2005 AMC 10B Problem 19

Percents are easier to handle as actual students, so pick a class size that makes every percent a whole number. $20$ students works, since each percent here is a multiple of $5\%$.

1. The missing group. $10 + 25 + 20 + 15 = 70$, so the other $30\%$ scored $95$.
2. Number of students with each score (out of $20$):

3. Median. Line up the $20$ scores from lowest to highest. The middle is between the 10th and 11th students. Counting up: students 1 to 2 have $70$, students 3 to 7 have $80$, students 8 to 11 have $85$. So the 10th and 11th both have $85$, and the median is $85$.
4. Mean. Add all the scores:
$$2(70) + 5(80) + 4(85) + 3(90) + 6(95) = 140 + 400 + 340 + 270 + 570 = 1720.$$
Divide by $20$: $1720 \div 20 = 86$.
5. Difference. $86 - 85 = 1$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

When a problem gives only percents, choosing a convenient class size turns them into real counts without changing the mean or median. The mean uses every score, weighted by how many students got it. The median is found by counting up from the bottom until you reach the middle of the class.

Work with the percents directly. The mean is $0.10(70) + 0.25(80) + 0.20(85) + 0.15(90) + 0.30(95) = 7 + 20 + 17 + 13.5 + 28.5 = 86$. For the median, add up percents from the bottom: $70$ covers $10\%$, $80$ brings the total to $35\%$, and $85$ brings it to $55\%$. The halfway point, $50\%$, is inside the $85$ group.