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Difference between revisions of "2024 AMC 12A Problems/Problem 5"

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==Problem 5==
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==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.
  
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Then, the mean of this new set 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{(A) }7}</math>.
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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>.
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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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== See Also ==
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{{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 15:21, 9 November 2024

Problem

A data set containing $20$ numbers, some of which are $6$, has mean $45$. When all the 6s are removed, the data set has mean $66$. How many 6s were in the original data set?

$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

Solution

Solution

Because the set has $20$ numbers and mean $45$, the sum of the terms in the set is $45\cdot 20=900$.

Let there be $s$ sixes in the set.

Then, the mean of the set without the sixes is $\frac{900-6s}{20-s}$. Equating this expression to $66$ and solving yields $s=7$, so we choose answer choice $\boxed{\textbf{(D) }7}$.

Solution 2

Let $S$ be the sum of the data set without the sixes and $x$ be the number of sixes. We are given that $\dfrac{S+6x}{20}=45$ and $\dfrac S{20-x}=66$; the former equation becomes $S+6x=900$ and the latter $S=1320-66x$. Since we want $x$, we equate the two equations and see that $900-6x=1320-66x\implies60x=420\implies x=\boxed{\textbf{(D) }7}$.

~Technodoggo

Solution 3

Suppose there are $x$ sixes. Then the sum of all the numbers can be written as $(20-x)\cdot 45+6x$

Then, the mean of this set is $\frac{(20-x)\cdot 66+6x}{20}=45$. Solving this, we get $x=\boxed{\textbf{(D) }7}$

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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