Difference between revisions of "2021 AMC 12B Problems/Problem 12"
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+ | {{duplicate|[[2021 AMC 10B Problems#Problem 19|2021 AMC 10B #19]] and [[2021 AMC 12B Problems#Problem 12|2021 AMC 12B #12]]}} | ||
+ | |||
+ | ==Problem== | ||
+ | Suppose that <math>S</math> is a finite set of positive integers. If the greatest integer in <math>S</math> is removed from <math>S</math>, then the average value (arithmetic mean) of the integers remaining is <math>32</math>. If the least integer in <math>S</math> is also removed, then the average value of the integers remaining is <math>35</math>. If the greatest integer is then returned to the set, the average value of the integers rises to <math>40.</math> The greatest integer in the original set <math>S</math> is <math>72</math> greater than the least integer in <math>S</math>. What is the average value of all the integers in the set <math>S?</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 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 | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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 | ||
+ | |||
+ | == 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== | ||
+ | {{AMC12 box|year=2021|ab=B|num-b=11|num-a=13}} | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:32, 1 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.
Contents
Problem
Suppose that is a finite set of positive integers. If the greatest integer in is removed from , then the average value (arithmetic mean) of the integers remaining is . If the least integer in is also removed, then the average value of the integers remaining is . If the greatest integer is then returned to the set, the average value of the integers rises to The greatest integer in the original set is greater than the least integer in . What is the average value of all the integers in the set
Solution 1
Let be the greatest integer, be the smallest, be the sum of the numbers in S excluding and , and be the number of elements in S.
Then,
Firstly, when the greatest integer is removed,
When the smallest integer is also removed,
When the greatest integer is added back,
We are given that
After you substitute , you have 3 equations with 3 unknowns , and .
This can be easily solved to yield , , .
average value of all integers in the set , D)
~ SoySoy4444
Solution 2
We should plug in and assume everything is true except the part. We then calculate that part and end up with . We also see with the formulas we used with the plug in that when you increase by the part decreases by . The answer is then . You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm
Solution 3
Let with We are given the following: Subtracting the third equation from the sum of the first two, we find that Furthermore, from the fourth equation, we have Combining like terms and simplifying, we have Thus, the sum of the elements in is and since there are 10 elements in the average of the elements in is
~peace09
Video Solution by OmegaLearn (System of equations)
~ 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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.