Difference between revisions of "2020 AMC 8 Problems/Problem 14"

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==Video Solution==
 
==Video Solution==
https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)
 
 
 
 
 
https://youtu.be/xjwDsaRE_Wo
 
https://youtu.be/xjwDsaRE_Wo
  

Revision as of 19:49, 31 December 2020

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 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 $4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}$.

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 $20\cdot 5{,}000 \approx 100{,}000$, which is closest to $\boxed{\textbf{(D) }95{,}000}$.

Video Solution

https://youtu.be/xjwDsaRE_Wo

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

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