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

Revision as of 12:00, 25 May 2009

$.4+.02+.006=$

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

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

1987 AJHSME (Problems • Answer Key • Resources)
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

@@ Line 7: / Line 7: @@
 ==Solution==
-<math>.4+.02+.006 = .400 + .020 + .006 = \boxed{.426}</math>
+<math>.4+.02+.006 = .400 + .020 + .006 = .426\rightarrow \boxed{\text{E}}</math>
 ==See Also==
-[[1987 AJHSME Problems]]
+{{AJHSME box|year=1987|before=First<br>Problem|num-a=2}}
+[[Category:Introductory Algebra Problems]]