Difference between revisions of "2020 AMC 8 Problems/Problem 14"
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==Solution 2 (estimation)== | ==Solution 2 (estimation)== | ||
The dashed line, which represents the average population of each city, is slightly below <math>5{,}000</math>. Since there are <math>20</math> cities, the answer is slightly less than <math>20\cdot 5{,}000 \approx 100{,}000</math>, which is closest to <math>\boxed{\textbf{(D) }95{,}000}</math>. | The dashed line, which represents the average population of each city, is slightly below <math>5{,}000</math>. Since there are <math>20</math> cities, the answer is slightly less than <math>20\cdot 5{,}000 \approx 100{,}000</math>, which is closest to <math>\boxed{\textbf{(D) }95{,}000}</math>. | ||
| + | |||
| + | ==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== | ==Video Solution by WhyMath== | ||
Revision as of 11:51, 19 June 2021
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] size(300); pen shortdashed=linetype(new real[] {6,6}); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); 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); [/asy]](http://latex.artofproblemsolving.com/e/e/b/eebdf6cf28d1dcd419871ffed62a6c02f8aab04b.png)
Solution 1
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 
.
Solution 2 (estimation)
The dashed line, which represents the average population of each city, is slightly below . Since there are cities, the answer is slightly less than 
, which is closest to 
.
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. 
