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Difference between revisions of "1989 AJHSME Problems/Problem 2"

(New page: ==Problem== <math>\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=</math> <math>\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246<...)
(No difference)

Revision as of 18:49, 26 May 2009

Problem

$\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=$

$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$

Solution

\begin{align*} \frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\ &= \frac{246}{1000} \\ &= .246 \rightarrow \boxed{\text{E}} \end{align*}

See Also

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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
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