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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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By the rules of [[exponentiation|exponents]], <math>10^{-49}=10*10^{-50}</math>, which clearly exceeds <math>2*10^{-50}</math>. <math>10^{-49}-2*10^{-50}=10*10^{-50}-2*10^{-50}=8*10^{-50}</math>, so we choose answer choice <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]]

Latest revision as of 08:15, 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

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

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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