For which natural numbers $a$ and $b$ do the conditions $90<a+b<100$ and $0.9<\frac{a}{b}<0.91$ hold?
Solution. Multiplying the second inequality by the positive number $b$, and then adding $b$ to it, we get $1.9 b \frac{90}{1.91} \approx 47.12$, and $b < \frac{100}{1.9} \approx 52.63$. Since $b$ is an integer, its values can be the integers between 48 and 52. Substituting these into the inequality $0.9 b < a < 0.91 b$ for $a$, we get the following values:

1. $43.2 < a < 43.68, \quad$. $\quad 45.9 < a < 46.41$,
2. $44.1 < a < 44.59, \quad$ 5. $46.8 < a < 47.32$.
3. $45 < a < 45.9$.

In the first three cases, $a$ cannot be an integer. In the fourth case, $a = 46$ and $b = 51$; in the fifth case, $a = 47$ and $b = 52$.

For these pairs of numbers, the condition $0.9 < \frac{a}{b} < 0.91$ naturally holds. Since $51 + 46 = 97$ and $47 + 52 = 99$, the condition $90 < a + b < 100$ also holds. Therefore, the sought pairs of numbers are $(46; 51)$ and $(47; 52)$.