2020 AMC 8 Problems/Problem 14

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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]

Diagram by sircalcsalot

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

Solution 1

The average is given to be $4750$. This is because the dotted line is halfway in between $4500$ and $5000$. There are $20$ cities, so our answer is simply \[4750\cdot20=95000==>\boxed{\textbf{(D) }95,000}\]

Solution 2

We know that the average ($a$) of these group of numbers is the sum ($s$) divided by $20$, so we can make the equation $a = \frac{s}{20}$. Since the average is $4750$, we can solve for $s$ to get $\boxed{\textbf{(D) } 95,000}$

~Pi_Pup

Solution 3

After reading the question, we notice that the dashed line is the average population of each city. Also, that dashed line is slightly less than $5\,000$. Since there are $20$ cities, the answer is slightly less than $20\cdot 5\,000\approx 100\,000$ which is closest to $\textbf{(D) }95{,}000$.

-franzliszt

Solution 4

we get that it is $4750$ multiplied by $20$,solve and get $\boxed{\textbf{(D) } 95,000}$

Video Solution

https://youtu.be/5y4uDwZEF0M

~savannahsolver

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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