2020 AMC 8 Problems/Problem 14

Problem

There are $20$ 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 $20$ 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]

$\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000$

Solution

We can see that the dotted line is exactly halfway between $4{,}500$ and $5{,}000$, so it is at $4{,}750$. As this is the average population of all $20$ cities, the total population is simply $4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}$.

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=gX3KbBHBdLQ4DJBT&t=2139

~Math-X

Video Solution (🚀ok Very Fast🚀)

https://youtu.be/2FYz_ze566I

~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

https://youtu.be/5y4uDwZEF0M

~savannahsolver

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=608

~Interstigation

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png