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Difference between revisions of "2024 AMC 10B Problems/Problem 15"
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } \text{infinitely many}</math>
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } \text{infinitely many}</math>
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==Solution==
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The sum of the six existing numbers is <math>24.8</math>. For the mean to be an integer, the sum of the remaining three numbers <math>x,y,z</math> and the remaining six must be divisible by <math>9</math>; thus <math>x+y+z</math> must equal either <math>2.2,11,2,20.2</math>. However, one of <math>x,y,z</math> must be the median since the middle four numbers in the original set are all non-integral. As a result, we can eliminate <math>2.2</math> since the sum would be too small to allow for one of <math>x,y,z</math> to be the median (notice that negative numbers are not permitted since the range would be greater than <math>7</math>). This leaves two cases.
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For the <math>x+y+z=20.2</math> case, notice that all three must be less than <math>8</math> due to the range restriction. However, also notice that <math>y</math> and <math>z</math> must be on the higher end of the range, meaning that the new median would have to fall between the new middle numbers <math>5.2</math> and <math>6.2</math>. Thus <math>x=6</math>, and we can manipulate the numbers to make the range <math>7</math> by making <math>z=8</math> and thus <math>y=6.2</math>, providing one case in <math>(6,6.2,8)</math>.
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For the <math>x+y+z=11.2</math> 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 <math>y</math>. 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 <math>3.2,5.2</math>. Then we find two triples <math>(0.1,4,7.1)</math> and <math>(0,5,6.2)</math>, resulting in <math>\boxed{\textbf{(C) }3}</math>.
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~eevee9406
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Revision as of 10:28, 14 November 2024

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

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

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

~ Pi Academy

See also

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

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