A path of length $14 \mathrm{~m}$ consists of 7 unshaded stripes, each of length $1 \mathrm{~m}$, alternating with 7 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_6ed09463f225f8ba1f07g-029.jpg?height=361&width=830&top_left_y=993&top_left_x=604)

The wheel makes exactly 1 complete revolution. 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) $16 \%$
(D) $24 \%$
(E) $22 \%$
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 label 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 $\pi \mathrm{m} \approx 3.14 \mathrm{~m}$.

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

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 $1 \mathrm{~m}$.

Therefore, the first shaded section touches shaded stripes between $1 \mathrm{~m}$ and $2 \mathrm{~m}$, and between $3 \mathrm{~m}$ and $\pi \mathrm{m}$.

The second shaded section touches shaded stripes between $7 \mathrm{~m}$ and $8 \mathrm{~m}$, and between $9 \mathrm{~m}$ and $3 \pi \mathrm{m}$.

Therefore, the total length of "shaded on shaded" is $1 \mathrm{~m}+(\pi-3) \mathrm{m}+1 \mathrm{~m}+(3 \pi-9) \mathrm{m}$ or $(4 \pi-10) \mathrm{m}$.

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

This means that the required percentage of time equals $\frac{(4 \pi-10) \mathrm{m}}{4 \pi \mathrm{m}} \times 100 \% \approx 20.4 \%$.

Of the given choices, this is closest to $20 \%$, or choice (A).

ANSWER: (A)