# Sunday Times Teaser 2486

*by H Bradley and C Higgins*

#### Published: 16 May 2010 (link)

Pat drove to the station at his regular speed to meet his brothers, John and Joe, whose train was due at 6pm. But John had caught an earlier train and, arriving half an hour early, decided to walk home at a steady pace. Pat waved as he passed John (a whole number of minutes before 6pm), but drove on to the station. Pat collected Joe on time and, a whole number of minutes later, they set off for home. Pat and Joe overtook John 13 minutes after Pat had waved to him.

At what time did Pat and Joe overtake John?

One Comment
Leave one →

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.