Difference between revisions of "2015 AMC 8 Problems/Problem 5"
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\textbf{(E) } \text{mid-range} | \textbf{(E) } \text{mid-range} | ||
</math> | </math> | ||
| − | ===Solution=== | + | |
| + | ==Solutions== | ||
| + | ===Solution 1=== | ||
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. | ||
| − | ==Solution 2== | + | ===Solution 2=== |
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. | ||
Revision as of 14:50, 16 January 2021
Billy's basketball team scored the following points over the course of the first 11 games of the season:
![\[42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73\]](http://latex.artofproblemsolving.com/4/c/2/4c24d4c07e2a72cf04fa5c16f6d7ec90faad13b5.png)
If his team scores 40 in the 12th game, which of the following statistics will show an increase?
Solutions
Solution 1
When they score a 
on the next game, the range increases from 
to 
. This means the 
increased.
Solution 2
Because 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 
, may increase.
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
