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Difference between revisions of "1987 AJHSME Problems/Problem 2"

 
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<math>\frac{2}{25}=</math>
 
<math>\frac{2}{25}=</math>
  
<math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)} 1.25 \qquad \text{(E)}\ 12.5</math>
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<math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5</math>
  
 
==Solution==
 
==Solution==
  
<math>\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = \boxed{0.08}</math>
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<math>\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{\text{B}}</math>
  
 
==See Also==
 
==See Also==
  
[[1987 AJHSME Problems]]
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{{AJHSME box|year=1987|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:52, 4 July 2013

Problem

$\frac{2}{25}=$

$\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5$

Solution

$\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{\text{B}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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