1957 AHSME Problems/Problem 20: Difference between revisions
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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 < | 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>. | ||
Revision as of 12:38, 14 July 2021
Suppose the first half of the trip's distance is called 
. Then the time for the first half is 
and the second half's time is 
, so the total time is 
, so the speed is 
, so the answer is 
.
