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1987 AJHSME Problems/Problem 1

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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 (Problems • 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
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