2019 AMC 10B Problems/Problem 3: Difference between revisions
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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== | ==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> | <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} | <cmath>x = 84\cdot\frac{10}{3} = 280</cmath> | ||
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> | 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>. | ||
~IronicNinja | |||
==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> | 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 | 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. | ||
Solving these system of equations give us <math>x = 280</math>, <math>y = 220</math>. | |||
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>. | |||
== Solution 3 (using the answer choices) == | |||
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>. | |||
==Solution 4== | |||
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 | |||
<cmath>46.8\% \cdot 500 = 46.8 \cdot 5 = \frac{468}{2} = 234</cmath> | |||
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. | |||
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: | |||
<cmath>0.3n + 0.6s = 234</cmath> | |||
<cmath>0.7n + 0.4s = 266</cmath> | |||
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>. | |||
The answer is <math>70</math> percent of <math>220</math>. This is equal to | |||
<cmath>0.7*220 = 7*22 = 154</cmath> | |||
Therefore, the answer is <math>\boxed{\textbf{(B) } 154}</math>. | |||
~TheGoldenRetriever | |||
==Video Solution== | |||
https://youtu.be/JBZ5AF-dxOc | |||
~Education, the Study of Everything | |||
==Video Solution== | |||
https://youtu.be/J8UdaSHyWJI | |||
~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}} | ||
[[Category: Introductory Algebra Problems]] | |||
Latest revision as of 16:15, 16 October 2025
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
Thus there are
non-seniors. Since
of the non-seniors play a musical instrument,
.
~IronicNinja
Solution 2
Let
be the number of seniors, and
be the number of non-seniors. Then
Multiplying both sides by
gives us
Also,
because there are 500 students in total.
Solving these system of equations give us
,
.
Since
of the non-seniors play a musical instrument, the answer is simply
of
, which gives us
.
Solution 3 (using the answer choices)
We can clearly deduce that
of the non-seniors do play an instrument, but, since the total percentage of instrument players is
, the non-senior population is quite low. By intuition, we can therefore see that the answer is around
or
. Testing both of these gives us the answer
.
Solution 4
We know that
of the seniors play a musical instrument, and
of the non-seniors do not. In addition, we know that the number of people who do not play a musical instrument is
We can also conclude that
of the seniors do not play an instrument,
of the non seniors do play an instrument, and
people do play an instrument.
We can now set up the following equations, where
is the number of seniors and
is the number of non-seniors:
By elimination, we get
to be equal to
. This means that
.
The answer is
percent of
. This is equal to
Therefore, the answer is
.
~TheGoldenRetriever
Video Solution
~Education, the Study of Everything
Video Solution
~savannahsolver
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 | ||
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