Difference between revisions of "1989 AJHSME Problems/Problem 4"

Revision as of 05:32, 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)
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

@@ Line 7: / Line 7: @@
 ==Solution==
-<math>401</math> is around <math>400</math> and <math>.205</math> is around $.2 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==