2022 AIME II Problem 1

Question

Adults made up

\frac{5}{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.

Answer

Let the original crowd be

12k

people, so that the adult count

\frac{5}{12} \cdot 12k = 5k

is a whole number

The new total

12k + 50

must be divisible by

25,

which forces

k

to be a multiple of

25

Let the original crowd have

12k

people, of whom

5k

are adults. After the bus arrives there are

12k + 50

people, and the number of adults is

\frac{11}{25}(12k + 50).

For this to be an integer,

25

must divide

12k + 50,

so

25

divides

12k,

and since

\gcd(12, 25) = 1

this means

k

is a multiple of

25.

The adult count

\frac{11}{25}(12k + 50)

increases with

k,

so the minimum occurs at

k = 25:

the new total is

350

and the number of adults is

\frac{11}{25} \cdot 350 = 154.

This is achievable, for example if the bus carries

29

adults and

21

non-adults, so the answer is

154.

Problem 1 in other years