2024 AMC 12A Problems/Problem 4

Problem

A data set containing $20$ numbers, some of which are $6$ , has mean $45$ . When all the $6$ s are removed, the data set has mean $66$ . How many $6$ s 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 1

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 $s = \boxed{\textbf{(D)}~7}$ .

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 $x=\boxed{\textbf{(D) }7}$

Video Solution by Math from my desk

https://www.youtube.com/watch?v=E_Cab6NsbUA&t=2s

Video Solution 2 (⚡️ 1 min solve ⚡️)

https://youtu.be/gGoqDf23XEk

~Education, the Study of Everything

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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