2024 AMC 10B Problems/Problem 15

Problem

A list of $9$ real numbers consists of $1$ , $2.2$ , $3.2$ , $5.2$ , $6.2$ , $7$ , as well as $x$ , $y$ , and $z$ with $x$ $\le$ $y$ $\le$ $z$ . The range of the list is $7$ , and the mean and the median are both positive integers. How many ordered triples ( $x$ , $y$ , $z$ ) are possible?

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } \text{infinitely many}$

Solution

The sum of the six existing numbers is $24.8$ . For the mean to be an integer, the sum of the remaining three numbers $x,y,z$ and the remaining six must be divisible by $9$ ; thus $x+y+z$ must equal either $2.2,11,2,20.2$ . However, one of $x,y,z$ must be the median since the middle four numbers in the original set are all non-integral. As a result, we can eliminate $2.2$ since the sum would be too small to allow for one of $x,y,z$ to be the median (notice that negative numbers are not permitted since the range would be greater than $7$ ). This leaves two cases.

For the $x+y+z=20.2$ case, notice that all three must be less than $8$ due to the range restriction. However, also notice that $y$ and $z$ must be on the higher end of the range, meaning that the new median would have to fall between the new middle numbers $5.2$ and $6.2$ . Thus $x=6$ , and we can manipulate the numbers to make the range $7$ by making $z=8$ and thus $y=6.2$ , providing one case in $(6,6.2,8)$ .

For the $x+y+z=11.2$ case, we note that the average of the three would fall near the median, signaling that one of the numbers would be the median, one would fall lesser than the median, and one would fall greater than the median. Thus we let the median be $y$ . Since adding a number to each side of the set would not change the median, we know that the median must fall between the two middle numbers $3.2,5.2$ . Then we find two triples $(0.1,4,7.1)$ and $(0,5,6.2)$ , resulting in $\boxed{\textbf{(C) }3}$ .

~eevee9406

Solution 2

We can start doing casework on the numbers. Notice that the median can either be four, five or six because there has to be one or two numbers on one side. We can start bashing everything out, starting from letting $x=a, y=4, z=7+a$ . Now plugging into the sequence, we get that $a, 1, 2.2, 3.2, 4, 5.2, 6.2, 7, 7+a$ . Notice that the sum of the numbers (not including $x, y, z$ is $24.8$ . To make this mean an integer, $x+y+z$ must have a decimal $0.2$ when added up. Now, plugging in our values into the mean gets us $35.8+2a=9k$ for some integer $k$ . Notice that $0\leq a\leq 1$ , so the only value for $a$ is $0.1$ , so we found our first ordered triplet, $(0.1, 4, 7.1)$ . Let’s make the median $5$ now. Then we have to put $x$ and one side and $z$ on the other to “balance” the numbers, or make it so that there are 4 numbers on each side of $5$ . Now, we can see that $x$ is $0$ , and solving for $z$ , we get that $z=6.2$ . We get our second triple is $(0, 5, 6.2)$ . Now coming to our third case scenario, we use $6$ as the median. Clearly it means that $x$ has to be the median, so that $y$ and $z$ can be on the same side as $x$ . $1, 2.2, 3.2, 5.2, 6, 6.2, y, 7, z$ We will use this diagram. Notice that $z$ has to be $8$ , and solving for $y$ yields $y=6.2$ , so we found our third and last triplet, $(6, 6.2, 8)$ , meaning that $\boxed{C}$ is the right answer.

~EaZ_Shadow

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

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2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2024 AMC 10B Problems/Problem 15

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Problem

Solution

Solution 2

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

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