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| − | ==Problem==
| + | #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>
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| − | <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>
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| − | ==Solution==
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| − | 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
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| − | ==Solution 2 (much more thorough version of Sol 1)==
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| − | Let <math>S = {x_1, x_2, \dots, x_n}</math> in increasing order. There are <math>n</math> integers in this set.
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| − | The greatest integer in <math>S</math> is <math>x_n</math>, so we need to remove this and get the average as <math>32</math>. There are <math>n-1</math> total terms. Thus, <cmath>\frac{x_1+x_2+\dots+x_{n-1}}{n-1} = 32</cmath><cmath>x_1+x_2+\dots+x_{n-1} = 32(n-1).</cmath>
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| − | The least integer is <math>x_1.</math> Removing it, we have <cmath>x_2+\dots+x_{n-1} = 35(n-2).</cmath>
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| − | Finally, adding back <math>x_n</math> we have <cmath>x_2+\dots+x_n = 40(n-1).</cmath>
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| − | Since the greatest integer in the original set <math>S</math> is <math>72</math> greater than the least integer in <math>S</math>, we have <cmath>x_n = x_1 + 72.</cmath>
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| − | Substituting this into the third equation, we have <math>x_2 + \dots + x_{n-1} + x_1 + 72 = 40(n-1).</math> Rearranging, this becomes <math>x_1 + \dots + x_{n-1} = 40(n-1) - 72.</math>
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| − | But the LHS of this equation is the same as the LHS of the first equation! Equating them, we have <math>40(n-1)-72 = 32(n-1)</math>, which means <math>n = 10.</math>
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| − | Let's plug <math>n = 10</math> into all of our previous equations. We have the following system:
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| − | <cmath>x_1+x_2+\dots+x_9 = 288</cmath><cmath>x_2+\dots+x_9 = 280</cmath><cmath>x_2+\dots+x_{10} = 360</cmath>
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| − | We're asked to find the average of <math>S</math>, which is <math>\frac{x_1+x_2+\dots+x_{10}}{10}.</math> Let's focus on the numerator. We can find <math>x_1</math> by subtracting the second equation from the first, to get <math>x_1 = 8.</math> We can also find <math>x_{10}</math> by subtracting the second equation from the third, to get <math>x_{10} = 80.</math>
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| − | We also know <math>x_2+\dots+x_9</math> to equal <math>280</math>. So when we sum up our three values, we get <cmath>x_1 + (x_2 + \dots + x_9) + x_{10} = 8 + 280 + 80 = 368.</cmath>
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| − | Therefore, the sum of the values is <math>368</math>, and finding the average in <math>10</math> terms we have <cmath>\frac{368}{10} = \boxed{(D)\text{ }36.8}.</cmath>
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| − | -PureSwag
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| − | {{AMC10 box|year=2021|ab=B|num-b=18|num-a=20}}
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