Question

mike runs four times as fast as jun. if mike gives jun a head start of 15 secs, how many secs will mike catch up with jun?

Answer 1

when mike starts jun will be 15*x far away then : // x=speed of jun
4xt=15x+t*x // t=time in secs
3xt=15x
t=5
then mike will catch up with jun after he runs for 5 seconds