1954 AHSME Problems/Problem 34

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Problem 34

The fraction $\frac{1}{3}$:

$\textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9}$

Solution

$\frac{333333333}{10^8}-\frac{1}{3}\implies \frac{3\cdot 10^7+3\cdot 10^6+\dots+3\cdot 10^0}{10^8}-\frac{1}{3}\implies\frac{3(10^{8}-1)}{9\cdot 10^{8}}-\frac{1}{3}\implies\frac{10^8-1}{3\cdot 10^8}-\frac{1}{3}\implies\frac{10^8}{3\cdot 10^8}-\frac{1}{3\cdot 10^8}-\frac{1}{3}\implies\frac{-1}{3\cdot 10^8}$ Because we did $.333333333-\frac{1}{3}$, it is $\fbox{B}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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