Maths Olympiad Prep

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, 2016

Algebra Difficulty 2.1 Junior Prove it Canada

When Esther and her older brother Paul race, Esther takes 5 steps every 2 seconds, and each of her steps is 0.4 m long. Paul also takes 5 steps every 2 seconds, but each of his steps is 1.2 m long.

In metres, how far does Esther travel in 2 seconds?
In metres per second, what is Paul’s speed?
If they both start a race at the same time, what distance ahead will Paul be after 2 minutes?
If Esther begins a race 3 minutes before Paul, how much time does it take Paul to catch Esther?

Solution

Every 2 seconds, Esther takes 5 steps and each of her steps is 0.4 m long.

Therefore, in 2 seconds Esther travels a distance of 5×0.4=25\times0.4=2 m.
Solution 1

Every 2 seconds, Paul takes 5 steps and each of his steps is 1.2 m long.

Therefore, every 2 seconds Paul travels a distance of 5×1.2=65\times1.2=6 m.

Since Paul travels 6 m every 2 seconds, his speed is 6÷2=36\div2=3 m/s.

Solution 2

Paul takes 5 steps every 2 seconds and so Paul takes 2.5 steps every second.

The length of each of Paul’s steps is 1.2 m, and so every second Paul travels a distance of 2.5×1.2=32.5\times1.2=3 m.

Thus, Paul travels at a speed of 3 m/s.
Solution 1

Paul travels at a speed of 3 m/s, and so in 120 seconds (2 minutes), Paul travels a distance of 3×120=3603\times120=360 m.

Esther travels 2 m every 2 seconds and so she travels at a speed of 1 m/s.

In 120 seconds, Esther travels a distance of 1×120=1201\times120=120 m.

If Paul and Esther start a race at the same time, then after 2 minutes, Paul will be 360120=240360-120=240 m ahead of Esther.

Solution 2

Each of Paul’s steps is 1.20.4=0.81.2-0.4=0.8 m longer than each of Esther’s steps.

Every 2 seconds, Paul and Esther each take 5 steps, and so in 2×60=1202\times60=120 seconds (2 minutes), Paul and Esther each take 5×60=3005\times60=300 steps.

If Paul and Esther start a race at the same time, then after 2 minutes Paul will be 300×0.8=240300\times0.8=240 m ahead of Esther.

Solution 3

Paul travels at a speed of 3 m/s, and Esther travels 2 m every 2 seconds, so she travels at a speed of 1 m/s.

This means that in 1 second, Paul travels 3 m and Esther travels 1 m.

Every second, Paul travels 31=23-1=2 m farther than Esther.

If Paul and Esther start a race at the same time, then after 120 seconds (2 minutes), Paul will be 120×2=240120\times2=240 m ahead of Esther.
Solution 1

Esther travels 2 m every 2 seconds or 1 m/s, and so in 180 seconds (3 minutes), Esther travels 180×1=180180\times1=180 m.

Paul travels at a speed of 3 m/s, which is 2 m/s faster than the speed at which Esther travels.

Therefore, every second, Paul travels 2 m farther than Esther travels.

Since Esther begins the race 180 m ahead of Paul, it will take Paul 180÷2=90180\div2=90 s to catch Esther.

Solution 2

Esther travels 2 m every 2 seconds or 1 m/s, and so in 180 seconds (3 minutes), Esther travels 180×1=180180\times1=180 m.

Each of Paul’s steps is 1.20.4=0.81.2-0.4=0.8 m longer than each of Esther’s steps.

Since Paul and Esther each step at the same rate (5 steps every 2 seconds), then it will take Paul 180÷0.8=225180\div0.8=225 steps to catch Esther.

Paul takes 5 steps every 2 seconds, and so it will take Paul 2255×2=45×2=90\frac{225}{5}\times2=45\times2=90 s to catch Esther.

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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.