NMOS 2012 Special Round question 16

Let's say they meet at time $T$ , and that the rest of Bondy's journey takes $T'$ (so that the rest of Abel's take $9T'$ ). If Bondy is $k$ times faster than Abel, then Abel's total journey time is $k$ times Bondy's; in other words $k=\frac{T+9T'}{T+T'}$ . We can plot their journeys on a distance-time graph as below:

The key observation is that the two blue triangles are similar. Using this, we see straight away that $\frac{T}{T'}=\frac{9T'}{T}$ ; solving gives $T=3T'$ and substituting into the above expression gives $k=3$ . So Bondy's speed is $3$ times Abel's, ie $\boxed{240}$ m/min.

Suppose the distance between the two towns is $D$ , and suppose Abel has traveled a distance $x$ when he encounters Bondy, who at this point has traveled a distance $D - x$ . Since their time of travel is the same at this point, with $v_{A}, v_{B}$ being their respective speeds we have that

$\dfrac{x}{v_{A}} = \dfrac{D - x}{v_{B}} \Longrightarrow \dfrac{v_{B}}{v_{A}} = \dfrac{D - x}{x} \space$ , (i).

Now after this point of encounter, Abel takes 9 times as long to travel the remaining distance $D - x$ to town B that Bondy takes to travel the remaining distance $x$ to town A, which translates to the equation

$\dfrac{D - x}{v_{A}} = 9 \times \dfrac{x}{v_{B}} \Longrightarrow \dfrac{v_{B}}{v_{A}} = \dfrac{9x}{D - x} \space$ , (ii)

Comparing equations (i) and (ii) we have that $\dfrac{D - x}{x} = \dfrac{9x}{D - x} \Longrightarrow \dfrac{(D - x)^{2}}{x^{2}} = 9 \Longrightarrow \dfrac{D - x}{x} = 3$ ,

and so from equation (i) we see that $v_{B} = 3v_{A} = 3 \times 80 = \boxed{240}$ m/min.