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>
  
==Solution==
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==Solution 1==
  
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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<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>
  
 
<cmath>\frac{3}{5}x + 150 - \frac{3}{10}x = 234</cmath>
 
<cmath>\frac{3}{5}x + 150 - \frac{3}{10}x = 234</cmath>
 
<cmath>\frac{3}{10}x = 84</cmath>
 
<cmath>\frac{3}{10}x = 84</cmath>
<cmath>x = 84\cdot\frac{10}{3} \Rightarrow 280</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 70% of the non-seniors play a musical instrument, <math>220 \cdot \frac{7}{10} = \boxed{\textbf{(B) } 154}</math>.
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~IronicNinja
  
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>
 
  
 
==Solution 2==
 
==Solution 2==
  
Let x be the number of seniors, and y 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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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>
  
Multiplying 10 to every term gives us
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Multiplying both sides by <math>10</math> gives us
 
<cmath>6x + 3y = 2340</cmath>
 
<cmath>6x + 3y = 2340</cmath>
  
 
Also, <math>x + y = 500</math> because there are 500 students in total.
 
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) ==
  
Solving these system of equations give us <math>x = 280</math>,   <math>y = 220</math>
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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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==Video Solution==
 +
https://youtu.be/J8UdaSHyWJI
  
Since 70% of the non-seniors play a musical instrument, we simply get 70% of 220, which gives us <math>\boxed{B) 154}</math>
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}
SUB2PEWDS
 

Latest revision as of 16:49, 16 June 2020

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

Video Solution

https://youtu.be/J8UdaSHyWJI

~savannahsolver

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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

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