Ana and Luíza train every day for the Big Race that will take place at the end of the year at school, each always at the same speed. The training starts at point $A$ and ends at point $B$, which are $3000 \mathrm{~m}$ apart. They start at the same time, but when Luíza finishes the race, Ana still has $120 \mathrm{~m}$ to go to reach point $B$. Yesterday, Luíza gave Ana a chance: "We start at the same time, but I start a few meters before point $A$ so we can finish together." How many meters before point $A$ should Luíza start?
We know that distance $=$ speed $\times$ time.

Let $v$ and $v^{\prime}$ be the speeds of Ana and Luíza, respectively, and $t$ the time Luíza takes to travel $3000 \mathrm{~m}$. Therefore, in the same time $t$, Ana travels
$3000-120=2880 \mathrm{~m}$. We have:

$$
3000=v \times t
$$

$$
3000-120=v^{\prime} t \Rightarrow t=\frac{3000}{v}=\frac{2880}{v^{\prime}}
$$

Therefore, $\frac{v^{\prime}}{v}=\frac{24}{25}$.

If we denote by $x$ the additional distance that Luíza will travel, we have:

$$
3000+x=v \times t
$$

and

$$
3000=v^{\prime} \times t \Rightarrow \frac{3000+x}{v}=\frac{3000}{v^{\prime}} \Rightarrow \frac{3000}{3000+x}=\frac{v^{\prime}}{v}
$$

It follows that

$$
\frac{3000}{3000+x}=\frac{v^{\prime}}{v}=\frac{24}{25} \Rightarrow x=125
$$

Therefore, the answer is $125 \mathrm{~m}$.

## List 7