Question 188·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

The mean is the total divided by the number of values, so A’s total is $25(8)$. Type $25\cdot8$ in Desmos; it shows $200$. That’s the fixed total for every data set that fits the question.

Write the values in order as $x_1\le\cdots\le x_{25}$. The median is the middle value. There are 12 positions on either side of position 13, so $x_{13}=7$.

B takes positions $14$ through $25$, leaving positions $1$ through $13$ outside it. So $\operatorname{sum}(B)=200-(x_1+\cdots+x_{13})$. The median stays outside B. To maximize B’s sum, minimize the sum outside B.

The first 12 values are positive integers, so each is at least $1$; position 13 is $7$. Type $12\cdot1+7$ in Desmos; it shows $19$. Repeats are allowed: the question says integers, not distinct integers, so all first 12 values can be $1$.

At least $19$ of the total $200$ must stay outside B. Type $200-19$; Desmos shows $181$. So B’s sum is at most $181$.

Put twelve $1$s in positions $1$ through $12$ and twelve $7$s in positions $13$ through $24$. Type $200-(12\cdot1+12\cdot7)$; Desmos shows $104$ for position $25$. This valid ordered list has 25 positive integers, total $200$, and median $7$. It reaches B’s sum of $181$.

B has 12 values, so its largest possible mean is $\frac{181}{12}$. Subtract A’s mean: type $\frac{181}{12}-8$. Desmos shows $7.0833\ldots$; its fraction button gives $\frac{85}{12}$. The maximum possible value of $\text{mean}(B)-\text{mean}(A)$ is $\frac{85}{12}$. Choice A.