A dog took a hare to the hunt, which has a 50-jump lead. The dog makes 5 jumps while the hare makes 6, but 7 of the dog's jumps equal 9 of the hare's jumps. How many more jumps can the hare make before the dog catches up?
If the hare makes $x$ more jumps, then the dog will catch up after $\frac{5}{6} x$ jumps, but this corresponds to $\frac{9}{7} \cdot \frac{5}{6} x$ jumps, so

$$
\frac{9}{7} \cdot \frac{5}{6} x - x = 50
$$

which means

$$
x = 700
$$

(Kálmán Forintos, Budapest.)