Difference between revisions of "2019 AMC 10B Problems/Problem 3"
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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? | ||
+ | |||
+ | <math>\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | <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}{10}x = 84</cmath> | ||
+ | <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{\textbf{(B) } 154}</math>. | ||
+ | |||
+ | ~IronicNinja | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 both sides by <math>10</math> gives us | ||
+ | <cmath>6x + 3y = 2340</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/J8UdaSHyWJI | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2019|ab=B|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:58, 11 January 2021
Contents
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 70% 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 .
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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