Difference between revisions of "2013 AMC 10B Problems/Problem 1"
m |
Mathguy5041 (talk | contribs) m |
||
| (One intermediate revision by the same user not shown) | |||
| Line 8: | Line 8: | ||
This expression is equivalent to <math>\frac{12}{9} - \frac{9}{12} = \frac{16}{12} - \frac{9}{12} = \boxed{\textbf{(C) }\frac{7}{12}}</math> | This expression is equivalent to <math>\frac{12}{9} - \frac{9}{12} = \frac{16}{12} - \frac{9}{12} = \boxed{\textbf{(C) }\frac{7}{12}}</math> | ||
| + | |||
| + | For the exact same problem but with answer choice D unsimplified, see http://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_1 | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=B|before=First Problem|num-a=2}} | {{AMC10 box|year=2013|ab=B|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:04, 13 February 2017
Problem
What is 
?
Solution
This expression is equivalent to 
For the exact same problem but with answer choice D unsimplified, see http://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_1
See Also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
