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

(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...)
 
(Solution)
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==Solution==
 
==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>
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<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 $\frac{401}{.205}$.

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

Solution

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

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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