Question 39·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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Data set A consists of 15 positive integers. The mean of data set A is 8, and the median of data set A is 7. Data set B consists of the 5 largest integers from data set A.

Which choice is the maximum possible value of the mean of data set B?

A fixed mean gives you the total, while the median fixes the middle position in an ordered list. To make the greatest values have the largest possible mean, minimize the sum of all the other values, then subtract from the total. Build a complete list to check that the limit is reachable. Don't treat the integers as distinct unless the problem says they are.

Hints

Hint 1
The mean is the total divided by the number of values. Reverse that calculation to find the fixed total before choosing any integers.
Hint 2
The median of an odd-length ordered list is its middle value. With fifteen entries, which position is the middle, and what does that require of the entries to its right?
Hint 3
The greatest five get the most when the other ten use as little of the total as possible. A positive integer can be $1$ , and repeats are allowed. How small can those ten values be?

Step-by-step

Minimize the other ten values
Step 1
Find the fixed total
The mean is the total divided by the number of values. So A's total is $15(8)$ . Type $15\cdot8$ in Desmos; it shows $120$ .
Step 2
Locate the median
Put the values in order and call them $x_1\le x_2\le\cdots\le x_{15}$ . The median is the middle value. Seven values sit on each side of position $8$ , so $x_8=7$ .
Step 3
Separate B from the other values
B contains positions $11$ through $15$ , the five largest. The first ten values and B share the fixed total:
$\underbrace{x_1+\cdots+x_{10}}_{\text{other ten}}+\underbrace{x_{11}+\cdots+x_{15}}_{\text{B}}=120.$
To maximize B's sum, minimize the sum of the other ten.
Step 4
Make the first ten as small as allowed
Positions $1$ through $7$ can each be $1$ , the smallest positive integer. Position $8$ is $7$ , and positions $9$ and $10$ must be at least $7$ because the list is ordered. Repeats are allowed: the problem doesn't require distinct integers. Type $7(1)+3(7)$ in Desmos; it shows $28$ , so
$x_1+\cdots+x_{10}\ge 7(1)+3(7)=28.$
Step 5
Bound B's sum
Subtract the least possible sum of the first ten from the fixed total. Type $120-28$ in Desmos; it shows $92$ . So B can add up to at most $92$ .
Step 6
Build a list that reaches the bound
Keep seven $1$ s and seven $7$ s, then put everything left in the last position. Type $120-7(1)-7(7)$ ; Desmos shows $64$ . The ordered list of seven $1$ s, seven $7$ s, and $64$ has $15$ positive integers. Its eighth value is $7$ , and its total is $120$ because $64$ is exactly what remained.
Step 7
Find the mean of the five largest
B contains four $7$ s and $64$ . Its mean is their sum divided by $5$ . Type $\frac{4(7)+64}{5}$ ; Desmos shows $18.4$ , or $\frac{92}{5}$ with the fraction button. This list reaches the bound, so the maximum possible mean of B is $\frac{92}{5}$ . Choice C.

Question 39·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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