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
  
<math>
+
<cmath> \frac{90+80+70+60+85+x}{6} = 80 </cmath>
  
</math>
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<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 $90, 80, 70, 60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?

$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 85 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 95$


Solution

Take the average of her current test scores, which is \[\frac{90+80+70+60+85}{5} = \frac{385}{5} = 77\]

This means that she wants her test average after the sixth test to be $80.$ Let $x$ be the score that Josanna receives on her sixth test. Thus, our equation is

\[\frac{90+80+70+60+85+x}{6} = 80\]

\[385+x = 480\]

\[x = \boxed{95\  \((E)}\] (Error compiling LaTeX. Unknown error_msg)

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions