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Difference between revisions of "1956 AHSME Problems/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
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<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
+
\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==
  
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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>
+
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>.
  
−
Therefore, our answer is <math>(D) 587 * 10^{12}</math>
+
Therefore, our answer is <math>\boxed{\textbf{(D) }587\cdot10^{12}}</math>.
  
 
==See Also==
 
==See Also==
 +
 +
{{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 16:02, 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
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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