Difference between revisions of "1987 AJHSME Problems/Problem 1"

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==Problem==
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== Problem ==
  
 
<math>.4+.02+.006=</math>
 
<math>.4+.02+.006=</math>
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<math>\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426</math>
 
<math>\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426</math>
  
==Solution==
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== Solution ==
  
<math>.4+.02+.006 = .400 + .020 + .006 = \boxed{.426}</math>
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<math>.4+.02+.006 = .400+.020+.006 = \boxed{\text{(E)} .426}</math>
  
==See Also==
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== See Also ==
  
[[1987 AJHSME Problems]]
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{{AJHSME box|year=1987|before=First<br>Problem|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:20, 4 March 2025

Problem

$.4+.02+.006=$

$\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426$

Solution

$.4+.02+.006 = .400+.020+.006 = \boxed{\text{(E)} .426}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
First
Problem
Followed by
Problem 2
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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