Difference between revisions of "1989 AJHSME Problems/Problem 4"
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Latest revision as of 23:57, 4 July 2013
Problem
Estimate to determine which of the following numbers is closest to 
.
Solution
is around 
and 
is around 
so the fraction is approximately ![\[\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}\]](http://latex.artofproblemsolving.com/3/c/3/3c3e40f02aa475637780a3d0cc61e858e0027587.png)
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
