7. (10 points) The correspondence stage of the Innopolis University Olympiad was held among 9th, 10th, and 11th grade students, and among them, $a \%$, $b \%$, and $c \%$ respectively advanced to the in-person stage. Moreover, among the 9th and 10th grade students together, $k \%$ advanced to the in-person stage, and among the 10th and 11th grade students together, $-l \%$ advanced. Determine the percentage of students from 9th and 11th grades together who advanced to the in-person stage $(a<k<b<l<c)$.
Solution: Let $x$ be the number of 9th grade students, $y$ be the number of 10th grade students, and $z$ be the number of 11th grade students. Then

$$
k=\frac{a x+b y}{x+y} \Longrightarrow \frac{y}{x}=\frac{k-a}{b-k} \quad \text { and } \quad l=\frac{b y+c z}{y+z} \Longrightarrow \frac{z}{y}=\frac{l-b}{c-l}
$$

Therefore,

$$
\frac{z}{x}=\frac{z}{y} \cdot \frac{y}{x}=\frac{(k-a)(l-b)}{(b-k)(c-l)}=\frac{m-a}{c-m}
$$

where $m$ is the percentage of students who advanced to the in-person stage from 9th and 11th grades combined. From here, we need to express $m$ and we get

$$
m=\frac{a(b-k)(c-l)+c(k-a)(l-b)}{(b-k)(c-l)+(k-a)(l-b)}
$$

Answers:

| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 26.6 | 21.2 | 19.8 | 33.8 | 15.6 | 28.5 | 28.5 | 37.8 | 19.5 | 14.8 |