Difference between revisions of "2021 AMC 12A Problems/Problem 8"

Revision as of 13:52, 11 February 2021

These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.

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

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 These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
 ==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
 ==Note==