Difference between revisions of "1987 AJHSME Problems/Problem 2"
5849206328x (talk | contribs) (New page: ==Problem== <math>\frac{2}{25}=</math> <math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)} 1.25 \qquad \text{(E)}\ 12.5</math> ==Solution== {{Solution...) |
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<math>\frac{2}{25}=</math> | <math>\frac{2}{25}=</math> | ||
| − | <math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)} 1.25 \qquad \text{(E)}\ 12.5</math> | + | <math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5</math> |
==Solution== | ==Solution== | ||
| − | {{ | + | <math>\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{\text{B}}</math> |
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1987|num-b=1|num-a=3}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:52, 4 July 2013
Problem
Solution
See Also
| 1987 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. 
