A6. Wouter is walking from his house to his sports club. He could have taken his racing bike; with that, the journey would be seven times faster. But he left it at home. After $1 \mathrm{~km}$, he has reached a point where it makes no difference in time whether he continues walking or goes back home to get his racing bike. How many km is he still away from his sports club at that moment?
(A) $\frac{8}{7}$
(B) $\frac{7}{6}$
(C) $\frac{6}{5}$
(D) $\frac{5}{4}$
(E) $\frac{4}{3}$
A6. Wouter is walking from his house to his sports club. He could have taken his racing bike; with that, the journey would be seven times faster. But he left it at home. After $1 \mathrm{~km}$, he has reached a point where it makes no difference in time whether he continues walking or goes back home to get his racing bike. How many km is he still away from his sports club? (E) $\frac{4}{3}$

Solution: Let's call the time Wouter spent on the first kilometer a 'quarter'; this might be more or less than 15 minutes, but it will turn out not to matter. Let $x$ be the distance in question. He has already spent one 'quarter' walking that one km. Continuing to walk will cost him $x$ 'quarters'. Going back home and then taking the bike will first cost another 1 'quarter' of walking and then $\frac{1+x}{7}$ 'quarters' of cycling; the bike is 7 times faster. Therefore, it must be that $x=1+\frac{1+x}{7}$, so $7x=7+(1+x)$, or $6x=8$. Conclusion: $x=\frac{8}{6}=\frac{4}{3}$.