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

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==Solution==
 
==Solution==
  
<math>.4+.02+.006 = .400 + .020 + .006 = \boxed{.426}</math>
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<math>.4+.02+.006 = .400 + .020 + .006 = .426\rightarrow \boxed{\text{E}}</math>
  
 
==See Also==
 
==See Also==
  
[[1987 AJHSME Problems]]
+
{{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 16:04, 12 May 2023

Problem

$.4+.02+.006=$

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

Solution

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

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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