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

m (Solution)
(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:  
 
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:  
  
<math>\textbf{(A)}\ 587 * 10^8\text{ miles}\qquad
+
<math>\textbf{(A)}\ 587\cdot10^8\text{ miles}\qquad
\textbf{(B)}\ 587 * 10^{10}\text{ miles}\qquad
+
\textbf{(B)}\ 587\cdot10^{10}\text{ miles}\qquad
\textbf{(C)}\ 587*10^{-10}\text{ miles} \\
+
\textbf{(C)}\ 587\cdot10^{-10}\text{ miles} \\
\textbf{(D)}\ 587 * 10^{12} \text{ miles} \qquad
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\textbf{(D)}\ 587\cdot10^{12} \text{ miles} \qquad
\textbf{(E)}\ 587* 10^{ - 12} \text{ miles} </math>
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\textbf{(E)}\ 587\cdot10^{ - 12} \text{ miles} </math>
 
 
  
 
==Solution==
 
==Solution==

Revision as of 16:01, 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 $587 * 10^{10}$. In 100 years, light will travel $(587 * 10^{10}) * 100 = 587 * 10^{12}$.

Therefore, our answer is $\boxed{\textbf{(D)}\quad  587 * 10^{12}}$.

See Also

1956 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AHSME Problems and Solutions

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