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

(Removed incorrect edits by user Vrajnash)
(Video Solution)
(19 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
 
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));
 
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) {
Line 37: Line 36:
 
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);
 +
 +
// axis
 +
draw((0,0)--(0,9300), linewidth(1.25));
 +
draw((0,0)--(11550,0), linewidth(1.25));
 
</asy>
 
</asy>
  
 
<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>
  
==Solution 1==
+
==Solution==
We can see that the dotted line is halfway between <math>4{,}500</math> and <math>5{,}000</math>, so 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>.
+
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>.
 +
 
 +
==Video Solution by North America Math Contest Go Go Go==
 +
https://www.youtube.com/watch?v=IqoLKBx20dQ
  
==Solution 2 (estimation)==
+
~North America Math Contest Go Go Go
The dashed line, which represents the average population of each city, is slightly below <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>\boxed{\textbf{(D) }95,000}</math>.
 
  
==Video Solution==
+
==Video Solution by WhyMath==
 
https://youtu.be/5y4uDwZEF0M
 
https://youtu.be/5y4uDwZEF0M
 +
 +
~savannahsolver
 +
 +
==Video Solution (CLEVER MANIPULATIONS!!!)==
 +
https://youtu.be/2FYz_ze566I
 +
 +
~Interstigation
 +
 +
==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}}

Revision as of 13:09, 2 March 2023

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

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 (CLEVER MANIPULATIONS!!!)

https://youtu.be/2FYz_ze566I

~Interstigation

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=608

~Interstigation

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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