Difference between revisions of "1998 AJHSME Problems/Problem 3"
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| − | ==Problem | + | ==Problem== |
<math>\dfrac{\dfrac{3}{8} + \dfrac{7}{8}}{\dfrac{4}{5}} = </math> | <math>\dfrac{\dfrac{3}{8} + \dfrac{7}{8}}{\dfrac{4}{5}} = </math> | ||
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==Solution== | ==Solution== | ||
| − | <math>\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}} = \frac{\frac{10}{8}}{\frac{4}{5}} = \frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4 | + | <math>\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}} = \frac{\frac{10}{8}}{\frac{4}{5}} = \frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \cdot \frac{5}{4} = \frac{25}{16} = \boxed{B}</math> |
== See also == | == See also == | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:09, 30 March 2015
Problem
Solution
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
