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Difference between revisions of "2019 AMC 10B Problems/Problem 3"
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<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>
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==Solution==
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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>
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<cmath>\frac{3}{5}x + 150 - \frac{3}{10}x = 234</cmath>
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<cmath>\frac{3}{10}x = 84</cmath>
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<cmath>x = 84\cdot\frac{10}{3} \Rightarrow 280</cmath>
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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{B) 154}</math>
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iron

Revision as of 12:32, 14 February 2019

Problem 3

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 \[\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500\]

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

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

iron

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