1957 AHSME Problems/Problem 20
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Suppose the first half of the trip's distance is called 
. Then the time for the first half is 
\dfrac{x}{45}
\dfrac{x}{50}+\dfrac{x}{45}
\dfrac{2x}{\frac{x}{50}+\frac{x}{45}}=\dfrac{2x}{\frac{19x}{450}}=2x \cdot \dfrac{450}{19x}=\dfrac{900}{19}=47\dfrac{7}{19}
\boxed{(A)}$.
