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

We can then say that $ \frac{A+S(n)}{n+1} = 32 $, $ \frac{S(n)}{n} = 35 $, and $ \frac{B+S(n)}{n+1} = 40 $.

Expanding gives us $ A+S(n) = 32n+32 $, $ S(n) = 35n $, and $ B+S(n) = 40n+40 $.

Substituting $ S(n) = 35n $ to all gives us $ A+35n=32n+32 $ and $ B+35n=40n+40 $.

Solving for $ A $ and $ B $ gives $ A=-3n+32 $ and $ B = 5n+40 $.

We now need to find $ \frac{S(n)+A+B}{n+2} $. We substitute everything to get $ \frac{35n+(-3n+32)+(5n+40)}{n+2} $, or $ \frac{37n+72}{n+2} $.

Say that the answer to this is $ Z $. Then, $ Z $ needs to be a number that makes $ n $ a positive integer. The only options that work is $\boxed{\textbf{(C) }36.6}$ and $\boxed{\textbf{(D) }36.8}$ .

However, if 36.6 is an option, we get $ n=3 $. So that means that $A$ is $23$ and $B$ is $55$, and $ S(n)=105 $. But if there is $3$ terms, then the middle number is $105$, but we said that $ B $ is the largest number in the set, so therefore our answer cannot be $\boxed{\textbf{(C) }36.6}$ and is instead $\boxed{\textbf{(D) }36.8}$ and now, we're finished!

~Pinotation

Solution 1 Alternative Ending

If A is smaller than B by 72 therefore from the equation on the top you can find out that $N = 8$ using substitution the plug it in to the equation $ \frac{37n+72}{n+2} $ then you will get that Z = $\boxed{\textbf{(D) }36.8}$ .

This was more efficient then the last part.

~LittleWavelet

~Pinotation (Grammar and Formatting)

Solution 2

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 3

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 4

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 5

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

Solution 6

Let $n$ be the number of elements in $S, m_l = 32, \Sigma_l = m_l \cdot (n-1), m_g = 40, \Sigma_g = m_g \cdot (n-1), m_{lg} = 35, \Sigma_lg = m_{lg} \cdot (n-2).$ $\Sigma_g - \Sigma_l = 72 = (n-1) \cdot (m_g - m_l) \implies n = 1 + \frac {72}{m_g - m_l} = 1 + \frac {72}{40-32}=10.$ $m = \frac {\Sigma_g + \Sigma_l - \Sigma_{lg}}{n} = \frac {40 \cdot 9 + 32 \cdot 9 - 35 \cdot 8}{10} = 36.8.$ vladimir.shelomovskii@gmail.com, vvsss

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)
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All AMC 12 Problems and Solutions

2021 AMC 10B (Problems • Answer Key • Resources)
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