An algebra problem by Ahmad Khamis

Let the length (in metres) and velocity (in metres per hour) of the Muizenberg-bound train be $l_m$ and $v_m$ respectively. For the Cape Town-bound train, we have $l_c$ and $v_c$ respectively. From the question, $\frac{l_m}{v_m} = \frac{8}{60} =\frac{2}{15}$ and $\frac{l_c}{v_c} = \frac{12}{60} =\frac{1}{5}$ . We know $l_c =150$ so we have $\frac{150}{v_c} =\frac{1}{5}$ and $v_c =750$ .

From the fact about the two trains passing each other, we deduce that $\frac{l_c + l_m}{v_c + v_m} = \frac{9}{60} = \frac{3}{20}$ . Substituting in the known values gives us $\frac{150 + l_m}{750 + v_m} = \frac{3}{20}$ which gives us $3v_m - 20l_m = 750$ upon simplification. Since $\frac{l_m}{v_m} = \frac{2}{15}$ , $v_m = \frac{15l_m}{2}$ so $3(\frac{15l_m}{2}) - 20l_m = 750$ .

Hence $45l_m - 40l_m = 1500$ and $5l_m =1500$ thus $l_m = \boxed{300}$

Let the train to Muizenberg have length $x$ m and speed $y$ m/min and let the train to Cape Town have speed $z$ m\min. Then $x = 8y$ and $150 = 12z$ giving $z = 12.5$ . Also $150 + x = 9(y+z)$ , so $8y +150 = 9y +112.5$ . This gives $y = 37.5$ and $x = 300$ .