Difference between revisions of "1975 AHSME Problems/Problem 1"

 
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Calculating, we find that <math>\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}} = \frac {1}{2 - \frac {1}{2 - \frac {2}{3}}} = \frac {1}{2 - \frac {3}{4}} = \frac {1}{\frac {5}{4}} = \boxed{\textbf{(B) } \dfrac{4}{5}}</math>.
 
Calculating, we find that <math>\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}} = \frac {1}{2 - \frac {1}{2 - \frac {2}{3}}} = \frac {1}{2 - \frac {3}{4}} = \frac {1}{\frac {5}{4}} = \boxed{\textbf{(B) } \dfrac{4}{5}}</math>.
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==See Also==
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{{AHSME box|year=1975|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 15:46, 19 January 2021

Problem

The value of $\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}}$ is

$\textbf{(A)}\ 3/4 \qquad  \textbf{(B)}\ 4/5 \qquad  \textbf{(C)}\ 5/6 \qquad  \textbf{(D)}\ 6/7 \qquad  \textbf{(E)}\ 6/5$


Solution

Solution by e_power_pi_times_i


Calculating, we find that $\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}} = \frac {1}{2 - \frac {1}{2 - \frac {2}{3}}} = \frac {1}{2 - \frac {3}{4}} = \frac {1}{\frac {5}{4}} = \boxed{\textbf{(B) } \dfrac{4}{5}}$.

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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 26 27 28 29 30
All AHSME Problems and Solutions

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