Difference between revisions of "2020 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
We can see that the dotted line is exactly halfway between <math>4{,}500</math> and <math>5{,}000</math>, so it is at <math>4{,}750</math>. As this is the average population of all <math>20</math> cities, the total population is simply <math>4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}</math>. | We can see that the dotted line is exactly halfway between <math>4{,}500</math> and <math>5{,}000</math>, so it is at <math>4{,}750</math>. As this is the average population of all <math>20</math> cities, the total population is simply <math>4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}</math>. | ||
| + | |||
| + | ==Video Solution by Math-X (First understand the problem!!!)== | ||
| + | https://youtu.be/UnVo6jZ3Wnk?si=6I2z2MCM6n7gWE8P&t=2139Zk&t=34 | ||
| + | |||
| + | ~Math-X | ||
==Video Solution (🚀Very Fast🚀)== | ==Video Solution (🚀Very Fast🚀)== | ||
Revision as of 10:05, 12 December 2023
Contents
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?
![[asy] // made by SirCalcsALot size(300); pen shortdashed=linetype(new real[] {6,6}); for (int i = 2000; i < 9000; i = i + 2000) { draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey); label(string(i), (0,i), W); } for (int i = 500; i < 9300; i=i+500) { draw((0,i)--(150,i),linewidth(1.25)); if (i % 2000 == 0) { draw((0,i)--(250,i),linewidth(1.25)); } } int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20; int r = 550; for (int i = 0; i < data_length; ++i) { fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey); draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); } draw((0,4750)--(11450,4750),shortdashed); label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); [/asy]](http://latex.artofproblemsolving.com/a/b/f/abffd65d59aacaaf81b6039c162d0898cca8b729.png)
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 Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=6I2z2MCM6n7gWE8P&t=2139Zk&t=34
~Math-X
Video Solution (🚀Very Fast🚀)
~Education, the Study of Everything
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 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. 
