Difference between revisions of "2015 AMC 8 Problems/Problem 5"

(Video Solution)
 
(7 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
Billy's basketball team scored the following points over the course of the first 11 games of the season:
+
Billy's basketball team scored the following points over the course of the first <math>11</math> games of the season. If his team scores <math>40</math> in the <math>12^{th}</math> game, which of the following statistics will show an increase?
 +
 
 
<cmath>42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73</cmath>
 
<cmath>42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73</cmath>
If his team scores 40 in the 12th game, which of the following statistics will show an increase?
 
  
 
+
<math>\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range} </math>
 
 
<math>
 
\textbf{(A) } \text{range} \qquad
 
\textbf{(B) } \text{median} \qquad
 
\textbf{(C) } \text{mean} \qquad
 
\textbf{(D) } \text{mode} \qquad
 
\textbf{(E) } \text{mid-range}
 
</math>
 
  
 
==Solutions==
 
==Solutions==
Line 19: Line 11:
  
 
When they score a <math>40</math> on the next game, the range increases from <math>73-42=31</math> to <math>73-40=33</math>.  This means the <math>\boxed{\textbf{(A) } \text{range}}</math> increased.
 
When they score a <math>40</math> on the next game, the range increases from <math>73-42=31</math> to <math>73-40=33</math>.  This means the <math>\boxed{\textbf{(A) } \text{range}}</math> increased.
 +
 +
Note: The range is defined to be the difference of the largest element and the smallest element of a set.
  
 
===Solution 2===
 
===Solution 2===
Line 24: Line 18:
 
Because <math>40</math> is less than the score of every game they've played so far, the measures of center will never rise. Only measures of spread, such as the <math>\boxed{\textbf{(A)}~\text{range}}</math>, may increase.
 
Because <math>40</math> is less than the score of every game they've played so far, the measures of center will never rise. Only measures of spread, such as the <math>\boxed{\textbf{(A)}~\text{range}}</math>, may increase.
  
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/3_D17-g3K-M
 +
 +
~Education, the Study of Everything
 +
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=QyBp_12cdGo&list=PL9t_YdPye18V_epxUUbwHTQNeHYBo7a1-&index=9&ab_channel=WhyMath
 +
 +
~savannahsolver
 +
 +
==See Also==
  
 
{{AMC8 box|year=2015|num-b=4|num-a=6}}
 
{{AMC8 box|year=2015|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:52, 12 May 2023

Problem

Billy's basketball team scored the following points over the course of the first $11$ games of the season. If his team scores $40$ in the $12^{th}$ game, which of the following statistics will show an increase?

\[42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73\]

$\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}$

Solutions

Solution 1

When they score a $40$ on the next game, the range increases from $73-42=31$ to $73-40=33$. This means the $\boxed{\textbf{(A) } \text{range}}$ increased.

Note: The range is defined to be the difference of the largest element and the smallest element of a set.

Solution 2

Because $40$ is less than the score of every game they've played so far, the measures of center will never rise. Only measures of spread, such as the $\boxed{\textbf{(A)}~\text{range}}$, may increase.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/3_D17-g3K-M

~Education, the Study of Everything


Video Solution

https://www.youtube.com/watch?v=QyBp_12cdGo&list=PL9t_YdPye18V_epxUUbwHTQNeHYBo7a1-&index=9&ab_channel=WhyMath

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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