2. On one island, frogs are always green or blue. Last year (compared to the year before last), the number of blue frogs increased by $60 \%$, and the number of green frogs decreased by $60 \%$. The new ratio of the number of blue frogs to the number of green frogs remained the same as the ratio of the year before last, but in reverse order (the number of green frogs to the number of blue frogs). By what percentage did the total number of frogs change last year compared to the year before last?
2. Let $p: z$ be the ratio of the number of blue to the number of green frogs from the year before last.

Last year, the number of blue frogs was $p_{1}=1.6 p$, and the number of green frogs was $z_{1}=0.4 z$.

1 POINT

Since the ratios of the number of blue to the number of green frogs last year and the year before last are the same but in reverse order, we have:

$$
\frac{1.6 p}{0.4 z}=\frac{z}{p} .
$$

From this equation, it follows that

$$
4 p^{2}=z^{2} \text {, hence } z=2 p \text { (we are dealing with positive integers). } 1 \text { POINT }
$$

The ratio of the total number of all frogs last year to the year before last is

$$
\frac{1.6 p+0.4 z}{p+z}=\frac{1.6 p+0.4 \cdot 2 p}{p+2 p}=\frac{2.4 p}{3 p}=\frac{8}{10}=80 \%
$$

Thus, the total number of frogs has decreased by 20% (last year compared to the year before last).