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Problem 568

AMC 10/12, early questions

Algebra Difficulty 3.7 Find the answer CEMC Fermat · Canada · 2012

Katie and Sarah run at different but constant speeds. They ran two races on a track that measured 100 m from start to finish. In the first race, when Katie crossed the finish line, Sarah was 5 m behind. In the second race, Katie started 5 m behind the original start line and they ran at the same speeds as in the first race. What was the outcome of the second race?

$\text{Katie and Sarah crossed the finish line at the same time.}$
$\text{When Katie crossed the finish line, Sarah was 0.25 m behind.}$
$\text{When Katie crossed the finish line, Sarah was 0.26 m behind.}$
$\text{When Sarah crossed the finish line, Katie was 0.25 m behind.}$
$\text{When Sarah crossed the finish line, Katie was 0.26 m behind.}$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

In the first race, Katie ran 100 m in the same time that Sarah ran 95 m.

This means that the ratio of their speeds is $100 : 95 = 20 : 19$ .

In other words, in the time that Sarah runs 1 m, Katie runs $\frac{20}{19} \approx 1.053$ m.

Put another way, in the time that Katie runs 1 m, Sarah runs $\frac{19}{20}=0.95$ m.

In the second race, Katie must run 105 m and Sarah must run 100 m.

If Sarah finishes first, then Katie must not have completed 105 m in the time that it takes Sarah to complete 100 m.

But Katie runs 1.053 m for 1 m that Sarah runs, so Katie will in fact run more than 105 m in the time that Sarah runs 100 m.

Therefore, Katie must finish first.

In the time that Katie runs 105 m, Sarah will run $105 \times \frac{19}{20} = \frac{1995}{20} = \frac{399}{4} = 99\frac{3}{4}$ .

Thus, Sarah was $100 - 99\frac{3}{4} = \frac{1}{4} = 0.25$ m behind.

Therefore, when Katie crossed the finish line, Sarah was 0.25 m behind.

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