Difference between revisions of "1998 AJHSME Problems/Problem 3"
(Created page with "==Problem 3== <math>\dfrac{\dfrac{3}{8} + \dfrac{7}{8}}{\dfrac{4}{5}} = </math> <math>\text{(A)}\ 1 \qquad \text{(B)} \dfrac{25}{16} \qquad \text{(C)}\ 2 \qquad \text{(D)}\ \df...") |
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== See also == | == See also == | ||
| − | {{AJHSME box|year=1998| | + | {{AJHSME box|year=1998|num-b=2|num-a=4}} |
* [[AJHSME]] | * [[AJHSME]] | ||
* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
Revision as of 23:05, 9 June 2011
Problem 3
Solution
$\frac{\frac{3}{8}+\frac{7}{8}}\frac{4}{5}}=\frac{\frac{10}{8}}{\frac{4}{5}=\frac{\frac{5}{4}}{\frac{4}{5}}=\frac{25}{16}$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
