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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>
  
−
==Solution==
+
==Solution 1==
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===Solution 1===
 
 
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 2==
 
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
  

Revision as of 12:09, 12 February 2021

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 great 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 $x$ 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 2

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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