Difference between revisions of "2000 AMC 10 Problems/Problem 8"
m (→Solution) |
Hashtagmath (talk | contribs) (→See Also) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 15: | Line 15: | ||
==Solution== | ==Solution== | ||
| − | Let <math>f</math> be the number of freshman and s be the number of sophomores. | + | Let <math>f</math> be the number of freshman and <math>s</math> be the number of sophomores. |
<math>\frac{2}{5}f=\frac{4}{5}s</math> | <math>\frac{2}{5}f=\frac{4}{5}s</math> | ||
| Line 29: | Line 29: | ||
{{AMC10 box|year=2000|num-b=7|num-a=9}} | {{AMC10 box|year=2000|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 15:36, 19 April 2021
Problem
At Olympic High School, 
of the freshmen and 
of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?
There are five times as many sophomores as freshmen.
There are twice as many sophomores as freshmen.
There are as many freshmen as sophomores.
There are twice as many freshmen as sophomores.
There are five times as many freshmen as sophomores.
Solution
Let 
be the number of freshman and 
be the number of sophomores.
There are twice as many freshmen as sophomores.
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
