Difference between revisions of "2011 AMC 12B Problems/Problem 2"
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Josanna's test scores to date are <math>90, 80, 70, 60,</math> and <math>85.</math> Her goal is to raise her test average at least <math>3</math> points with her next test. What is the minimum test score she would need to accomplish this goal? | Josanna's test scores to date are <math>90, 80, 70, 60,</math> and <math>85.</math> Her goal is to raise her test average at least <math>3</math> points with her next test. What is the minimum test score she would need to accomplish this goal? | ||
| − | <math> | + | <math>\textbf{(A)}\ 80 \qquad |
| − | \textbf{(A)}\ 80 \qquad | ||
\textbf{(B)}\ 82 \qquad | \textbf{(B)}\ 82 \qquad | ||
\textbf{(C)}\ 85 \qquad | \textbf{(C)}\ 85 \qquad | ||
\textbf{(D)}\ 90 \qquad | \textbf{(D)}\ 90 \qquad | ||
\textbf{(E)}\ 95 </math> | \textbf{(E)}\ 95 </math> | ||
| − | |||
== Solution == | == Solution == | ||
Latest revision as of 13:12, 19 January 2021
Problem
Josanna's test scores to date are 
and 
Her goal is to raise her test average at least 
points with her next test. What is the minimum test score she would need to accomplish this goal?
Solution
Take the average of her current test scores, which is
![\[\frac{90+80+70+60+85}{5} = \frac{385}{5} = 77\]](http://latex.artofproblemsolving.com/c/1/5/c1521cab2e19aee7a6ad80fb78c54c4757fb4bf8.png)
This means that she wants her test average after the sixth test to be 
Let 
be the score that Josanna receives on her sixth test. Thus, our equation is
![\[\frac{90+80+70+60+85+x}{6} = 80\]](http://latex.artofproblemsolving.com/8/c/6/8c6e1f7450052953bc25a0006ad917b7eb71b0c8.png)
![\[385+x = 480\]](http://latex.artofproblemsolving.com/8/9/1/8916fc02a40bbe5cdfa286489582de6ede8e4e5f.png)
![\[x = \boxed{95 \textbf{(E)}}\]](http://latex.artofproblemsolving.com/5/0/c/50c77746db4141356460dd69fbc15a713e186852.png)
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
