Difference between revisions of "2021 AMC 12A Problems/Problem 8"
Icematrix2 (talk | contribs) (Created page with "==Problem== These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. ==Solution== The solutions will be posted once the problems are...") |
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These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. | These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. | ||
==Solution== | ==Solution== | ||
| − | + | It is known that <math>0.\overline{ab}=\frac{ab}{99}</math> and <math>0.ab=\frac{ab}{100}</math>. Let <math>\overline {ab} = x</math>. We have that <math>66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})</math>. Solving gives that <math>100x-75=99x</math> so <math>x=75</math>. ~aop2014 | |
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==Note== | ==Note== | ||
See [[2021 AMC 12A Problems/Problem 1|problem 1]]. | See [[2021 AMC 12A Problems/Problem 1|problem 1]]. | ||
Revision as of 13:51, 11 February 2021
Contents
Problem
These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
Solution
It is known that 
and 
. Let 
. We have that 
. Solving gives that 
so 
. ~aop2014
Note
See problem 1.
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
