1998 AJHSME Problems/Problem 3
Revision as of 23:18, 9 June 2011 by Smallpeoples343 (talk | contribs) (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...")
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 1997 AJHSME |
Followed by 1999 AMC 8 | |
| 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 | ||
