2024 AMC 12A Problems/Problem 5

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

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

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 5

Contents

Problem

Solution

Solution 2

Solution 3

See Also