$\mathrm{Al}$ and Bert must arrive at a town $22.5 \mathrm{~km}$ away. They have one bicycle between them and must arrive at the same time. Bert sets out riding at $8 \mathrm{~km} / \mathrm{h}$, leaves the bicycle and then walks at $5 \mathrm{~km} / \mathrm{h}$. Al walks at $4 \mathrm{~km} / \mathrm{h}$, reaches the bicycle and rides at $10 \mathrm{~km} / \mathrm{h}$. For how many minutes was the bicycle not in motion?
(A) 60
(B) 75
(C) 84
(D) 94
(E) 109
$\mathrm{Al}$ and Bert must arrive at a town $22.5 \mathrm{~km}$ away. They have one bicycle between them and must arrive at the same time. Bert sets out riding at $8 \mathrm{~km} / \mathrm{h}$, leaves the bicycle and then walks at $5 \mathrm{~km} / \mathrm{h}$. Al walks at $4 \mathrm{~km} / \mathrm{h}$, reaches the bicycle and rides at $10 \mathrm{~km} / \mathrm{h}$. For how many minutes was the bicycle not in motion?
(A) 60
(B) 75
(C) 84
(D) 94
(E) 109

## Solution

Let $x$ represent the distance that Bert rides his bicycle.

Therefore, he walks for $(22.5-x) \mathrm{km}$.

Bert's total time for the trip is $\left(\frac{x}{8}+\frac{22.5-x}{5}\right)$ hours and Al's is $\left(\frac{x}{4}+\frac{22.5-x}{10}\right)$ hours.

Since their times are equal,

$$
\begin{aligned}
\frac{x}{8}+\frac{22.5-x}{5} & =\frac{x}{4}+\frac{22.5-x}{10} \\
\frac{22.5-x}{5}-\frac{22.5-x}{10} & =\frac{x}{4}-\frac{x}{8} \\
\frac{2(22.5-x)}{10}-\frac{22.5-x}{10} & =\frac{2 x}{8}-\frac{x}{8} \\
\frac{22.5-x}{10} & =\frac{x}{8} \\
10 x & =180-8 x \\
18 x & =180 \\
x & =10 .
\end{aligned}
$$

This means that Bert rode for $1.25 \mathrm{~h}$ before he left the bicycle and $\mathrm{Al}$ walked for $2.5 \mathrm{~h}$ before he picked it up. The bicycle was thus not in motion for $1.25 \mathrm{~h}$ or 75 minutes.

ANSWER: (B)