Question 24·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

No Desmos needed. The symmetric pair and the bounds settle every possible list. The mean is the total divided by the count, so the $12$ values in $L$ total $12m$. Add the new values: $(m-8)+(m+8)=2m$. Their average is $m$, the same as $L$’s. Combining two groups that both average $m$ leaves $M$ with mean $m$, so statement I holds.

The given inclusive bound means every old value is at most $4$ from $m$. Its squared distance from $m$ is at most $4^2=16$. Each new value is $8$ from $m$, with squared distance $8^2=64$. So both new squared distances exceed every old one.

Standard deviation is the square root of the average squared distance from the mean. Since the mean remains $m$, the old squared distances average at most $16$, and the two new ones are each $64$. Adding values above the old average raises that average; taking its square root preserves the increase. With the mean fixed, adding values farther from it than every old value increases standard deviation. Statement II holds.

Call the old values in increasing order $x_1,x_2,\ldots,x_{12}$. Because every old value lies between $m-4$ and $m+4$, the new values go outside all the old values. The ordered list for $M$ is

The median averages the two middle entries when the count is even. In $L$, the $6$th and $7$th entries are $x_6$ and $x_7$. In $M$, the middle positions are $7$ and $8$; the new smallest value shifts $x_6$ and $x_7$ into those positions. The same two entries determine both medians, so statement III holds. All three statements must be true. Choice B.