Difference between revisions of "1993 AJHSME Problems/Problem 16"
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==Solution== | ==Solution== | ||
<cmath>\frac{1}{1+\frac{1}{2+\frac13}} = \frac{1}{1+\frac{1}{\frac73}} = \frac{1}{1+\frac37} = \frac{1}{\frac{10}{7}} = \boxed{\text{(C)}\ \frac{7}{10}} </cmath> | <cmath>\frac{1}{1+\frac{1}{2+\frac13}} = \frac{1}{1+\frac{1}{\frac73}} = \frac{1}{1+\frac37} = \frac{1}{\frac{10}{7}} = \boxed{\text{(C)}\ \frac{7}{10}} </cmath> | ||
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==See Also== | ==See Also== | ||
{{AJHSME box|year=1993|num-b=15|num-a=17}} | {{AJHSME box|year=1993|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:59, 10 October 2016
Problem
Solution
![\[\frac{1}{1+\frac{1}{2+\frac13}} = \frac{1}{1+\frac{1}{\frac73}} = \frac{1}{1+\frac37} = \frac{1}{\frac{10}{7}} = \boxed{\text{(C)}\ \frac{7}{10}}\]](http://latex.artofproblemsolving.com/2/0/4/2049d00fe79ffb05b164c4a606d39f0b6e11449b.png)
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
