Difference between revisions of "1993 AJHSME Problems/Problem 16"
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<math> \frac{1}{1+\frac{1}{2+\frac{1}{3}}}= </math> | <math> \frac{1}{1+\frac{1}{2+\frac{1}{3}}}= </math> | ||
− | <math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{3}{10} \qquad \text{(C)}\ \dfrac{7}{10} \qquad \text{(D)}\ \dfrac{5}{6} \qquad \text{(E)}\ \dfrac{10}{3}</math> | + | <math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{3}{10} \qquad \text{(C)}\ \dfrac{7}{10} \qquad \text{(D)}\ \dfrac{5}{6} \qquad \text{(E)}\ \dfrac{10}{3}</math> |
==Solution== | ==Solution== | ||
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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}} |
Latest revision as of 18:59, 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.