22. Lewis and Geraint left Acaster to travel to Beetown at the same time. Lewis stopped for an hour in Beetown and then drove back towards Acaster. He drove at a constant 70 $\mathrm{km} / \mathrm{h}$. He met Geraint, who was cycling at a constant $30 \mathrm{~km} / \mathrm{h}, 105 \mathrm{~km}$ from Beetown. How far is it from Acaster to Beetown?
A $315 \mathrm{~km}$
B $300 \mathrm{~km}$
C $250 \mathrm{~km}$
D $210 \mathrm{~km}$
E $180 \mathrm{~km}$
Solution
A

Let the distance from Acaster to Beetown be $x \mathrm{~km}$. Lewis met Geraint $105 \mathrm{~km}$ from Beetown. Therefore, when they met, Lewis had travelled $(x+105) \mathrm{km}$ and Geraint had travelled $(x-105) \mathrm{km}$. Since time $=\frac{\text { distance }}{\text { speed }}$, when they met, Geraint had been travelling for $\left(\frac{x-105}{30}\right)$ hours and Lewis had been travelling for $\left(\frac{x+105}{70}+1\right)$ hours, where the extra ' +1 ' represents the hour Lewis spends in Beetown. These two times are equal and hence $\frac{x-105}{30}=\frac{x+105}{70}+1$. When we multiply each term in this equation by 210 , we get $7(x-105)=3(x+105)+210$. This can be simplified to $4 x=1260$, which has solution $x=315$.