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

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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>.
 
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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== See Also ==
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{{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 17:30, 8 November 2024

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

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 $\boxed{\textbf{(A) }7}$.

See Also

2024 AMC 12A (ProblemsAnswer KeyResources)
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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