Question

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

$1$

$2$

$3$

$4$

infinitely many

Solution:

The six fixed numbers sum to $24.8.$ The mean $\dfrac{24.8 + x + y + z}{9}$ is an integer exactly when $x + y + z$ has fractional part $0.2.$ The fixed numbers span $[1, 7],$ so to make the range $7$ the extremes must be pushed apart by one more unit.

Checking the possibilities gives exactly three valid triples:

$(x, y, z) = (0, 5, 6.2)$ with mean $4$ and median $5;$ $(x, y, z) = (0.1, 4, 7.1)$ with mean $4$ and median $4;$ and $(x, y, z) = (6, 6.2, 8)$ with mean $5$ and median $6.$ Each has range $7$ and integer mean and median.

Thus, the correct answer is C.

2024 AMC 12B Problem 10

Solution:

Problem 10 in Other Years