Question 191·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

The mean is the total divided by the number of values. Since every value in $A$ increases by $4$, its mean increases by $4$ too. Add $4$ to the given mean:

So the mean of $C$ is $14$.

The added number is $14$, equal to $C$’s mean. Adding the current mean leaves a mean unchanged: the total gains one mean’s worth as the count gains one value. So $\operatorname{mean}(B)=14$, and statement I is true.

To disprove a claim that must be true, one valid counterexample is enough. The range is the largest value minus the smallest. Choose $4$ as the smallest value, so $16=4+12$ is the largest. Five values with mean $10$ must total $5\times10=50$. Choose two $8$s to keep the middle low, and use $14$ for the remaining value so the total is $50$. Try $A=[4,8,8,14,16]$; type the list, then $\operatorname{mean}(A)$ and $\max(A)-\min(A)$. Desmos shows $10$ and $12$, so this list meets both conditions.

Type $C=A+4$ to apply the same shift to every value in your test list. Desmos displays $[8,12,12,18,20]$. The extra $14$ belongs between the last $12$ and $18$ when you put the new list in order.

Type $B=[8,12,12,14,18,20]$ and $\operatorname{median}(B)$. Desmos shows $13$. With six values, the median averages the two middle values, here $12$ and $14$. Adding the mean doesn’t guarantee it becomes the median, so statement II does not have to be true.

Let $S$ be the sum of the five squared distances from $A$’s mean. The standard deviation is the square root of the average of those squares. Shifting every value and its mean by $4$ leaves each distance unchanged, so $C$ still has sum $S$. The added $14$ is zero distance from $B$’s mean, so the sum stays $S$ while the count grows from $5$ to $6$:

The range is $12$, so the values aren’t all equal and $S>0$; the decrease is strict. Adding a value at the mean keeps the mean but lowers a nonzero standard deviation. Statements I and III must be true, but II needn’t be. Choice B.