Difference between revisions of "1989 AJHSME Problems/Problem 4"
5849206328x (talk | contribs) (New page: ==Problem== Estimate to determine which of the following numbers is closest to <math>\frac{401}{.205}</math>. <math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text...) |
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==Solution== | ==Solution== | ||
− | <math>401</math> is around <math>400</math> and <math>.205</math> is around | + | <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== |
Revision as of 04:32, 25 April 2010
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 |