2. The sides of a rectangle were reduced: the length - by $10 \%$, the width - by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$?
Answer: by $18 \%$.

Solution. Let $a$ and $b$ be the length and width of the rectangle. After decreasing the length by $10 \%$ and the width by $20 \%$, we get a rectangle with sides $0.9 a$ and $0.8 b$, the perimeter of which is $0.88$ of the perimeter of the original rectangle. Therefore, $2(0.9 a + 0.8 b) = 0.88 \cdot 2(a + b)$, from which $a = 4 b$. If now the length is decreased by $20 \%$ and the width by $10 \%$, we get a rectangle with a perimeter of $2(0.8 \cdot 4 b + 0.9 b) = 8.2 b$, which is $8.2 b : 10 b = 0.82$ or $82 \%$ of the perimeter of the original rectangle. Therefore, the perimeter decreased by $18 \%$ the second time.

Remark. Correct solution with arithmetic errors - 5 points, solution with specific side lengths - 2 points.