2011 AMC 10B Problems/Problem 2: Difference between revisions
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== Problem== | |||
(A) 80 | Josanna's test scores to date are <math>90, 80, 70, 60,</math> and <math>85</math>. Her goal is to raise here 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> \textbf{(A)}\ 80 \qquad\textbf{(B)}\ 82 \qquad\textbf{(C)}\ 85 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 95 </math> | |||
==Solution== | |||
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> | |||
==Video Solution== | |||
https://youtu.be/HkrHiIeBucU | |||
~savannahsolver | |||
== See Also== | |||
{{AMC10 box|year=2011|ab=B|num-b=1|num-a=3}} | |||
{{MAA Notice}} | |||
Latest revision as of 18:59, 17 November 2020
Problem
Josanna's test scores to date are 
and 
. Her goal is to raise here test average at least 
points with her next test. What is the minimum test score she would need to accomplish this goal?
Solution
The average of her current scores is 
. To raise it points, she needs an average of 
, and so after her 
tests, a sum of 
. Her current sum is 
, so she needs a 
Video Solution
~savannahsolver
See Also
| 2011 AMC 10B (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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
