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Difference between revisions of "1956 AHSME Problems/Problem 3"

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==Problem #3==
 
==Problem #3==
  
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The distance light travels in one year is approximately <math>5,870,000,000,000</math> miles. The distance light travels in <math>100</math> years is:
  
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<math>\textbf{(A)}\ 587\cdot10^8\text{ miles}\qquad
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\textbf{(B)}\ 587\cdot10^{10}\text{ miles}\qquad
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\textbf{(C)}\ 587\cdot10^{-10}\text{ miles} \\
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\textbf{(D)}\ 587\cdot10^{12} \text{ miles} \qquad
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\textbf{(E)}\ 587\cdot10^{ - 12} \text{ miles} </math>
  
 
==Solution==
 
==Solution==
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The distance light travels in one year can also be written as <math>587\cdot10^{10}</math>. In 100 years, light will travel <math>(587\cdot10^{10})\cdot100=587\cdot10^{12}</math>.
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Therefore, our answer is  <math>\boxed{\textbf{(D) }587\cdot10^{12}}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME 50p box|year=1956|num-b=2|num-a=4}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:02, 14 March 2023

Problem #3

The distance light travels in one year is approximately $5,870,000,000,000$ miles. The distance light travels in $100$ years is:

$\textbf{(A)}\ 587\cdot10^8\text{ miles}\qquad \textbf{(B)}\ 587\cdot10^{10}\text{ miles}\qquad \textbf{(C)}\ 587\cdot10^{-10}\text{ miles} \\ \textbf{(D)}\ 587\cdot10^{12} \text{ miles} \qquad \textbf{(E)}\ 587\cdot10^{ - 12} \text{ miles}$

Solution

The distance light travels in one year can also be written as $587\cdot10^{10}$. In 100 years, light will travel $(587\cdot10^{10})\cdot100=587\cdot10^{12}$.

Therefore, our answer is $\boxed{\textbf{(D) }587\cdot10^{12}}$.

See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AHSME Problems and Solutions

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