Difference between revisions of "1957 AHSME Problems/Problem 3"

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== Problem 3==
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The simplest form of <math>1 - \frac{1}{1 + \frac{a}{1 - a}}</math> is:
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<math>\textbf{(A)}\ {a}\text{ if }{a\not= 0} \qquad \textbf{(B)}\ 1\qquad
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\textbf{(C)}\ {a}\text{ if }{a\not=-1}\qquad
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\textbf{(D)}\ {1-a}\text{ with not restriction on }{a}\qquad
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\textbf{(E)}\ {a}\text{ if }{a\not= 1}  </math>
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==Solution==
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We have <math>1 - \frac{1}{1 + \frac{a}{1 - a}} = 1 - \frac{1}{\frac{1}{1-a}} = 1 - \frac{1-a}{1} = a</math> for almost all <math>a</math>. However, the first step is invalid when <math>a=1</math>, and each step is valid otherwise, so the answer is (E).
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==See Also==
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{{AHSME box|year=1957|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 17:40, 22 May 2019

Problem 3

The simplest form of $1 - \frac{1}{1 + \frac{a}{1 - a}}$ is:

$\textbf{(A)}\ {a}\text{ if }{a\not= 0} \qquad \textbf{(B)}\ 1\qquad  \textbf{(C)}\ {a}\text{ if }{a\not=-1}\qquad \textbf{(D)}\ {1-a}\text{ with not restriction on }{a}\qquad \textbf{(E)}\ {a}\text{ if }{a\not= 1}$

Solution

We have $1 - \frac{1}{1 + \frac{a}{1 - a}} = 1 - \frac{1}{\frac{1}{1-a}} = 1 - \frac{1-a}{1} = a$ for almost all $a$. However, the first step is invalid when $a=1$, and each step is valid otherwise, so the answer is (E).

See Also

1957 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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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