Geraint always cycles to work, leaving at $$8$$am every morning. When he averages $$15{\text{km}}/{\text{h}}\textbackslash;$$,he arrives $$10$$ minutes late. However, when he averages $$30{\text{km}}/{\text{h}}\textbackslash;$$, he arrives $$10$$ minutes early. What speed should he average to arrive on time? 
Overseas Competition->Knowledge Point->Word Problem Modules
Let $$x\text{km}$$ be the distance Geraint cycles and let $$t $$ hours be the time his journey should take if he is to be on time. Since $$\frac{\text{distance}}{\text{speed}}\text{=time}$$, the information in the question tells us that  $$\frac{x}{15}=t+\frac{1}{6}$$ and that $$\frac{x}{30}=t-\frac{1}{6}$$. When we subtract the second equation from the first, we obtain  $$\frac{x}{30}=\frac{2}{6}$$ and so $$x = 10$$. Hence, from the second equation, $$\frac{10}{30}=t-\frac{1}{6}$$ and so $$t=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$$. Therefore, to arrive on time, Geraint needs to travel $$10\text{km}$$ in~ $$\frac{1}{2}$$ hour, which is an average speed of $$20{\text{km}}/{\text{h}}\textbackslash;$$. 