Difference between revisions of "1957 AHSME Problems/Problem 12"

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Since <math>10^{-50}=\dfrac{1}{10^{50}}</math>,
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== Problem 12==
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Comparing the numbers <math>10^{-49}</math> and <math>2\cdot 10^{-50}</math> we may say:
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<math>\textbf{(A)}\ \text{the first exceeds the second by }{8\cdot 10^{-1}}\qquad\\
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\textbf{(B)}\ \text{the first exceeds the second by }{2\cdot 10^{-1}}\qquad\\
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\textbf{(C)}\ \text{the first exceeds the second by }{8\cdot 10^{-50}}\qquad\\
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\textbf{(D)}\ \text{the second is five times the first}\qquad\\
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\textbf{(E)}\ \text{the first exceeds the second by }{5}  </math>
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== Solution ==
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<math>\boxed{\textbf{(C)}}</math>
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==See Also==
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{{AHSME 50p box|year=1957|num-b=11|num-a=13}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Revision as of 09:12, 25 July 2024

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

$\boxed{\textbf{(C)}}$

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AHSME Problems and Solutions

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