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 "2019 AMC 10B Problems/Problem 3"

m (Reverted edits by Bgav (talk) to last revision by Toastybaker)
(Tag: Rollback)
(Cleaned up the solutions)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
60% of seniors do not play a musical instrument. If we denote x as the number of seniors, then <cmath>\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500</cmath>
+
<math>60\%</math> of seniors do not play a musical instrument. If we denote <math>x</math> as the number of seniors, then <cmath>\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500</cmath>
  
 
<cmath>\frac{3}{5}x + 150 - \frac{3}{10}x = 234</cmath>
 
<cmath>\frac{3}{5}x + 150 - \frac{3}{10}x = 234</cmath>
 
<cmath>\frac{3}{10}x = 84</cmath>
 
<cmath>\frac{3}{10}x = 84</cmath>
−
<cmath>x = 84\cdot\frac{10}{3} \Rightarrow 280</cmath>
+
<cmath>x = 84\cdot\frac{10}{3} = 280</cmath>
  
−
Thus there are <math>500-x = 220</math> non-seniors. Since 70% of the non-seniors play a musical instrument, <math>220 \cdot \frac{7}{10} = \boxed{B) 154}</math>
+
Thus there are <math>500-x = 220</math> non-seniors. Since 70% of the non-seniors play a musical instrument, <math>220 \cdot \frac{7}{10} = \boxed{\textbf{(B) } 154}</math>
  
 
==Solution 2==
 
==Solution 2==
  
−
Let x be the number of seniors, and y be the number of non-seniors. Then <cmath>\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234</cmath>
+
Let <math>x</math> be the number of seniors, and <math>y</math> be the number of non-seniors. Then <cmath>\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234</cmath>
  
−
Multiplying 10 to every term gives us
+
Multiplying both sides by <math>10</math> gives us
 
<cmath>6x + 3y = 2340</cmath>
 
<cmath>6x + 3y = 2340</cmath>
  
 
Also, <math>x + y = 500</math> because there are 500 students in total.
 
Also, <math>x + y = 500</math> because there are 500 students in total.
  
 +
Solving this system of equations give us <math>x = 280</math>,  <math>y = 220</math>.
  
−
Solving these system of equations give us <math>x = 280</math>,  <math>y = 220</math>
+
Since <math>70\%</math> of the non-seniors play a musical instrument, the answer is simply <math>70\%</math> of <math>220</math>, which gives us <math>\boxed{\textbf{(B) } 154}</math>
−
 
 
−
 
 
−
Since 70% of the non-seniors play a musical instrument, we simply get 70% of 220, which gives us <math>\boxed{B) 154}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 19:01, 17 February 2019

Problem

In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?

$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution

$60\%$ of seniors do not play a musical instrument. If we denote $x$ as the number of seniors, then \[\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500\]

\[\frac{3}{5}x + 150 - \frac{3}{10}x = 234\] \[\frac{3}{10}x = 84\] \[x = 84\cdot\frac{10}{3} = 280\]

Thus there are $500-x = 220$ non-seniors. Since 70% of the non-seniors play a musical instrument, $220 \cdot \frac{7}{10} = \boxed{\textbf{(B) } 154}$

Solution 2

Let $x$ be the number of seniors, and $y$ be the number of non-seniors. Then \[\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234\]

Multiplying both sides by $10$ gives us \[6x + 3y = 2340\]

Also, $x + y = 500$ because there are 500 students in total.

Solving this system of equations give us $x = 280$, $y = 220$.

Since $70\%$ of the non-seniors play a musical instrument, the answer is simply $70\%$ of $220$, which gives us $\boxed{\textbf{(B) } 154}$

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_3&oldid=103177"