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

1955 AHSME Problems/Problem 6

Revision as of 10:17, 28 April 2019 by Brudder (talk | contribs) (Created page with "Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges. That means...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Since we are buying at $3$ for $10$ cents and $5$ for $20$ cents, let's assume that together, we are buying 15 oranges. That means that we are getting a total of $30$ oranges for $(10*5) + (20*3)$ cents. That comes to a total of $30$ oranges for $110$ cents. $110/30$ = $11/3$ and we are done. -Brudder

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_6&oldid=105524"