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

(Problem #3)
(Solution)
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\textbf{(D)}\ 587 * 10^{12} \text{ miles} \qquad
 
\textbf{(D)}\ 587 * 10^{12} \text{ miles} \qquad
 
\textbf{(E)}\ 587* 10^{ - 12} \text{ miles} </math>
 
\textbf{(E)}\ 587* 10^{ - 12} \text{ miles} </math>
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==Solution==
 
==Solution==
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The distance light travels in one year can also be written as <math>587 * 10^10</math>. In 100 years, light will travel <math>(587 * 10^10) * 100 = 587 * 10^12</math>
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Therefore, our answer is <math>\fbox{(D) 587 * 10^{12}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 17:17, 31 December 2015

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:

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


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 $\fbox{(D) 587 * 10^{12}}$ (Error compiling LaTeX. Unknown error_msg)

See Also