Difference between revisions of "2000 AMC 10 Problems/Problem 8"
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==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> | ||
+ | |||
+ | <math>2f = 4s</math> | ||
<math>f=2s</math> | <math>f=2s</math> | ||
− | There are twice as many | + | There are twice as many freshmen as sophomores. |
<math>\boxed{\text{D}}</math> | <math>\boxed{\text{D}}</math> | ||
Line 27: | 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]] |
Latest 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 |
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