Elina and Gustavo leave Cayley H.S. at 3:00 p.m. Elina runs north at a constant speed of $12 \mathrm{~km} / \mathrm{h}$. Gustavo walks east at a constant speed of $5 \mathrm{~km} / \mathrm{h}$. After 12 minutes, Elina and Gustavo change direction and travel directly towards each other, still at $12 \mathrm{~km} / \mathrm{h}$ and $5 \mathrm{~km} / \mathrm{h}$, respectively. The time that they will meet again is closest to
(A) 3:24 p.m.
(B) 3:35 p.m.
(C) $3: 25$ p.m.
(D) 3:29 p.m.
(E) 3:21 p.m.
Elina and Gustavo start by running and walking for 12 minutes.

Since there are 60 minutes in 1 hour, 12 minutes equals $\frac{1}{5}$ of an hour.

When Elina runs at $12 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{5}$ of an hour, she runs $12 \mathrm{~km} / \mathrm{h} \times \frac{1}{5} \mathrm{~h}=2.4 \mathrm{~km}$ to the north.

When Gustavo walks at $5 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{5}$ of an hour, he walks $5 \mathrm{~km} / \mathrm{h} \times \frac{1}{5} \mathrm{~h}=1 \mathrm{~km}$ to the east. At this point, Elina and Gustavo start to travel directly towards each other.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-030.jpg?height=307&width=369&top_left_y=1630&top_left_x=932)

As they change direction, we use the Pythagorean Theorem to calculate their distance from each other. Using this, we obtain $\sqrt{(2.4 \mathrm{~km})^{2}+(1 \mathrm{~km})^{2}}=2.6 \mathrm{~km}$.

Since Elina continues to travel at $12 \mathrm{~km} / \mathrm{h}$, Gustavo continues to travel at $5 \mathrm{~km} / \mathrm{h}$, and they travel directly towards each other, they close the gap at a rate of $12 \mathrm{~km} / \mathrm{h}+5 \mathrm{~km} / \mathrm{h}=17 \mathrm{~km} / \mathrm{h}$.

Thus, it takes $\frac{2.6 \mathrm{~km}}{17 \mathrm{~km} / \mathrm{h}} \approx 0.153 \mathrm{~h}$ for them to meet.

Since there are 60 minutes in an hour, $0.153 \mathrm{~h}$ is equivalent to roughly 9.18 minutes.

Since Elina and Gustavo leave at 3:00 p.m. and travel for 12 minutes and then for an additional 9 minutes, they meet again at approximately 3:21 p.m.

ANSWER: (E)