Sunday Times Teaser 2486

by H Bradley and C Higgins

1. 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.