1998 AJHSME Problems/Problem 3

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 (ProblemsAnswer KeyResources)
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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