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Difference between revisions of "2020 AMC 8 Problems/Problem 14"
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==Problem==
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There are <math>20</math> 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 <math>20</math> cities?
 
There are <math>20</math> 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 <math>20</math> cities?
  
 
<asy>
 
<asy>
 +
// made by SirCalcsALot
 +
 
size(300);
 
size(300);
  
 
pen shortdashed=linetype(new real[] {6,6});
 
pen shortdashed=linetype(new real[] {6,6});
−
 
−
// axis
 
−
draw((0,0)--(0,9300), linewidth(1.25));
 
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draw((0,0)--(11550,0), linewidth(1.25));
 
  
 
for (int i = 2000; i < 9000; i = i + 2000) {
 
for (int i = 2000; i < 9000; i = i + 2000) {
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label("Cities", (11450*0.5,0), S);
 
label("Cities", (11450*0.5,0), S);
 
label(rotate(90)*"Population", (0,9000*0.5), 10*W);
 
label(rotate(90)*"Population", (0,9000*0.5), 10*W);
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 +
// axis
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draw((0,0)--(0,9300), linewidth(1.25));
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draw((0,0)--(11550,0), linewidth(1.25));
 
</asy>
 
</asy>
−
 
−
Diagram by sircalcsalot
 
  
 
<math>\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000</math>
 
<math>\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000</math>
  
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==Solution 1==
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==Solution==
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The average is given to be <math>4750</math>. This is because the dotted line is halfway in between <math>4500</math> and <math>5000</math>. There are <math>20</math> cities, so our answer is simply <cmath>4750\cdot20=95000==>\boxed{\textbf{(D) }95,000}</cmath>
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We can see that the dotted line is exactly halfway between <math>4{,}500</math> and <math>5{,}000</math>, so it is at <math>4{,}750</math>. As this is the average population of all <math>20</math> cities, the total population is simply <math>4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}</math>.
  
 
==Solution 2==
 
==Solution 2==
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We know that the average (<math>a</math>) of these group of numbers is the sum (<math>s</math>) divided by <math>20</math>, so we can make the equation <math>a = \frac{s}{20}</math>. Since the average is <math>4750</math>, we can solve for <math>s</math> to get <math>\boxed{\textbf{(D) } 95,000}</math>
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The dotted line is slightly below <math>5000</math>. Since <math>5000\cdot20=100000</math>, the answer is slightly below this, so <math>\boxed{\textbf{(D) }95{,}000}</math> would be the best choice to guess.
  
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~Pi_Pup
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=bHNrBwwUCMI
  
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==Solution 3==
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~NiuniuMaths
  
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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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==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=gX3KbBHBdLQ4DJBT&t=2139
  
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-franzliszt
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~Math-X
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==Solution 4==
 
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we get that it is <math>4750</math> multiplied by <math>20</math>,solve and get <math>\boxed{\textbf{(D) } 95,000}</math>
 
  
−
==Video Solution==
+
==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
 
https://youtu.be/5y4uDwZEF0M
  
 
~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=608
 +
 +
~Interstigation
  
 
==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}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 19:53, 4 June 2025

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

Solution 2

The dotted line is slightly below $5000$. Since $5000\cdot20=100000$, the answer is slightly below this, so $\boxed{\textbf{(D) }95{,}000}$ would be the best choice to guess.

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 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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