Difference between revisions of "2020 AMC 8 Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
| − | We can see that the dotted line is halfway between <math>4{,}500</math> and <math>5{,}000</math>, so 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>. This is | + | We can see that the dotted line is halfway between <math>4{,}500</math> and <math>5{,}000</math>, so 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>. |
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| + | We can see that between 4000 and 6000 There are three number ticks. This means that the number is slightly below 5000, or it is 4750. To find the opposite of an average, you do the opposite algorithm. We multiply 4750 by 20 and we get 10500 so the answer is E | ||
==Solution 2 (estimation)== | ==Solution 2 (estimation)== | ||
Revision as of 12:35, 24 November 2020
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 halfway between 
and 
, so is at 
. As this is the average population of all cities, the total population is simply 
.
We can see that between 4000 and 6000 There are three number ticks. This means that the number is slightly below 5000, or it is 4750. To find the opposite of an average, you do the opposite algorithm. We multiply 4750 by 20 and we get 10500 so the answer is E
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 
Video Solution
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. 
