Anita goes to College at 20 km/h and reaches college 4 minutes late. Next time she goes at 25 km/h and reaches the college 2 minutes earlier than the scheduled time. What is the distance of her school?

Question

Anita goes to College at 20 km/h and reaches college 4 minutes late. Next time she goes at 25 km/h and reaches the college 2 minutes earlier than the scheduled time. What is the distance of her school?

A. 12 km

B. 16 km

C. 10 km

D. 15 km

Answer: C

Solution:

Speed = Distance/ time

The distance is common here. So let’s equate the distances.

Case of going late

D = 20 * (T + 4 / 60) ——- Equation 1

Case of going early

D = 25 * (T – 2 / 60) ——-Equation 2

Equating the Distances,

20 * (T + 4 / 60) = 25 * (T – 2 / 60)

20T + 80 / 60 = 25T – 50 / 60

130 / 60 = 5T

T = 130 / (60 * 5) = 13 / 30

Now, let’s calculate the distance from Equation 1.

D = 20 * (13 / 30 + 4 / 60) = 20 * (26 / 60 + 4 / 60) = 20 * 30 / 60

Therefore, D = 10 km