B1. At a gala, there are different male-female couples dancing, and specifically, $\frac{2}{3}$ of the present women are dancing with $\frac{3}{5}$ of the present men.

What fraction of the attendees are dancing?
B1.

$\frac{12}{19} \quad$ Let the number of women present be $v$ and the number of men present be $m$. The problem tells us that $\frac{2}{3} v=\frac{3}{5} m$ and thus that $v=\frac{9}{10} m$. The number of people dancing is exactly twice the number of men dancing, namely $\frac{6}{5} m$.

The number of people present is of course $m+v=m+\frac{9}{10} m=\frac{19}{10} m$. Therefore, we see that the fraction of those present who are dancing is

$$
\frac{\frac{6}{5} m}{\frac{19}{10} m}=\frac{6}{5} \cdot \frac{10}{19}=\frac{12}{19}
$$