1989 AJHSME Problems/Problem 2

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

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

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1989 AJHSME Problems/Problem 2

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