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Difference between revisions of "1998 AJHSME Problems/Problem 3"
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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}} \cdot \frac{5}{4}} = \frac{25}{16} = \boxed{B}</math>
+
<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 ==

Latest revision as of 23:09, 30 March 2015

Problem

$\dfrac{\dfrac{3}{8} + \dfrac{7}{8}}{\dfrac{4}{5}} =$

$\text{(A)}\ 1 \qquad \text{(B)} \dfrac{25}{16} \qquad \text{(C)}\ 2 \qquad \text{(D)}\ \dfrac{43}{20} \qquad \text{(E)}\ \dfrac{47}{16}$

Solution

$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}} = \frac{\frac{10}{8}}{\frac{4}{5}} = \frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \cdot \frac{5}{4} = \frac{25}{16} = \boxed{B}$

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

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