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Difference between revisions of "2019 AMC 10B Problems/Problem 3"
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Oh honey, if I get restless
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==Problem==
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In a high school with <math>500</math> students, <math>40\%</math> of the seniors play a musical instrument, while <math>30\%</math> of the non-seniors do not play a musical instrument. In all, <math>46.8\%</math> of the students do not play a musical instrument. How many non-seniors play a musical instrument?
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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>
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==Solution 1==
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<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>
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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} = 280</cmath>
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Thus there are <math>500-x = 220</math> non-seniors. Since <math>70\%</math> of the non-seniors play a musical instrument, <math>220 \cdot \frac{7}{10} = \boxed{\textbf{(B) } 154}</math>.
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~IronicNinja
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==Solution 2==
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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>
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Multiplying both sides by <math>10</math> gives us
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<cmath>6x + 3y = 2340</cmath>
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Also, <math>x + y = 500</math> because there are 500 students in total.
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Solving these system of equations give us <math>x = 280</math>,  <math>y = 220</math>.
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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>.
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== Solution 3 (using the answer choices) ==
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We can clearly deduce that <math>70\%</math> of the non-seniors do play an instrument, but, since the total percentage of instrument players is <math>46.8\%</math>, the non-senior population is quite low. By intuition, we can therefore see that the answer is around <math>\text{B}</math> or <math>\text{C}</math>. Testing both of these gives us the answer <math>\boxed{\textbf{(B) } 154}</math>.
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==Solution 4==
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We know that <math>40\%</math> of the seniors play a musical instrument, and <math>30\%</math> of the non-seniors do not. In addition, we know that the number of people who do not play a musical instrument is
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<cmath>46.8\% \cdot 500 = 46.8 \cdot 5 = \frac{468}{2} = 234</cmath>
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We can also conclude that <math>60\%</math> of the seniors do not play an instrument, <math>70\%</math> of the non seniors do play an instrument, and <math>500-234 = 266</math> people do play an instrument.
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We can now set up the following equations, where <math>s</math> is the number of seniors and <math>n</math> is the number of non-seniors:
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<cmath>0.3n + 0.6s = 234</cmath>
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<cmath>0.7n + 0.4s = 266</cmath>
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By elimination, we get <math>1.5n</math> to be equal to <math>330</math>. This means that <math>n = \frac{330}{1.5} = 220</math>.
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The answer is <math>70</math> percent of <math>220</math>. This is equal to
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<cmath>0.7*220 = 7*22 = 154</cmath>
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Therefore, the answer is <math>\boxed{\textbf{(B) } 154}</math>.
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~TheGoldenRetriever
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==Video Solution==
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https://youtu.be/JBZ5AF-dxOc
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/J8UdaSHyWJI
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~savannahsolver
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==See Also==
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{{AMC10 box|year=2019|ab=B|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 08:59, 16 July 2024

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 1

$60\%$ 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, $220 \cdot \frac{7}{10} = \boxed{\textbf{(B) } 154}$.

~IronicNinja

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 these 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 $\boxed{\textbf{(B) } 154}$.

Solution 3 (using the answer choices)

We can clearly deduce that $70\%$ of the non-seniors do play an instrument, but, since the total percentage of instrument players is $46.8\%$, the non-senior population is quite low. By intuition, we can therefore see that the answer is around $\text{B}$ or $\text{C}$. Testing both of these gives us the answer $\boxed{\textbf{(B) } 154}$.

Solution 4

We know that $40\%$ of the seniors play a musical instrument, and $30\%$ of the non-seniors do not. In addition, we know that the number of people who do not play a musical instrument is \[46.8\% \cdot 500 = 46.8 \cdot 5 = \frac{468}{2} = 234\] We can also conclude that $60\%$ of the seniors do not play an instrument, $70\%$ of the non seniors do play an instrument, and $500-234 = 266$ people do play an instrument.

We can now set up the following equations, where $s$ is the number of seniors and $n$ is the number of non-seniors: \[0.3n + 0.6s = 234\] \[0.7n + 0.4s = 266\] By elimination, we get $1.5n$ to be equal to $330$. This means that $n = \frac{330}{1.5} = 220$. The answer is $70$ percent of $220$. This is equal to \[0.7*220 = 7*22 = 154\] Therefore, the answer is $\boxed{\textbf{(B) } 154}$.

~TheGoldenRetriever

Video Solution

https://youtu.be/JBZ5AF-dxOc

~Education, the Study of Everything

Video Solution

https://youtu.be/J8UdaSHyWJI

~savannahsolver

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
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Problem 4
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All AMC 10 Problems and Solutions

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