Katie and Sarah run at different but constant speeds. They ran two races on a track that measured $100 \mathrm{~m}$ from start to finish. In the first race, when Katie crossed the finish line, Sarah was $5 \mathrm{~m}$ behind. In the second race, Katie started $5 \mathrm{~m}$ 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?

(A) Katie and Sarah crossed the finish line at the same time.

(B) When Katie crossed the finish line, Sarah was $0.25 \mathrm{~m}$ behind.

(C) When Katie crossed the finish line, Sarah was $0.26 \mathrm{~m}$ behind.

(D) When Sarah crossed the finish line, Katie was $0.25 \mathrm{~m}$ behind.

(E) When Sarah crossed the finish line, Katie was $0.26 \mathrm{~m}$ behind.

## Part C: Each correct answer is worth 8.
In the first race, Katie ran $100 \mathrm{~m}$ in the same time that Sarah ran $95 \mathrm{~m}$.

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

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

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

In the second race, Katie must run $105 \mathrm{~m}$ and Sarah must run $100 \mathrm{~m}$.

If Sarah finishes first, then Katie must not have completed $105 \mathrm{~m}$ in the time that it takes Sarah to complete $100 \mathrm{~m}$.

But Katie runs $1.053 \mathrm{~m}$ for $1 \mathrm{~m}$ that Sarah runs, so Katie will in fact run more than $105 \mathrm{~m}$ in the time that Sarah runs $100 \mathrm{~m}$.

Therefore, Katie must finish first.

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

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

Therefore, when Katie crossed the finish line, Sarah was $0.25 \mathrm{~m}$ behind.

ANSWER: (B)