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

Revision as of 08:04, 20 November 2020

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?

$[asy] size(300); pen shortdashed=linetype(new real[] {6,6}); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); for (int i = 2000; i < 9000; i = i + 2000) { draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey); label(string(i), (0,i), W); } for (int i = 500; i < 9300; i=i+500) { draw((0,i)--(150,i),linewidth(1.25)); if (i % 2000 == 0) { draw((0,i)--(250,i),linewidth(1.25)); } } int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20; int r = 550; for (int i = 0; i < data_length; ++i) { fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey); draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); } draw((0,4750)--(11450,4750),shortdashed); label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W); [/asy]$

$\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000$

Solution 1

We can see that the dotted line is halfway between $4{,}500$ and $5{,}000$ , so is $4{,}750$ . 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 dashed line, which represents the average population of each city, is slightly below $5{,}000$ . Since there are $20$ cities, the answer is slightly less than $20\cdot 5{,}000 \approx 100{,}000$ , which is closest to $\boxed{\textbf{(D) }95,000}$ .

Video Solution

https://youtu.be/5y4uDwZEF0M

@@ Line 1: / Line 1: @@
+==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 37: / Line 38: @@
 label(rotate(90)*"Population", (0,9000*0.5), 10*W);
 </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>
 ==Solution 1==
-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>
+We can see that the dotted line is halfway between <math>4{,}500</math> and <math>5{,}000</math>, so is <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==
-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>
+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>.
-~Pi_Pup
-==Solution 3==
-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>.
--franzliszt
-==Solution 4==
-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==
 https://youtu.be/5y4uDwZEF0M
-~savannahsolver
 ==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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