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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==
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==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 <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
 
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 1==
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==Solution 2==
 
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 <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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~ SoySoy4444
 
~ SoySoy4444
  
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==Solution 2==
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==Solution 3==
 
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{(D) 36.8}</math>. You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm
 
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{(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 3==
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==Solution 4==
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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>
  

Revision as of 20:45, 19 June 2021

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 $Z-L=32n+40$, which gives $n=10$. Turning the second into $Z-2L=35n+2$ we see $L=8$ and $Z=368$ so the average is $\frac{Z}{n}=\boxed{(D)36.8}$ ~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, $S=x+y+z$

Firstly, 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 $=S/k = 368/10 = 36.8$, D)

~ 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 $\boxed{(D) 36.8}$. 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

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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