Let's write down the fractions between $7 \frac{1}{3}$ and $7 \frac{1}{2}$, which in their simplest form have a two-digit (integer) numerator and denominator.
The sought fractions (two-digit) denominator is $7 \frac{1}{3}=22 / 3$ times smaller than the numerator, which is also two-digit, thus always less than 99. Therefore, the denominator is less than $3 \cdot 99 / 22=27 / 2$, so the denominator is at least 10 and at most 13.

Let's consider the possible denominators one by one. The numerator can only be a two-digit number between the denominator's 22/3 times and 15/2 times, not allowing the boundaries, since those (if integers) would only provide a simplified form of 22/3 or 15/2.

If the denominator is 10, the numerator must be between $220 / 3=73 \frac{1}{3}$ and $150 / 2=75$, so it could only be 74, but this leads to a reducible fraction.

For the denominator 11, the numerator must be between $242 / 3=80 \frac{2}{3}$ and $165 / 2=82 \frac{1}{2}$, so 81 and 82 are suitable numerators.

For the denominator 12, the numerator must be between $264 / 3=88$ and $180 / 2=90$, so 89 is a suitable numerator.

Finally, for the denominator 13, the numerator must be between $286 / 3=95 \frac{1}{3}$ and $195 / 2=97 \frac{1}{2}$, so 96 and 97 are suitable numerators. Thus, the fractions that meet the criteria are

$$
\frac{81}{11}, \quad \frac{82}{11}, \quad \frac{89}{12}, \quad \frac{96}{13}, \quad \frac{97}{13}
$$

(The order by size is obtained by swapping the second and fourth fractions.)

Judit Gaizler (Budapest, VI., Bajza u. primary school VIII. grade)

Note. 8 contestants wrote as a second solution nearly 500 mixed numbers within the given limits, whose fractional part in their simplest form has a two-digit numerator and denominator. However, these contestants unnecessarily performed this tedious and unproductive work. The problem speaks of fractions, which consist exclusively of a numerator and a denominator.