Difference between revisions of "2024 AMC 12A Problems/Problem 5"
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Let there be <math>s</math> sixes in the set. | Let there be <math>s</math> sixes in the set. | ||
| − | Then, the mean of | + | 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>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2024|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:31, 8 November 2024
Problem
A data set containing 
numbers, some of which are 
, has mean 
. When all the 6s are removed, the data set has mean 
. How many 6s were in the original data set?
Solution
Because the set has numbers and mean , the sum of the terms in the set is 
.
Let there be 
sixes in the set.
Then, the mean of the set without the sixes is 
. Equating this expression to and solving yields 
, so we choose answer choice 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
