2020 AMC 8 Problems/Problem 14

Revision as of 12:33, 24 November 2020 by Vrajshah (talk | contribs) (Solution 2 (estimation))

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] 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]

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

Solution 1

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

Solution 2 (estimation)

The dashed line, which represents the average population of each city, is slightly below $5{,}000$. Since there are $20$ cities, the answer is slightly less than $20\cdot 5{,}000 \approx 100{,}000$

Video Solution

https://youtu.be/5y4uDwZEF0M

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

Invalid username
Login to AoPS