Jack walked from school to his house at a rate of 240 ft/min. his sister jill left school 2 minutes after jack, walking home at a rate of 360 ft/min. how long after jack leaves school does jill catch up to him?

6 minutes

Explanation:

Let



Note that, Jill only catches Jack when she is faster (given in the problem) and has traveled the same distance with Jack. Mathematically, 

    (1)

Moreover, since  represents the distance elapsed by Jill when she catches Jack and it takes time t for Jill to do that, 

   (2)

The speed of Jack is already known so we need to figure out the time he already traveled from school when Jill catches up on him to know the value of . 

Since Jack already walked for 2 minutes before Jill walked and it will take time t for Jill to catch up on him, Jack has already walked for time (t + 2) when he is caught up by Jill. So,

   (3)

Then, we substitute the values of  and  derived in equations (2) and (3) and use equation (1) so that 



Hence, the time spent by Jack after he leaves school is given by

t + 2 = 6 minutes

Jim has run 5/6 of a mile so far. i hope that answers your question! good luck : )