A hare is 30 meters in front of a hound and he runs 3 meters

A hare is 30 meters in front of a hound., and he runs 3 meters while the hound runs 6 meters, how many meters must the hound run to catch the hare?

Solution

let n= intervals, periods of time
.
6n = 30 +3n,,,,,hound runs 6 / interval,,hare runs 3/ interval but has 30 head start
.
3n =30
.
n=10
.
therefore hound runs,,,10*6=60
.
hare runs 10*3 = 30,,,plus head start of 30 = 60