Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 10B Problems/Problem 19"

(Created page with "wat a waste of time")
 
Line 1: Line 1:
−
wat a waste of time
+
==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 is <math>S</math> is [i]also[/i] 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 of <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==
 +
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

Revision as of 14:09, 11 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 is $S$ is [i]also[/i] 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 of $40$. 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_19&oldid=144889"