A path of length $38 \mathrm{~m}$ consists of 19 unshaded stripes, each of length $1 \mathrm{~m}$, alternating with 19 shaded stripes, each of length $1 \mathrm{~m}$. A circular wheel of radius $2 \mathrm{~m}$ is divided into four quarters which are alternately shaded and unshaded. The wheel rolls at a constant speed along the path from the starting position shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-029.jpg?height=369&width=830&top_left_y=1019&top_left_x=604)

The wheel makes exactly 3 complete revolutions. The percentage of time during which a shaded section of the wheel is touching a shaded part of the path is closest to
(A) $20 \%$
(B) $18 \%$
(C) $24 \%$
(D) $22 \%$
(E) $26 \%$
Since the wheel turns at a constant speed, then the percentage of time when a shaded part of the wheel touches a shaded part of the path will equal the percentange of the total length of the path where there is "shaded on shaded" contact.

Since the wheel has radius $2 \mathrm{~m}$, then its circumference is $2 \pi \times 2 \mathrm{~m}$ which equals $4 \pi \mathrm{m}$.

Since the wheel is divided into four quarters, then the portion of the circumference taken by each quarter is $\pi \mathrm{m}$.

We call the left-hand end of the path $0 \mathrm{~m}$.

As the wheel rotates once, the first shaded section of the wheel touches the path between $0 \mathrm{~m}$ and $\pi \approx 3.14 \mathrm{~m}$.

As the wheel continues to rotate, the second shaded section of the wheel touches the path between $2 \pi \approx 6.28 \mathrm{~m}$ and $3 \pi \approx 9.42 \mathrm{~m}$.

While the wheel makes 3 complete rotations, a shaded quarter will be in contact with the path over 6 intervals (2 intervals per rotation).

The path is shaded for $1 \mathrm{~m}$ starting at each odd multiple of $1 \mathrm{~m}$, and unshaded for $1 \mathrm{~m}$ starting at each even multiple of $1 \mathrm{~m}$.

We make a chart of the sections where shaded quarters touch the path and the parts of these intervals that are shaded:

| Beginning of quarter $(\mathrm{m})$ | End of quarter $(\mathrm{m})$ | Shaded parts of path $(\mathrm{m})$ |
| :---: | :---: | :---: |
| 0 | $\pi \approx 3.14$ | 1 to $2 ; 3$ to $\pi$ |
| $2 \pi \approx 6.28$ | $3 \pi \approx 9.42$ | 7 to $8 ; 9$ to $3 \pi$ |
| $4 \pi \approx 12.57$ | $5 \pi \approx 15.71$ | 13 to $14 ; 15$ to $5 \pi$ |
| $6 \pi \approx 18.85$ | $7 \pi \approx 21.99$ | 19 to $20 ; 21$ to $7 \pi$ |
| $8 \pi \approx 25.13$ | $9 \pi \approx 28.27$ | $8 \pi$ to $26 ; 27$ to 28 |
| $10 \pi \approx 31.42$ | $11 \pi \approx 34.56$ | $10 \pi$ to $32 ; 33$ to 34 |

Therefore, the total length of "shaded on shaded", in metres, is

$$
1+(\pi-3)+1+(3 \pi-9)+1+(5 \pi-15)+1+(7 \pi-21)+(26-8 \pi)+1+(32-10 \pi)+1
$$

which equals $(16-2 \pi) \mathrm{m}$.

The total length of the path along which the wheel rolls is $3 \times 4 \pi \mathrm{m}$ or $12 \pi \mathrm{m}$.

This means that the required percentage of time equals $\frac{(16-2 \pi) \mathrm{m}}{12 \pi \mathrm{m}} \times 100 \% \approx 25.8 \%$.

Of the given choices, this is closest to $26 \%$, or choice (E).

ANSWER: (E)