Difference between revisions of "2020 AMC 8 Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
The average is given to be <math>4750</math>. This is because the dotted line is halfway in between <math>4500</math> and <math>5000</math>. There are <math>20</math> cities, so our answer is simply <cmath>4750\cdot20=95000==>\boxed{\textbf{(D) }95,000}</cmath> | The average is given to be <math>4750</math>. This is because the dotted line is halfway in between <math>4500</math> and <math>5000</math>. There are <math>20</math> cities, so our answer is simply <cmath>4750\cdot20=95000==>\boxed{\textbf{(D) }95,000}</cmath> | ||
| + | |||
| + | ==Solution 2== | ||
| + | We know that the average (<math>a</math>) of these group of numbers is the sum (<math>s</math>) divided by <math>20</math>, so we can make the equation <math>a = \frac{s}{20}</math>. Since the average is <math>4750</math>, we can solve for <math>s</math> to get <math>\boxed{\textbf{(D) } 95,000}</math> | ||
| + | |||
| + | ~Pi_Pup | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=13|num-a=15}} | {{AMC8 box|year=2020|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:37, 18 November 2020
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)
Diagram by sircalcsalot
Solution 1
The average is given to be 
. This is because the dotted line is halfway in between 
and 
. There are cities, so our answer is simply ![\[4750\cdot20=95000==>\boxed{\textbf{(D) }95,000}\]](http://latex.artofproblemsolving.com/e/9/1/e91f2778d67e299af89d20c1259223feab658d81.png)
Solution 2
We know that the average (
) of these group of numbers is the sum (
) divided by , so we can make the equation 
. Since the average is , we can solve for to get 
~Pi_Pup
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. 
