Difference between revisions of "2011 AMC 10B Problems/Problem 2"

 
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The average of her current scores is <math>77</math>. To raise it <math>3</math> points, she needs an average of <math>80</math>, and so after her <math>6</math> tests, a sum of <math>480</math>. Her current sum is <math>385</math>, so she needs a <math>480 - 385 = \boxed{\mathrm{(E) \ } 95}</math>
 
The average of her current scores is <math>77</math>. To raise it <math>3</math> points, she needs an average of <math>80</math>, and so after her <math>6</math> tests, a sum of <math>480</math>. Her current sum is <math>385</math>, so she needs a <math>480 - 385 = \boxed{\mathrm{(E) \ } 95}</math>
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==Video Solution==
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https://youtu.be/HkrHiIeBucU
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~savannahsolver
  
 
== See Also==
 
== See Also==

Latest revision as of 19:59, 17 November 2020

Problem

Josanna's test scores to date are $90, 80, 70, 60,$ and $85$. Her goal is to raise here 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

The average of her current scores is $77$. To raise it $3$ points, she needs an average of $80$, and so after her $6$ tests, a sum of $480$. Her current sum is $385$, so she needs a $480 - 385 = \boxed{\mathrm{(E) \ } 95}$

Video Solution

https://youtu.be/HkrHiIeBucU

~savannahsolver

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
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 10 Problems and Solutions

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