1. Mike is doing a one-hour cycling challenge. He has a computer which predicts how far he will cycle in the rest of the hour based on his average speed so far. After cycling $1 \mathrm{~km}$ in $t$ minutes, he checks the distance the computer predicts he will cycle in the remaining time and it shows $d \mathrm{~km}$.
In the next 36 minutes, he cycles $15 \mathrm{~km}$. He checks the computer again and finds it still predicts he will cycle $d \mathrm{~km}$ in the remaining time.
Find the distance shown on the computer each time he looks.
Solution
Predicted distance is calculated by average speed so far multiplied by the time remaining.
$$
\begin{array}{c}
d=\frac{60}{t} \times \frac{60-t}{60} \\
d=\frac{16 \times 60}{t+36} \times \frac{60-t-36}{60}
\end{array}
$$

These lead to the following equation:
$$
\frac{60-t}{t}=\frac{16(24-t)}{t+36}
$$

This can be multiplied up and simplified:
$$
2160+24 t-t^{2}=384 t-16 t^{2}
$$

This simplifies down to the equation below:
$$
(t-12)^{2}=0 .
$$

This leads to the only solution being $t=12$. Substituting this into either of the original equations gives $d=4$ so the computer shows a predicted distance of $4 \mathrm{~km}$ each time.