Difference between revisions of "2011 AMC 12B Problems/Problem 2"
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== Solution == | == Solution == | ||
+ | Take the average of her current test scores, which is | ||
+ | <cmath> \frac{90+80+70+60+85}{5} = \frac{385}{5} = 77 </cmath> | ||
+ | This means that she wants her test average after the sixth test to be <math>80.</math> Let <math>x</math> be the score that Josanna receives on her sixth test. Thus, our equation is | ||
− | < | + | <cmath> \frac{90+80+70+60+85+x}{6} = 80 </cmath> |
− | </ | + | <cmath> 385+x = 480 </cmath> |
+ | |||
+ | <cmath> x = \boxed{95\ \((E)} </cmath> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|before=Problem 1|num-a=3|ab=B}} | {{AMC12 box|year=2011|before=Problem 1|num-a=3|ab=B}} |
Revision as of 15:58, 27 February 2011
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
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
\[x = \boxed{95\ \((E)}\] (Error compiling LaTeX. Unknown error_msg)
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 |