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

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

@@ Line 1: / Line 1: @@
-==Problem 14==
+==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?
@@ Line 8: / Line 9: @@
 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) {
@@ Line 39: / Line 36: @@
 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>
 <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==
-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{D}</cmath>
+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
+~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== {{AMC8 box|year=2020|before=First problem|num-a=2}} {{MAA Notice}}
+==See also==
+{{AMC8 box|year=2020|num-b=13|num-a=15}}
+{{MAA Notice}}

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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