Difference between revisions of "2000 AMC 10 Problems/Problem 8"

m
 
(6 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
 +
At Olympic High School, <math>\frac{2}{5}</math> of the freshmen and <math>\frac{4}{5}</math> 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?
 +
 +
<math>\textbf{(A)}</math> There are five times as many sophomores as freshmen.
 +
 +
<math>\textbf{(B)}</math> There are twice as many sophomores as freshmen.
 +
 +
<math>\textbf{(C)}</math> There are as many freshmen as sophomores.
 +
 +
<math>\textbf{(D)}</math> There are twice as many freshmen as sophomores.
 +
 +
<math>\textbf{(E)}</math> There are five times as many freshmen as sophomores.
  
 
==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 freshman as sophomores.
+
There are twice as many freshmen as sophomores.
 
<math>\boxed{\text{D}}</math>
 
<math>\boxed{\text{D}}</math>
 +
 +
==Video Solution by Daily Dose of Math==
 +
 +
https://youtu.be/qFwbXU-guuA?si=Lm84VudzdQPJ301d
 +
 +
~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==
  
 
{{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 23:41, 14 July 2024

Problem

At Olympic High School, $\frac{2}{5}$ of the freshmen and $\frac{4}{5}$ 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?

$\textbf{(A)}$ There are five times as many sophomores as freshmen.

$\textbf{(B)}$ There are twice as many sophomores as freshmen.

$\textbf{(C)}$ There are as many freshmen as sophomores.

$\textbf{(D)}$ There are twice as many freshmen as sophomores.

$\textbf{(E)}$ There are five times as many freshmen as sophomores.

Solution

Let $f$ be the number of freshman and $s$ be the number of sophomores.

$\frac{2}{5}f=\frac{4}{5}s$

$2f = 4s$

$f=2s$

There are twice as many freshmen as sophomores. $\boxed{\text{D}}$

Video Solution by Daily Dose of Math

https://youtu.be/qFwbXU-guuA?si=Lm84VudzdQPJ301d

~Thesmartgreekmathdude

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
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. AMC logo.png