Difference between revisions of "1989 AJHSME Problems/Problem 4"
(→Solution) |
|||
(One intermediate revision by one other user not shown) | |||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | <math>401</math> is around <math>400</math> and <math>.205</math> is around <math>.2 </math>so the [[fraction]] is approximately <cmath>\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}</cmath> | + | <math>401</math> is around <math>400</math> and <math>.205</math> is around <math>.2 </math> so the [[fraction]] is approximately <cmath>\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}</cmath> |
==See Also== | ==See Also== | ||
Line 13: | Line 13: | ||
{{AJHSME box|year=1989|num-b=3|num-a=5}} | {{AJHSME box|year=1989|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
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
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.