1957 AHSME Problems/Problem 20

Revision as of 13:38, 14 July 2021 by Mr.sharkman (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{(A)}$.

Invalid username
Login to AoPS