12. Chris planned a $210 \mathrm{~km}$ bike ride. However, he rode $5 \mathrm{~km} / \mathrm{h}$ faster than he planned and finished his ride 1 hour earlier than he planned. His average speed for the ride was $x \mathrm{~km} / \mathrm{h}$. What is the value of $x$ ?
12. 35 Chris's time for the ride when he rode at an average speed of $x \mathrm{~km} / \mathrm{h}$ was $\frac{210}{x}$ hours. His planned speed was $(x-5) \mathrm{km} / \mathrm{h}$ when his time would have been $\frac{210}{x-5}$ hours. The question tells us that he completed the ride 1 hour earlier than planned, so $\frac{210}{x-5}-\frac{210}{x}=1$.
Therefore $210 x-210(x-5)=x(x-5)$ and hence $1050=x^{2}-5 x$. Thus $x^{2}-5 x-1050=0$ and hence $(x-35)(x+30)=0$. Therefore, since $x$ is positive, $x=35$.