11.1. The price of a product was first increased by $x$ percent, and then decreased by $y$ percent. As a result, the price remained the same. Find all values that the difference $\frac{1}{x}-\frac{1}{y}$ can take.
Solution: Let the item cost 1 monetary unit. Then after the price increase, its cost became $a=1+\frac{x}{100}$, and after the decrease, it became

$$
a-\frac{a y}{100}=a\left(1-\frac{y}{100}\right)=\left(1+\frac{x}{100}\right)\left(1-\frac{y}{100}\right)
$$

The condition of the problem, therefore, is described by the equation

$$
\left(1+\frac{x}{100}\right)\left(1-\frac{y}{100}\right)=1
$$

From this, $(100+x)(100-y)=10000$ or $100 x-100 y=x y$. The expression $\frac{1}{x}-\frac{1}{y}$ is then equal to $\frac{y-x}{x y}=-\frac{1}{100}$.

Note: Of course, the initial price of the item can be set to any other non-negative number or denoted by an unknown. In the latter case, the unknown will cancel out in the equation.

Answer: $-\frac{1}{100}$.

Recommendations for checking:

| is in the work | points |
| :--- | :---: |
| Correct and fully justified answer | 7 points |
| The answer is incorrect due to an arithmetic error, but the solution approach is correct | 6 points |
| Answer without justification and/or incorrect answer | 0 points |