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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/HkrHiIeBucU | ||
| + | |||
| + | ~savannahsolver | ||
== See Also== | == See Also== | ||
Latest revision as of 18:59, 17 November 2020
Contents
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
