# Sunday Times Teaser 2588 – On-Line

*by H Bradley and C Higgins*

#### Published: 29 April 2012 (link)

Linesmen Alf and Bert were checking the railway line from Timesborough to Teaserton: Alf started at Timesborough, Bert at Teaserton and they walked towards each other at their own steady speeds. A train from Timesborough, travelling at constant speed, reached Alf at noon and passed him ten seconds later. At 12.06 the train reached Bert and passed him eight seconds later.

At what time did the two men meet?

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No program for this one since the analysis needed to write a program makes the resulting maths is trivial.

Let the length of the train be \(l\) and its speed \(v_t\); let the speeds of the two linesmen be \(v_1\) and \(v_2\). From the time it takes the train to pass the linesmen we have: \[l = 10(v_t – v_1) = 8(v_t + v_2)\] and hence \[v_t = 5v_1 + 4v_2\] Let \(L\) be the distance between the linemen When the train passes the first linesman at 12 noon (\(t = 0\) and distance \(x = 0\)). Let \(t_2\) and \(x_2\) be the time and distance at which the train passes the second linesman at 12:06, six minutes later (hence \(t_2=6\) minutes): \[x_2 = v_t t_2\] \[L – x_2 = v_2 t_2\] \[L=(v_2+v_t)t_2 = 5(v_1+v_2)t_2\] Let \(t_3\) and \(x_3\) be the time and distance at which the linesmen meet: \[x_3=v_1 t_3\] \[L-x_3=v_2 t_3\] \[L=(v_1+v_2)t_3\] Hence \(t_3=5t_2\), which gives \(t_3\) as 30 minutes. So the linesmen meet at 12:30.