Difference between revisions of "2024 AMC 12A Problems/Problem 5"
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| − | ==Problem | + | ==Problem== |
A data set containing <math>20</math> numbers, some of which are <math>6</math>, has mean <math>45</math>. When all the 6s are removed, the data set has mean <math>66</math>. How many 6s were in the original data set? | A data set containing <math>20</math> numbers, some of which are <math>6</math>, has mean <math>45</math>. When all the 6s are removed, the data set has mean <math>66</math>. How many 6s were in the original data set? | ||
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Let there be <math>s</math> sixes in the set. | Let there be <math>s</math> sixes in the set. | ||
| − | Then, the mean of | + | 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 | ||
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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> | ||
== 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}} | ||
Latest revision as of 15:21, 9 November 2024
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
Solution 3
Suppose there are sixes. Then the sum of all the numbers can be written as 
Then, the mean of this set is 
. Solving this, we get 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
