2021 AMC 12B Problems/Problem 12

The following problem is from both the 2021 AMC 10B #19 and 2021 AMC 12B #12, so both problems redirect to this page.

Problem

Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then returned to the set, the average value of the integers rises to $40$ . The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ . What is the average value of all the integers in the set $S$ ?

$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$

Solution 1

Let the lowest value be $L$ and the highest $G$ , and let the sum be $Z$ and the amount of numbers $n$ . We have $\frac{Z-G}{n-1}=32$ , $\frac{Z-L-G}{n-2}=35$ , $\frac{Z-L}{n-1}=40$ , and $G=L+72$ . Clearing denominators gives $Z-G=32n-32$ , $Z-L-G=35n-70$ , and $Z-L=40n-40$ . We use $G=L+72$ to turn the first equation into $Z-L=32n+40$ . Since $Z-L=40(n-1)$ we substitute it into the equation which gives $n=10$ . Turning the second into $Z-2L=35n+2$ using $G=L+72$ we see $L=8$ and $Z=368$ so the average is $\frac{Z}{n}=\boxed{\textbf{(D) }36.8}$ ~aop2014

Solution 2

Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$ , and $k$ be the number of elements in S.

Then, $S=x+y+z$

First, when the greatest integer is removed, $\frac{S-x}{k-1}=32$

When the smallest integer is also removed, $\frac{S-x-y}{k-2}=35$

When the greatest integer is added back, $\frac{S-y}{k-1}=40$

We are given that $x=y+72$

After you substitute $x=y+72$ , you have 3 equations with 3 unknowns $S,$ , $y$ and $k$ .

$S-y-72=32k-32$

$S-2y-72=35k-70$

$S-y=40k-40$

This can be easily solved to yield $k=10$ , $y=8$ , $S=368$ .

$\therefore$ average value of all integers in the set $=S/k = 368/10 = \boxed{\textbf{(D) }36.8}$

~ SoySoy4444

Solution 3

We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$ . We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$ . The answer is then $\boxed{\textbf{(D) }36.8}$ . You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm

Solution 4

Let $S = \{a_1, a_2, a_3, \hdots, a_n\}$ with $a_1 < a_2 < a_3 < \hdots < a_n.$ We are given the following: ${\begin{cases} \sum_{i=1}^{n-1} a_i = 32(n-1) = 32n-32, \\ \sum_{i=2}^n a_i = 40(n-1) = 40n-40, \\ \sum_{i=2}^{n-1} a_i = 35(n-2) = 35n-70, \\ a_n-a_1 = 72 \implies a_1 + 72 = a_n. \end{cases}}$ Subtracting the third equation from the sum of the first two, we find that $\sum_{i=1}^n a_i = \left(32n-32\right) + \left(40n-40\right) - \left(35n-70\right) = 37n - 2.$ Furthermore, from the fourth equation, we have $\sum_{i=2}^{n} a_i - \sum_{i=1}^{n-1} a_i = \left[\left(a_1 + 72\right) + \sum_{i=2}^{n-1} a_i\right] - \left[\left(a_1\right) + \sum_{i=2}^{n-1} a_i\right] = \left(40n-40\right)-\left(32n-32\right).$ Combining like terms and simplifying, we have $72 = 8n-8 \implies 8n = 80 \implies n=10.$ Thus, the sum of the elements in $S$ is $37 \cdot 10 - 2 = 368,$ and since there are 10 elements in $S,$ the average of the elements in $S$ is $\tfrac{368}{10}=\boxed{\textbf{(D) }36.8}$

~peace09

Video Solution by OmegaLearn (System of equations)

https://youtu.be/dRdT9gzm-Pg

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

Video Solution by TheBeautyofMath

https://youtu.be/FV9AnyERgJQ?t=676

~IceMatrix

Video Solution by Interstigation

https://youtu.be/TbHluJQoy8s

~Interstigation

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 10 Problems and Solutions

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