Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1957 AHSME Problems/Problem 20"
 
(No difference)

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 $\boxed{(A)}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_20&oldid=158473"