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

Latest revision as of 16:04, 12 May 2023

$.4+.02+.006=$

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

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

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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