A $400 \mathrm{~m}$ track is constructed so that the points $A, B, C$, and $D$ divide the track into four segments of equal length. The Start is half-way between $A$ and $B$. Andrew begins at the Start and walks at a steady rate of $1.4 \mathrm{~m} / \mathrm{s}$ in a counter-clockwise direction. After exactly 30 minutes, to what point will Andrew

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-111.jpg?height=257&width=415&top_left_y=1926&top_left_x=1294)
be closest?
(A) $A$
(B) $B$
(C) $C$
(D) $D$
(E) Start
If Andrew walks 1.4 metres per second, then he walks $60 \times 1.4=84$ metres per minute.

Since Andrew is walking for 30 minutes, then he walks a total of $30 \times 84=2520 \mathrm{~m}$.

Now the total length of track is $400 \mathrm{~m}$, so after walking $2400 \mathrm{~m}$, Andrew is back at the Start line.

Since the points $A, B, C$, and $D$ are equally spaced, then consecutive points are $100 \mathrm{~m}$ apart. Since the Start is half-way between $A$ and $B$, then the Start is $50 \mathrm{~m}$ from $B$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-204.jpg?height=266&width=434&top_left_y=409&top_left_x=889)

Therefore, after walking $2450 \mathrm{~m}$, Andrew is at $B$.

After walking $70 \mathrm{~m}$ more to get to his total of $2520 \mathrm{~m}$, Andrew will be $70 \mathrm{~m}$ beyond $B$ and $30 \mathrm{~m}$ from $C$, so he will be closest to $C$.

ANSWER: $(\mathrm{C})$