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

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~Pi_Pup
 
~Pi_Pup
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==Solution 3==
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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 <math>5\,000</math>. Since there are <math>20</math> cities, the answer is slightly less than <math>20\cdot 5\,000\approx 100\,000</math> which is closest to <math>\textbf{(D) }95{,}000</math>.
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-franzliszt
  
 
==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 13:58, 18 November 2020

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

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

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