Difference between revisions of "1985 AJHSME Problems/Problem 3"
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So the answer is <math>\boxed{\text{D}}</math> | So the answer is <math>\boxed{\text{D}}</math> | ||
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| + | ==Video Solution by BoundlessBrain!== | ||
| + | https://youtu.be/BfFv227egOg | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/KW5HexBjEHU | ||
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| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
| − | + | {{AJHSME box|year=1985|num-b=2|num-a=4}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 10:52, 24 June 2023
Problem
Solution
We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: ![\[\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}\]](http://latex.artofproblemsolving.com/d/c/7/dc7eff4ce39984d2952c9d3bfa23ad8686805055.png)
We know that 
, so
So the answer is 
Video Solution by BoundlessBrain!
Video Solution
~savannahsolver
See Also
| 1985 AJHSME (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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
