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

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==Problem==
 
==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?
 
  
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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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    draw((0,i)--(150,i),linewidth(1.25));
 
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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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int data_length = 20;
 
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int r = 550;
 
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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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label("Cities", (11450*0.5,0), S);
 
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label(rotate(90)*"Population", (0,9000*0.5), 10*W);
 
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</asy>
 
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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>
 
  
 
==Solution==
 
==Solution==

Revision as of 09:26, 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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