Difference between revisions of "2010 AMC 10B Problems/Problem 12"
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The maximum possible value occurs when <math>60\%</math> of the students (which accounts for <math>30\%</math> of the overall student population) who originally answered "Yes." answer "No." and the <math>100\%</math> of the students (which accounts for <math>50\%</math> of the overall student population) who originally answered "No." answer "Yes." In this case, <math>x=50+30=80</math> | The maximum possible value occurs when <math>60\%</math> of the students (which accounts for <math>30\%</math> of the overall student population) who originally answered "Yes." answer "No." and the <math>100\%</math> of the students (which accounts for <math>50\%</math> of the overall student population) who originally answered "No." answer "Yes." In this case, <math>x=50+30=80</math> | ||
| − | Subtract <math>80-20</math> to obtain an answer of <math>\boxed{\textbf{( | + | Subtract <math>80-20</math> to obtain an answer of <math>\boxed{\textbf{(Z)}\ 60}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2010|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:01, 28 September 2016
Problem
At the beginning of the school year, 
of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and answered "No." At the end of the school year, 
answered "Yes" and 
answered "No." Altogether, 
of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of 
?
Solution
The minimum possible value occurs when 
of the students who originally answered "No." answer "Yes." In this case, 
The maximum possible value occurs when 
of the students (which accounts for of the overall student population) who originally answered "Yes." answer "No." and the 
of the students (which accounts for of the overall student population) who originally answered "No." answer "Yes." In this case, 
Subtract 
to obtain an answer of 
See Also
| 2010 AMC 10B (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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
