Difference between revisions of "1993 AJHSME Problems/Problem 16"
m (→Solution) |
(Undo revision 80587 by Ivanthekingiii (talk)) |
||
Line 7: | Line 7: | ||
==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> | ||
− | |||
==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 19:59, 10 October 2016
Problem
Solution
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.