Art of Problem Solving

1993 AJHSME Problems/Problem 16: Difference between revisions

Gina (talk | contribs)
No edit summary
Nathan wailes (talk | contribs)
No edit summary
Line 10: Line 10:
==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}}

Revision as of 23:11, 4 July 2013

Problem

$\frac{1}{1+\frac{1}{2+\frac{1}{3}}}=$

$\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{3}{10} \qquad \text{(C)}\ \dfrac{7}{10} \qquad \text{(D)}\ \dfrac{5}{6} \qquad \text{(E)}\ \dfrac{10}{3}$ 84.3 $

Solution

\[\frac{1}{1+\frac{1}{2+\frac13}} = \frac{1}{1+\frac{1}{\frac73}} = \frac{1}{1+\frac37} = \frac{1}{\frac{10}{7}} = \boxed{\text{(C)}\ \frac{7}{10}}\]

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America.