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Difference between revisions of "1957 AHSME Problems/Problem 20"

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

Latest revision as of 13:38, 14 July 2021

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 $\boxed{(A)}$.

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