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<...) |
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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> | ||
| Line 17: | Line 17: | ||
{{AJHSME box|year=1989|num-b=1|num-a=3}} | {{AJHSME box|year=1989|num-b=1|num-a=3}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:57, 4 July 2013
Problem
Solution
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
