Difference between revisions of "2021 AMC 10B Problems/Problem 19"

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==Problem==
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#redirect [[2021 AMC 12B Problems/Problem 12]]
Suppose that <math>S</math> is a finite set of positive integers. If the greatest integer in <math>S</math> is removed from <math>S</math>, then the average value (arithmetic mean) of the integers remaining is <math>32</math>. If the least integer is <math>S</math> is [i]also[/i] removed, then the average value of the integers remaining is <math>35</math>. If the greatest integer is then returned to the set, the average value of the integers rises of <math>40</math>. The greatest integer in the original set <math>S</math> is <math>72</math> greater than the least integer in <math>S</math>. What is the average value of all the integers in the set <math>S ?</math>
 
 
 
<math>\textbf{(A)} ~36.2 \qquad\textbf{(B)} ~36.4 \qquad\textbf{(C)} ~36.6 \qquad\textbf{(D)} ~36.8 \qquad\textbf{(E)} ~37</math>
 
 
 
==Solution==
 
Let the lowest value be L and the highest G, and let the sum be Z and the amount of numbers n. We have <math>\frac{Z-G}{n-1}=32</math>, <math>\frac{Z-L-G}{n-2}=35</math>, <math>\frac{Z-L}{n-1}=40</math>, and <math>G=L+72</math>. Clearing denominators gives <math>Z-G=32n-32</math>, <math>Z-L-G=35n-70</math>, and <math>Z-L=40n-40</math>. We use <math>G=L+72</math> to turn the first equation into <math>Z-L=32n+40</math>, which gives <math>n=10</math>. Turning the second into <math>Z-2L=35n+2</math> we see <math>L=8</math>  and <math>Z=368</math> so the average is <math>\frac{Z}{n}=\boxed{(D)36.8}</math> ~aop2014
 

Latest revision as of 20:18, 12 February 2021

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