# 2022 AIME II Problems/Problem 1

## Problem

Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

## Solution 1

Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$, and since we are looking for the minimum amount of people, $x$ is $300$. That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party.

~eamo

## Solution 2 (Kind of lame)

Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12.

Call the amount of people in the beginning $x$.Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12. Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5.

Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0.

Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.

So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\frac{11}{25}=154$. Therefore the answer is $\boxed{154}$.

I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though.

~hastapasta