Let the distance between home and the station be \(l\) and let the speeds of John and the car be \(v_j\) and \(v_c\) respectively. Assume that Pat starts driving to the station at \(d\) minutes past 5:30 and passes John at \(t_1\) minutes paqst 5:30. We then have: \[l=v_c(30-d) = v_j t_1 + v_c(t_1 – d)\] We can eliminate \(l\) and \(d\) from these two equations to obtain:\[t_1=\frac{30}{1+v_j/v_c}\] Assume that the return journey from the station starts at \(w\) minutes past 6:00 (\(w+30\) after 5:30) and that John is passed again at \(t_2\) minutes after 5:30. We hence have: \[v_j t_2=v_c(t_2 – w – 30)\] from which we obtain:\[t_2=\frac{w+30}{1-v_j/v_c}\] We can now eliminate \(v_j/v_c\) to give (we know that \(t_2=t_1+13\)): \[t_2=\frac{(w+30)t_1}{2t_1-30}=t_1+13\] Simplifying gives a quadratic equation for \(t_1\):\[2t_1^2-(w+34)t_1-390=0\] We can now “complete the square” to show that:\[\{4t_1-(w+34)\}^2=(w+34)^2+3120\] The values of \(w\) that give integer solution are 3, 13, 19, 34, … the first of which gives \[4t_1-37=\pm 67\] Hence \(t_1\) and \(t_2\) are 26 and 39 minutes respectively. So John is passed by Pat and Joe at 6:09.