1993 AJHSME Problems/Problem 2

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Problem

When the fraction $\dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be

$\text{(A)}\ 11 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 133$

Solution

\begin{align*} \dfrac{49}{84} &= \dfrac{7^2}{2^2\cdot 3\cdot 7} \\ &= \dfrac{7}{2^2\cdot 3} \\ &= \dfrac{7}{12}. \end{align*}

The sum of the numerator and denominator is $7+12=19\rightarrow \boxed{\text{C}}$.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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