1998 AJHSME Problems/Problem 3

Problem 3

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

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

Solution

$\frac{\frac{3}{8}+\frac{7}{8}}\frac{4}{5}}=\frac{\frac{10}{8}}{\frac{4}{5}=\frac{\frac{5}{4}}{\frac{4}{5}}=\frac{25}{16}$ (Error compiling LaTeX. Unknown error_msg)

$\boxed{B}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1998 AJHSME Problems/Problem 3

Problem 3

Solution

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