Arvind goes to his school by cycle at a speed of 4 km/h and reaches his school 15 minutes late. If he moves at a speed of 4
2
1
​
km/h, he reaches the school 10 minutes early. What is the distance (in km) of his school from his home?

Let the distance to the school be D km.
Speed 1 (S
1
​
) = 4 km/h
Speed 2 (S
2
​
) = 4.5=
2
9
​
km/h

The difference in time between being 15 minutes late and 10 minutes early is:

Δt=15−(−10)=25 minutes=
60
25
​
 hours=
12
5
​
 hours
Using the distance-speed relationship formula:

D=
∣S
1
​
−S
2
​
∣
S
1
​
×S
2
​

​
×Δt
D=
∣4−4.5∣
4×4.5
​
×
12
5
​

D=
0.5
18
​
×
12
5
​

D=36×
12
5
​
=3×5=15 km