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

Latest revision as of 15:34, 26 January 2024

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 NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=gX3KbBHBdLQ4DJBT&t=2139

~Math-X

Video Solution (🚀ok Very Fast🚀)

https://youtu.be/2FYz_ze566I

~Education, the Study of Everything

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 by Interstigation

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

~Interstigation

@@ 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?
 <asy>
+// made by SirCalcsALot
 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) {
@@ Line 36: / 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>
-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>
-<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==
+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 NiuniuMaths (Easy to understand!)==
+https://www.youtube.com/watch?v=bHNrBwwUCMI
+~NiuniuMaths
+==Video Solution by Math-X (First understand the problem!!!)==
+https://youtu.be/UnVo6jZ3Wnk?si=gX3KbBHBdLQ4DJBT&t=2139
+~Math-X
+==Video Solution (🚀ok Very Fast🚀)==
+https://youtu.be/2FYz_ze566I
+~Education, the Study of Everything
-==Solution 1==
+==Video Solution by North America Math Contest Go Go Go==
-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>
+https://www.youtube.com/watch?v=IqoLKBx20dQ
-==Solution 2==
+~North America Math Contest Go Go Go
-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>
-~Pi_Pup
+==Video Solution by WhyMath==
+https://youtu.be/5y4uDwZEF0M
-==Solution 3==
+~savannahsolver
-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>.
+==Video Solution by Interstigation==
+https://youtu.be/YnwkBZTv5Fw?t=608
--franzliszt
+~Interstigation
 ==See also==
 {{AMC8 box|year=2020|num-b=13|num-a=15}}
 {{MAA Notice}}

2020 AMC 8 (Problems • Answer Key • Resources)
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