1989 AJHSME Problems/Problem 4: Difference between revisions

Revision as of 04:32, 25 April 2010

Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$ .

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

$401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately $\frac{400}{.2}=2000\rightarrow \boxed{\text{E}}$

1989 AJHSME (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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 ==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==