Difference between revisions of "2019 AMC 10B Problems/Problem 3"

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

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

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

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 10 to every term gives us $$6x + 3y = 2340$$

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

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

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