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 | + | <math>\textbf{(A)}\ 587\cdot10^8\text{ miles}\qquad |
− | \textbf{(B)}\ 587 | + | \textbf{(B)}\ 587\cdot10^{10}\text{ miles}\qquad |
− | \textbf{(C)}\ 587 | + | \textbf{(C)}\ 587\cdot10^{-10}\text{ miles} \\ |
− | \textbf{(D)}\ 587 | + | \textbf{(D)}\ 587\cdot10^{12} \text{ miles} \qquad |
− | \textbf{(E)}\ 587 | + | \textbf{(E)}\ 587\cdot10^{ - 12} \text{ miles} </math> |
==Solution== | ==Solution== | ||
+ | |||
+ | 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>\boxed{\textbf{(D) }587\cdot10^{12}}</math>. | ||
==See Also== | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1956|num-b=2|num-a=4}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:02, 14 March 2023
Problem #3
The distance light travels in one year is approximately miles. The distance light travels in
years is:
Solution
The distance light travels in one year can also be written as . In 100 years, light will travel
.
Therefore, our answer is .
See Also
1956 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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