1957 AHSME Problems/Problem 12

Problem 12

Comparing the numbers $10^{-49}$ and $2\cdot 10^{-50}$ we may say:

$\textbf{(A)}\ \text{the first exceeds the second by }{8\cdot 10^{-1}}\qquad\\ \textbf{(B)}\ \text{the first exceeds the second by }{2\cdot 10^{-1}}\qquad\\ \textbf{(C)}\ \text{the first exceeds the second by }{8\cdot 10^{-50}}\qquad\\ \textbf{(D)}\ \text{the second is five times the first}\qquad\\ \textbf{(E)}\ \text{the first exceeds the second by }{5}$

Solution

By the rules of exponents, $10^{-49}=10*10^{-50}$ , which clearly exceeds $2*10^{-50}$ . $10^{-49}-2*10^{-50}=10*10^{-50}-2*10^{-50}=8*10^{-50}$ , so we choose answer choice $\boxed{\textbf{(C)}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1957 AHSME Problems/Problem 12

Problem 12

Solution

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