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

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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?
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==Problem==
  
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<asy>
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In how many ways can the letters in BEEKEEPER be rearranged so that two or more Es do not appear together?
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size(300);
 
  
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pen shortdashed=linetype(new real[] {6,6});
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<math>\textbf{(A)} ~1\qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~24\qquad\textbf{(E)} ~120\qquad</math>
  
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// axis
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==Solution==
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draw((0,0)--(0,9300), linewidth(1.25));
 
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draw((0,0)--(11550,0), linewidth(1.25));
 
  
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for (int i = 2000; i < 9000; i = i + 2000) {
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For two or more Es to not appear together, the letters would have to be arranged in the form of E_E_E_E_E, where the blanks are one of each of the letters B, K, P, R. There are <math>4! = \boxed{24}</math> ways you can arrange the letter in the blanks, so that is our answer.
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    draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);
 
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    label(string(i), (0,i), W);
 
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}
 
  
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~mahaler
  
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for (int i = 500; i < 9300; i=i+500) {
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==Solution==
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    draw((0,i)--(150,i),linewidth(1.25));
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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>.
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    if (i % 2000 == 0) {
 
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        draw((0,i)--(250,i),linewidth(1.25));
 
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    }
 
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}
 
  
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int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};
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==Video Solution by North America Math Contest Go Go Go==
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int data_length = 20;
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https://www.youtube.com/watch?v=IqoLKBx20dQ
  
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int r = 550;
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~North America Math Contest Go Go Go
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for (int i = 0; i < data_length; ++i) {
 
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    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);
 
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    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));
 
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}
 
  
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draw((0,4750)--(11450,4750),shortdashed);
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==Video Solution by WhyMath==
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https://youtu.be/5y4uDwZEF0M
  
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label("Cities", (11450*0.5,0), S);
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~savannahsolver
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label(rotate(90)*"Population", (0,9000*0.5), 10*W);
 
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</asy>
 
  
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Diagram by sircalcsalot
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==Video Solution==
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https://youtu.be/xjwDsaRE_Wo
  
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<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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==Video Solution by Interstigation==
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https://youtu.be/YnwkBZTv5Fw?t=608
  
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==Solution 1==
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~Interstigation
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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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==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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~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 10:26, 28 January 2022

Problem

In how many ways can the letters in BEEKEEPER be rearranged so that two or more Es do not appear together?

$\textbf{(A)} ~1\qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~24\qquad\textbf{(E)} ~120\qquad$

Solution

For two or more Es to not appear together, the letters would have to be arranged in the form of E_E_E_E_E, where the blanks are one of each of the letters B, K, P, R. There are $4! = \boxed{24}$ ways you can arrange the letter in the blanks, so that is our answer.

~mahaler

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

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

https://youtu.be/xjwDsaRE_Wo

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
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All AJHSME/AMC 8 Problems and Solutions

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