Difference between revisions of "1993 AJHSME Problems/Problem 16"

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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}}
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{{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

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