1957 AHSME Problems/Problem 20

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Suppose the first half of the trip's distance is called $x$. Then the time for the first half is $\dfrac{x}{50}, and the second half's time is$\dfrac{x}{45}$, so the total time is$\dfrac{x}{50}+\dfrac{x}{45}$, so the speed is$\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}$, so the answer is$\boxed{(A)}$.