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

(New page: ==Problem== <math>.4+.02+.006=</math> <math>\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426</math> ==Solution== {{Solution...)
 
(2 intermediate revisions by 2 users not shown)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
{{Solution}}
+
<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}}
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 23:52, 4 July 2013

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png