Difference between revisions of "2006 AMC 8 Problems/Problem 12"
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Adding them up gets <math> 7+16+27=50 </math>. The overall percentage correct would be <math> \frac{50}{60}=\frac{5}{6}=5 \cdot 16.\overline{6}=83.\overline{3} \approx \boxed{\textbf{(D)}\ 83} </math>. | Adding them up gets <math> 7+16+27=50 </math>. The overall percentage correct would be <math> \frac{50}{60}=\frac{5}{6}=5 \cdot 16.\overline{6}=83.\overline{3} \approx \boxed{\textbf{(D)}\ 83} </math>. | ||
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| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/6GetTV9Mnmo | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=11|num-a=13}} | {{AMC8 box|year=2006|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:12, 30 October 2024
Problem
Antonette gets 
on a 10-problem test, 
on a 20-problem test and 
on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
Solution
Adding them up gets 
. The overall percentage correct would be 
.
Video Solution by WhyMath
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
