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

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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
 
  
 
==Note==
 
==Note==

Revision as of 14:52, 11 February 2021

Problem

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

Solution

Note

See problem 1.

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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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