20. Laura and Dina have a running race. Laura runs at constant speed and Dina runs $n$ times as fast where $n>1$. Laura starts $s \mathrm{~m}$ in front of Dina.
What distance, in metres, does Dina run before she overtakes Laura?
A $\frac{n s}{n-1}$
B $n s$
C $\frac{s}{n-1}$
D $\frac{n s}{n+1}$
E $\frac{s}{n}$
SolUtion
A
Suppose that Dina has run a distance of $d$ metres when she overtakes Laura. Because Laura has a start of $s$ metres, at this time Laura has run a distance of $d-s$ metres.

Because they have been running for the same amount of time when Dina overtakes Laura, at this time the ratio of the distances they have run is the same as the ratio of their speeds. That is
$$
d: d-s=n: 1 \text {. }
$$

It follows that
$$
\frac{d}{d-s}=\frac{n}{1} .
$$

Hence
$$
d=n(d-s) \text {. }
$$

This last equation may be rearranged as
$$
n s=d(n-1) .
$$

Therefore
$$
d=\frac{n s}{n-1} .
$$