Difference between revisions of "1989 AJHSME Problems/Problem 2"
5849206328x (talk | contribs) (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<...) |
SuicideJoe (talk | contribs) (→Solution) |
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\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\ | \frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\ | ||
&= \frac{246}{1000} \\ | &= \frac{246}{1000} \\ | ||
| − | &= .246 \rightarrow \boxed{\text{ | + | &= .246 \rightarrow \boxed{\text{D}} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
