A student goes to school at the rate of 2.5 km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is

When speed is 2.5 km/hour time = t + (6/60) hours
When speed is 3 km/hour time is t - (10/60) hours
Distance is same.Hence the product of speed and time must be same.
2.5 (t + (6/60)) = 3(t - (10/60)
2.5t + (15/60) = 3t - (30/60)
0.5t=(15/60)+(30/60)
0.5t = 45/60 hours
t = 3/2 hours
Distance = 3 ( t - (10/60) )
=3t - (30/60)
=3 (3/2) - (30/60) = (9/2)-(1/2) = (8/2)=4 kms
Hence Option B is the correct answer.