Difference between revisions of "2019 AMC 10B Problems/Problem 3"
Sevenoptimus (talk | contribs) m (Fixed punctuation) |
Sevenoptimus (talk | contribs) m (Fixed solution numbering) |
||
| Line 5: | Line 5: | ||
<math>\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266</math> | <math>\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266</math> | ||
| − | ==Solution== | + | ==Solution 1== |
<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> | <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> | ||
Revision as of 19:55, 17 February 2019
Contents
Problem
In a high school with 
students, 
of the seniors play a musical instrument, while 
of the non-seniors do not play a musical instrument. In all, 
of the students do not play a musical instrument. How many non-seniors play a musical instrument?
Solution 1
of seniors do not play a musical instrument. If we denote 
as the number of seniors, then ![\[\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500\]](http://latex.artofproblemsolving.com/3/9/b/39beb27287fd9606cb72594254ed23d8f84d808a.png)
![\[\frac{3}{5}x + 150 - \frac{3}{10}x = 234\]](http://latex.artofproblemsolving.com/3/b/b/3bbf60a030fd69addddedd21963193e62f2f4e54.png)
![\[\frac{3}{10}x = 84\]](http://latex.artofproblemsolving.com/a/e/0/ae0b9942f3104e50463964898645e3071ad35d75.png)
![\[x = 84\cdot\frac{10}{3} = 280\]](http://latex.artofproblemsolving.com/e/9/1/e917d53d7f0e260365f5b83bbee0d622ee11793d.png)
Thus there are 
non-seniors. Since 70% of the non-seniors play a musical instrument, 
.
iron
Solution 2
Let be the number of seniors, and 
be the number of non-seniors. Then ![\[\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234\]](http://latex.artofproblemsolving.com/c/7/f/c7f0504b3bfa775132523a004e290ee0d48efc5e.png)
Multiplying both sides by 
gives us
![\[6x + 3y = 2340\]](http://latex.artofproblemsolving.com/e/e/d/eed3fa790604393aa68809319e61bca0eba6aee2.png)
Also, 
because there are 500 students in total.
Solving this system of equations give us 
, 
.
Since 
of the non-seniors play a musical instrument, the answer is simply of 
, which gives us 
.
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. 
