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

Revision as of 15:49, 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

Solution 2 (much more thorough version of Sol 1)

Let $S = {x_1, x_2, \dots, x_n}$ in increasing order. There are $n$ integers in this set.

The greatest integer in $S$ is $x_n$ , so we need to remove this and get the average as $32$ . There are $n-1$ total terms. Thus, $\frac{x_1+x_2+\dots+x_{n-1}}{n-1} = 32$ $x_1+x_2+\dots+x_{n-1} = 32(n-1).$

The least integer is $x_1.$ Removing it, we have $x_2+\dots+x_{n-1} = 35(n-2).$

Finally, adding back $x_n$ we have $x_2+\dots+x_n = 40(n-1).$

Since the greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ , we have $x_n = x_1 + 72.$

Substituting this into the third equation, we have $x_2 + \dots + x_{n-1} + x_1 + 72 = 40(n-1).$ Rearranging, this becomes $x_1 + \dots + x_{n-1} = 40(n-1) - 72.$

But the LHS of this equation is the same as the LHS of the first equation! Equating them, we have $40(n-1)-72 = 32(n-1)$ , which means $n = 10.$

Let's plug $n = 10$ into all of our previous equations. We have the following system: $x_1+x_2+\dots+x_9 = 288$ $x_2+\dots+x_9 = 280$ $x_2+\dots+x_{10} = 360$

We're asked to find the average of $S$ , which is $\frac{x_1+x_2+\dots+x_{10}}{10}.$ Let's focus on the numerator. We can find $x_1$ by subtracting the second equation from the first, to get $x_1 = 8.$ We can also find $x_{10}$ by subtracting the second equation from the third, to get $x_{10} = 80.$

We also know $x_2+\dots+x_9$ to equal $280$ . So when we sum up our three values, we get $x_1 + (x_2 + \dots + x_9) + x_{10} = 8 + 280 + 80 = 368.$

Therefore, the sum of the values is $368$ , and finding the average in $10$ terms we have $\frac{368}{10} = \boxed{(D)\text{ }36.8}.$

-PureSwag

@@ Line 7: / Line 7: @@
 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
-==Solution 2==
+==Solution 2 (much more thorough version of Sol 1)==
 Let <math>S = {x_1, x_2, \dots, x_n}</math> in increasing order. There are <math>n</math> integers in this set.

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

Revision as of 15:49, 11 February 2021

Problem

Solution

Solution 2 (much more thorough version of Sol 1)