In math class, $2 / 3$ of the students had a problem set, and $4 / 5$ of them brought a calculator. Among those who brought a calculator, the same proportion did not have a problem set as those who did not bring a calculator. What fraction of the students had both a problem set and a calculator?
Let $n(>0)$ be the number of students in the class. According to the problem statement,

$\frac{2}{3} n$ students had a problem collection, $\frac{1}{3} n$ students did not have a problem collection,

$\frac{4}{5} n$ students had a calculator, $\frac{1}{5} n$ students did not have a calculator.

Let $x$ be the common ratio, that is,

$\frac{4}{5} n x$ students brought a calculator but did not bring a problem collection;

and

$\frac{1}{5} n x$ students did not bring a problem collection and did not bring a calculator either.

Then the following equality holds:

$$
\frac{4}{5} n x+\frac{1}{5} n x=\frac{1}{3} n
$$

that is, $n\left(x-\frac{1}{3}\right)=0$, but $n>0$, and thus $x=\frac{1}{3}$.

The $\frac{1}{3}$ of the students who brought a calculator did not have a problem collection, so $\frac{2}{3}$ of them did; this is a total of $\frac{4}{5} \cdot \frac{2}{3}=\frac{8}{15}$ of the class.

Németh László (Jászapáti, Mészáros L. Gymnasium, 4th grade)