Difference between revisions of "2024 AMC 12A Problems/Problem 5"
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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=7</math>, so we choose answer choice <math>\boxed{\textbf{(D) }7}</math>. | 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=7</math>, so we choose answer choice <math>\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 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:35, 8 November 2024
Contents
[hide]Problem
A data set containing numbers, some of which are , has mean . When all the 6s are removed, the data set has mean . How many 6s were in the original data set?
Solution
Because the set has numbers and mean , the sum of the terms in the set is .
Let there be sixes in the set.
Then, the mean of the set without the sixes is . Equating this expression to and solving yields , so we choose answer choice .
Solution 2
Let be the sum of the data set without the sixes and be the number of sixes. We are given that and ; the former equation becomes and the latter . Since we want , we equate the two equations and see that .
~Technodoggo
See Also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 12 Problems and Solutions |
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