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| − | ==Problem==
| + | #redirect[[2024 AMC 10A Problems/Problem 5]] |
| − | A data set containing <math>20</math> numbers, some of which are <math>6</math>, has mean <math>45</math>. When all the <math>6</math>s are removed, the data set has mean <math>66</math>. How many <math>6</math>s were in the original data set?
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| − | <math>\textbf{(A)}~4\qquad\textbf{(B)}~5\qquad\textbf{(C)}~6\qquad\textbf{(D)}~7\qquad\textbf{(E)}~8</math>
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| − | ==Solution 1==
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| − | Because the set has <math>20</math> numbers and mean <math>45</math>, the sum of the terms in the set is <math>45\cdot 20=900</math>.
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| − | Let there be <math>s</math> sixes in the set.
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| − | Then, the mean of the set without the sixes is <math>\frac{900-6s}{20-s}</math>. Equating this expression to <math>66</math> and solving yields <math>s = \boxed{\textbf{(D)}~7}</math>.
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| − | ==Solution 2==
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| − | Let <math>S</math> be the sum of the data set without the sixes and <math>x</math> be the number of sixes. We are given that <math>\dfrac{S+6x}{20}=45</math> and <math>\dfrac S{20-x}=66</math>; the former equation becomes <math>S+6x=900</math> and the latter <math>S=1320-66x</math>. Since we want <math>x</math>, we equate the two equations and see that <math>900-6x=1320-66x\implies60x=420\implies x=\boxed{\textbf{(D) }7}</math>.
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| − | ~Technodoggo
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| − | ==Solution 3==
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| − | Suppose there are <math>x</math> sixes. Then the sum of all the numbers can be written as <math>(20-x)\cdot 45+6x</math>
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| − | Then, the mean of this set is <math>\frac{(20-x)\cdot 66+6x}{20}=45</math>. Solving this, we get <math>x=\boxed{\textbf{(D) }7}</math>
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| − | == Video Solution by Math from my desk ==
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| − | https://www.youtube.com/watch?v=E_Cab6NsbUA&t=2s
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| − | == Video Solution 2 (⚡️ 1 min solve ⚡️) ==
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| − | https://youtu.be/gGoqDf23XEk
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| − | <i>~Education, the Study of Everything</i>
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| − | ==See also==
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| − | {{AMC12 box|year=2024|ab=A|num-b=3|num-a=5}}
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| − | {{MAA Notice}}
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