Maths Olympiad Prep

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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 15\frac{1}{5} of an hour.

When Elina runs at 12 km/h for 15\frac{1}{5} of an hour, she runs 12 km/h 1 5 h = 2.4\text{12 km/h 1 5 h = 2.4} km to the north.

When Gustavo walks at 5 km/h for 15\frac{1}{5} of an hour, he walks 5 km/h 1 5 h = 1\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.

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As they change direction, we use the Pythagorean Theorem to calculate their distance from each other. Using this, we obtain (2.4 km ) 2 + (1 km ) 2 = 2.6 km\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 12 km/h + 5 km/h = 17 km/h\text{12 km/h + 5 km/h = 17 km/h}.

Thus, it takes 2.6 km 17 km/h 0.153 h\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.

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