Problem 11.1. Petya and Vasya participated in the election for the position of president of the chess club. By noon, Petya had $25 \%$ of the votes, and Vasya had $45 \%$. After noon, only Petya's friends came to vote (and, accordingly, voted only for him). In the end, Vasya was left with only $27 \%$ of the votes. What percentage of the votes did Petya receive?
Answer: $55 \%$.

Solution. Let $x$ be the number of people who voted before noon, and $y$ be the number of people who voted after. Then, $0.45 x$ people voted for Vasya, which constitutes $27 \%$ of $x+y$. Thus, we have the equation $0.45 x = 0.27(x + y)$, from which $2 x = 3 y$. According to the condition, Petya received $0.25 x + y$ votes. Let's calculate what fraction of $x + y$ this number represents:

$$
\frac{0.25 x + y}{x + y} = \frac{0.25 + y / x}{1 + y / x} = \frac{1 / 4 + 2 / 3}{1 + 2 / 3} = \frac{11}{20} = 55 \%.
$$

## Criteria

One of the largest suitable criteria is used:

7 6. Any correct solution to the problem.

6 6. Correct overall solution, but the answer indicates the difference between the desired quantity and the initial percentage of votes for Petya.

5 6. Correct in other respects, but with an incorrect answer due to an arithmetic error.

1 6. The correct answer is given.

Other intermediate steps in the solution are not evaluated.

## Kurchatov School Olympiad in Mathematics - 2019 Final Stage.