Difference between revisions of "2019 AMC 10B Problems/Problem 3"
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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>. | 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>. | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 19:56, 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, 
.
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. 
