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?

Answer: 6 minutes

Explanation:

Let

[tex]t = \text{time it takes for Jill to catch up with Jack} \newline \indent d_1 = \text{distance traveled by Jill when she catches up with Jack} \newline \indent d_2 = \text{distance traveled by Jack when Jill catches him}[/tex]

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

[tex]d_1 = d_2[/tex]    (1)

Moreover, since [tex]d_1[/tex] represents the distance elapsed by Jill when she catches Jack and it takes time t for Jill to do that, 

[tex]d_1 = 360t[/tex]   (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 [tex]d_2[/tex]. 

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,

[tex]d_2 = 240(t + 2)[/tex]   (3)

Then, we substitute the values of [tex]d_1[/tex] and [tex]d_2[/tex] derived in equations (2) and (3) and use equation (1) so that 

[tex]360t = 240(t + 2) \\ \indent 360t = 240t + 480 \\ \indent 360t - 240t = 480 \\ \indent 120t = 480 \\ \indent t = 4 [/tex]

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

t + 2 = 6 minutes