CEMC Cayley, 2021

Algebra Difficulty 4.6 AIME Find the answer Canada

Elina and Gustavo leave Cayley H.S. at 3:00 p.m. Elina runs north at a constant speed of 12 km/h. Gustavo walks east at a constant speed of 5 km/h. After 12 minutes, Elina and Gustavo change direction and travel directly towards each other, still at 12 km/h and 5 km/h, respectively. The time that they will meet again is closest to

Pick one

Solution

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 km/h for $\frac{1}{5}$ of an hour, she runs $\text{12 km/h 1 5 h = 2.4}$ km to the north.

When Gustavo walks at 5 km/h for $\frac{1}{5}$ of an hour, he walks $\text{5 km/h 1 5 h = 1}$ km to the east.

At this point, Elina and Gustavo start to travel directly towards each other.

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

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

Thus, it takes $\text{2.6 km 17 km/h 0.153 h}$ for them to meet.

Since there are 60 minutes in an hour, 0.153 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.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.