2024 AMC 12A Problems/Problem 5

Problem 5

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 this new set is $\frac{900-6s}{20-s}$ . Equating this expression to $66$ and solving yields $s=7$ , so we choose answer choice $\boxed{\textbf{(A) }7}$ .

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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2024 AMC 12A Problems/Problem 5

Problem 5

Solution

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