2025 AMC 8 Problems/Problem 17
Contents
[hide]Problem
In the land of Markovia, there are three cities: ,
, and
. There are 100 people who live in
, 120 who live in
, and 160 who live in
. Everyone works in one of the three cities, and a person may work in the same city where they live. In the figure below, an arrow pointing from one city to another is labeled with the fraction of people living in the first city who work in the second city. (For example,
of the people who live in
work in
.) How many people work in
?
Solution 1+2
THere are people who do not work in city
that live in city
, meaning that
people who live in city
work in city
. There are
people who live in city
and work in
, as well as
people who live in city
that work in city
. Therefore, the answer is
.
~ alwaysgonnagiveyouup
We could also make an equation. Let's denote the number of people that live in city as
,
as
, and
as
. If
denotes the number of people working in city
, then
. This is because
of the people from City
and
of the people from City
work in city
as shown in the image. We also know that
of the people living in
work in other cities. We are already given the values of variables
,
, and
as
,
, and
respectively. Plug the values it into the main equation like this:
. We solve for it and get our answer
.
~sushipie2025
Remark
- Main article: Markov Chains
This model is known as the Markov Chain, a type of stochastic process that models systems where the next state depends only on the current state, not on the sequence of events that preceded it. This is known as the Markov property (memoryless property).
Video Solution by Pi Academy
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
Video Solution(Quick, fast, easy!)
~MC
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=fV-dPbMPVzWTkSV3&t=2020 ~hsnacademy
Video Solution by Thinking Feet
See Also
2025 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.