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Difference between revisions of "2021 AMC 12B Problems/Problem 12"

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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>
 
<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 1==
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==Solution 1 (Thorough)==
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Let the lowest value be <math>L</math> and the highest <math>G</math>, and let the sum be <math>Z</math> and the amount of numbers <math>n</math>. 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>. Since <math>Z-L=40(n-1)</math> we substitute it into the equation which gives <math>n=10</math>. Turning the second into <math>Z-2L=35n+2</math> using <math>G=L+72</math> we see <math>L=8</math> and <math>Z=368</math> so the average is <math>\frac{Z}{n}=\boxed{\textbf{(D) }36.8}</math> ~aop2014
+
 
 +
Let \( A \) denote the smallest positive integer in the set, \( B \) denote the largest, and \( n \) be the other numbers in the set, with \( S(n) \) being their sum.
 +
 
 +
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 \( B+S(n) = 40n+40 \).  
 +
 
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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 <math>\boxed{\textbf{(C) }36.6}</math> and <math>\boxed{\textbf{(D) }36.8}</math>. However, if 36.6 is an option, we get \( n=3 \). So that means that one number is \(23\) and the other 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 <math>\boxed{\textbf{(C) }36.6}</math> and is instead \boxed{\textbf{(D) }36.8}<math> and now, we're finished!
 +
 
 +
~Pinotation
  
 
==Solution 2==
 
==Solution 2==
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Let <math>x</math> be the greatest integer, <math>y</math> be the smallest, <math>z</math> be the sum of the numbers in S excluding <math>x</math> and <math>y</math>, and <math>k</math> be the number of elements in S.
+
Let the lowest value be </math>L<math> and the highest </math>G<math>, and let the sum be </math>Z<math> and the amount of numbers </math>n<math>. 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>. Since </math>Z-L=40(n-1)<math> we substitute it into the equation which gives </math>n=10<math>. Turning the second into </math>Z-2L=35n+2<math> using </math>G=L+72<math> we see </math>L=8<math> and </math>Z=368<math> so the average is </math>\frac{Z}{n}=\boxed{\textbf{(D) }36.8}<math> ~aop2014
 +
 
 +
==Solution 3==
 +
Let </math>x<math> be the greatest integer, </math>y<math> be the smallest, </math>z<math> be the sum of the numbers in S excluding </math>x<math> and </math>y<math>, and </math>k<math> be the number of elements in S.
  
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Then, <math>S=x+y+z</math>
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Then, </math>S=x+y+z<math>
  
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First, when the greatest integer is removed, <math>\frac{S-x}{k-1}=32</math>
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First, when the greatest integer is removed, </math>\frac{S-x}{k-1}=32<math>
  
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When the smallest integer is also removed, <math>\frac{S-x-y}{k-2}=35</math>
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When the smallest integer is also removed, </math>\frac{S-x-y}{k-2}=35<math>
  
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When the greatest integer is added back, <math>\frac{S-y}{k-1}=40</math>
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When the greatest integer is added back, </math>\frac{S-y}{k-1}=40<math>
  
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We are given that <math>x=y+72</math>
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We are given that </math>x=y+72<math>
  
  
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After you substitute <math>x=y+72</math>, you have 3 equations with 3 unknowns <math>S,</math>, <math>y</math> and <math>k</math>.
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After you substitute </math>x=y+72<math>, you have 3 equations with 3 unknowns </math>S,<math>, </math>y<math> and </math>k<math>.
  
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<math>S-y-72=32k-32</math>
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</math>S-y-72=32k-32<math>
  
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<math>S-2y-72=35k-70</math>
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</math>S-2y-72=35k-70<math>
  
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<math>S-y=40k-40</math>
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</math>S-y=40k-40<math>
  
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This can be easily solved to yield <math>k=10</math>, <math>y=8</math>, <math>S=368</math>.
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This can be easily solved to yield </math>k=10<math>, </math>y=8<math>, </math>S=368<math>.
  
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<math>\therefore</math> average value of all integers in the set <math>=S/k = 368/10 = \boxed{\textbf{(D) }36.8}</math>
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</math>\therefore<math> average value of all integers in the set </math>=S/k = 368/10 = \boxed{\textbf{(D) }36.8}<math>
  
 
~ SoySoy4444
 
~ SoySoy4444
  
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==Solution 3==
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==Solution 4==
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We should plug in <math>36.2</math> and assume everything is true except the <math>35</math> part. We then calculate that part and end up with <math>35.75</math>. We also see with the formulas we used with the plug in that when you increase by <math>0.2</math> the <math>35.75</math> part decreases by <math>0.25</math>. The answer is then <math>\boxed{\textbf{(D) }36.8}</math>. You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm
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We should plug in </math>36.2<math> and assume everything is true except the </math>35<math> part. We then calculate that part and end up with </math>35.75<math>. We also see with the formulas we used with the plug in that when you increase by </math>0.2<math> the </math>35.75<math> part decreases by </math>0.25<math>. The answer is then </math>\boxed{\textbf{(D) }36.8}<math>. You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm
  
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==Solution 4==
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==Solution 5==
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Let <math>S = \{a_1, a_2, a_3, \hdots, a_n\}</math> with <math>a_1 < a_2 < a_3 < \hdots < a_n.</math> We are given the following: <cmath>{\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}}</cmath> Subtracting the third equation from the sum of the first two, we find that <cmath>\sum_{i=1}^n a_i = \left(32n-32\right) + \left(40n-40\right) - \left(35n-70\right) = 37n - 2.</cmath> Furthermore, from the fourth equation, we have <cmath>\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).</cmath> Combining like terms and simplifying, we have <cmath>72 = 8n-8 \implies 8n = 80 \implies n=10.</cmath> Thus, the sum of the elements in <math>S</math> is <math>37 \cdot 10 - 2 = 368,</math> and since there are 10 elements in <math>S,</math> the average of the elements in <math>S</math> is <math>\tfrac{368}{10}=\boxed{\textbf{(D) }36.8}</math>
+
Let </math>S = \{a_1, a_2, a_3, \hdots, a_n\}<math> with </math>a_1 < a_2 < a_3 < \hdots < a_n.<math> We are given the following: <cmath>{\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}}</cmath> Subtracting the third equation from the sum of the first two, we find that <cmath>\sum_{i=1}^n a_i = \left(32n-32\right) + \left(40n-40\right) - \left(35n-70\right) = 37n - 2.</cmath> Furthermore, from the fourth equation, we have <cmath>\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).</cmath> Combining like terms and simplifying, we have <cmath>72 = 8n-8 \implies 8n = 80 \implies n=10.</cmath> Thus, the sum of the elements in </math>S<math> is </math>37 \cdot 10 - 2 = 368,<math> and since there are 10 elements in </math>S,<math> the average of the elements in </math>S<math> is </math>\tfrac{368}{10}=\boxed{\textbf{(D) }36.8}<math>
  
 
~peace09
 
~peace09
  
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==Solution 5==
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==Solution 6==
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Let <math>n</math> be the number of elements in <math>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).</math>
+
Let </math>n<math> be the number of elements in </math>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).$
 
<cmath>\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.</cmath>
 
<cmath>\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.</cmath>
 
<cmath>m = \frac {\Sigma_g + \Sigma_l - \Sigma_{lg}}{n} = \frac {40 \cdot 9 + 32 \cdot 9 - 35 \cdot 8}{10} = 36.8.</cmath>
 
<cmath>m = \frac {\Sigma_g + \Sigma_l - \Sigma_{lg}}{n} = \frac {40 \cdot 9 + 32 \cdot 9 - 35 \cdot 8}{10} = 36.8.</cmath>

Revision as of 18:33, 7 September 2025

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 (Thorough)

Let \( A \) denote the smallest positive integer in the set, \( B \) denote the largest, and \( n \) be the other numbers in the set, with \( S(n) \) being their sum.

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 one number is \(23\) and the other 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 2== Let the lowest value be$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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,$ (Error compiling LaTeX. Unknown error_msg)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$$ (Error compiling LaTeX. Unknown error_msg)S-2y-72=35k-70$$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)S = \{a_1, a_2, a_3, \hdots, a_n\}$with$a_1 < a_2 < a_3 < \hdots < a_n.$We are given the following: <cmath>{\begin{cases} \sum_{i=1}^{n-1} a_i = 32(n-1) = 32n-32, \\ \sum_{i=2}^n a_i = 40(n-1) = 40n-40, \\ \sum_{i=2}^{n-1} a_i = 35(n-2) = 35n-70, \\ a_n-a_1 = 72 \implies a_1 + 72 = a_n. \end{cases}}</cmath> Subtracting the third equation from the sum of the first two, we find that <cmath>\sum_{i=1}^n a_i = \left(32n-32\right) + \left(40n-40\right) - \left(35n-70\right) = 37n - 2.</cmath> Furthermore, from the fourth equation, we have <cmath>\sum_{i=2}^{n} a_i - \sum_{i=1}^{n-1} a_i = \left[\left(a_1 + 72\right) + \sum_{i=2}^{n-1} a_i\right] - \left[\left(a_1\right) + \sum_{i=2}^{n-1} a_i\right] = \left(40n-40\right)-\left(32n-32\right).</cmath> Combining like terms and simplifying, we have <cmath>72 = 8n-8 \implies 8n = 80 \implies n=10.</cmath> Thus, the sum of the elements in$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$ (Error compiling LaTeX. Unknown error_msg)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

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 10 Problems and Solutions

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