Henry Ernest Dudeney/Modern Puzzles/35 - Sharing a Bicycle/Solution

by : $35$

 * Sharing a Bicycle

Solution
Anderson cycles $11 \tfrac 1 9$ miles, drops the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.

Thus he cycles for $1 \tfrac 1 9$ hours and walks for $2 \tfrac 2 9$ hours.

Brown thus walks $11 \tfrac 1 9$ miles, picks up the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.

Thus he also walks for $2 \tfrac 2 9$ hours and cycles for $1 \tfrac 1 9$ hours.

Proof
Let Anderson and Brown be denoted by $A$ and $B$ respectively.

To keep it simple, we will assume one changeover, and that $A$ starts by cycling.

Let $d$ miles be the distance from the start to where $A$ dismounts to start walking.

Let $t$ hours be the time taken to do the total journey.

Let $t_a$ hours be the time taken by $A$ to travel $d$.

Let $t_b$ hours be the time taken by $B$ to travel $d$.

We have: