Difference between revisions of "2020 AMC 8 Problems/Problem 14"
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==Problem== | ==Problem== | ||
− | + | There are <math>20</math> cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all <math>20</math> cities? | |
− | <math>\textbf{(A)} | + | |
+ | https://latex.artofproblemsolving.com/0/c/b/0cbe70f3983021d466417bf77c034a81f92c9894.png | ||
+ | |||
+ | <math>\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000</math> | ||
==Solution== | ==Solution== |
Revision as of 09:33, 8 July 2022
Contents
[hide]Problem
There are cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all cities?
https://latex.artofproblemsolving.com/0/c/b/0cbe70f3983021d466417bf77c034a81f92c9894.png
Solution
For two or more Es to not appear together, the letters would have to be arranged in the form of E_E_E_E_E, where the blanks are one of each of the letters B, K, P, R. There are ways you can arrange the letter in the blanks, so that is our answer.
~mahaler
Solution
We can see that the dotted line is exactly halfway between and , so it is at . As this is the average population of all cities, the total population is simply .
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=IqoLKBx20dQ
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=608
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.