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Difference between revisions of "1988 AJHSME Problems/Problem 6"

(New page: ==Problem== <math>\frac{(.2)^3}{(.02)^2} =</math> <math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20</math> ==Solution== Convert...)
 
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Line 6: Line 6:
  
 
==Solution==
 
==Solution==
Converting the decimals to fractions gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>.
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Converting the [[decimal|decimals]] to [[fraction|fractions]] gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>.
  
 
==See Also==
 
==See Also==
  
[[1988 AJHSME Problems]]
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{{AJHSME box|year=1988|num-b=5|num-a=7}}
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[[Category:Introductory Algebra Problems]]

Revision as of 17:29, 1 June 2009

Problem

$\frac{(.2)^3}{(.02)^2} =$

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20$

Solution

Converting the decimals to fractions gives us $\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}$.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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
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