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

(Created page with "==Problem== <math> \frac{1}{1+\frac{1}{2+\frac{1}{3}}}= </math> <math> \text{(A)}\ 81.0\qquad\text{(B)}\ 82.7\qquad\text{(C)}\ 83.0\qquad\text{(D)}\ 84.0\qquad\text{(E)}\ 84.3 ...")
 
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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)}\ 81.0\qquad\text{(B)}\ 82.7\qquad\text{(C)}\ 83.0\qquad\text{(D)}\ 84.0\qquad\text{(E)}\ 84.3 </math>
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<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> 84.3 $
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==Solution==
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<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==
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{{AJHSME box|year=1993|num-b=15|num-a=17}}

Revision as of 22:10, 22 December 2012

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
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All AJHSME/AMC 8 Problems and Solutions
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