MathDB
1987 AJHSME Problem 1

Source:

June 25, 2011

Problem Statement

.4+.02+.006=.4+.02+.006=
(A) .012(B) .066(C) .12(D) .24(E).426\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426
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