1. State the problem. Problem: Create and solve several advanced percentage problems suitable for grade 7 and show full step-by-step solutions. 2. Formulas and rules. Percentage of a whole: $\text{Percentage}=\frac{\text{part}}{\text{whole}}\times100$. Successive percentage change rule: multiply factors $1+\frac{p}{100}$ for an increase of $p$ and $1-\frac{q}{100}$ for a decrease of $q$, then subtract 1 to get the net change. Overlapping-group rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Important rules: always convert percentages to decimal factors when multiplying, simplify fractions before converting when possible, and state whether a change is an increase or decrease. 3. Problem 1: Score percentage. Problem: A student scored 78 out of 92 in a test and the class average is 85\%. Find the student's percentage score and say how it compares to the class average. Solution: Compute the raw percentage using the percentage formula. Compute $\frac{78}{92}\times100$. Simplify the fraction: $\frac{78}{92}=\frac{39}{46}$. Compute the decimal: $\frac{39}{46}\approx0.84782608696$. Convert to percent: $0.84782608696\times100\approx84.782608696\%$. Round to two decimal places: $84.78\%$. Difference from class average: $85\% - 84.782608696\% \approx0.217391304\%$. Interpretation: The student scored about $84.78\%$, which is approximately $0.22$ percentage points below the class average. 4. Problem 2: Successive percentage changes. Problem: An item's price is increased by 15\% and then discounted by 20\%. Find the net percentage change from the original price. Solution: Use multiplicative factors for successive changes. Increase factor: $1+0.15=1.15$. Decrease factor: $1-0.20=0.80$. Multiply factors: $1.15\times0.80=0.92$. Net change factor minus 1: $0.92-1=-0.08$. Convert to percent: $-0.08\times100=-8\%$. Interpretation: The final price is 8\% lower than the original price, so there is an 8\% net decrease. 5. Problem 3: Overlapping groups and totals. Problem: In a school, 40\% of students play soccer, 30\% play basketball, and 15\% play both sports. Find the percentage who play at least one of the two sports and, if there are 600 students total, find how many play neither sport. Solution: Use the union formula for overlapping groups. Compute the percent who play at least one: $40\% + 30\% - 15\% = 55\%$. Compute the number who play at least one: $0.55\times600=330$. Compute the number who play neither: $600-330=270$. Interpretation: $55\%$ of students, i.e. 330 students, play at least one of the two sports, and 270 students play neither. 6. Final answers summary. Problem 1 answer: Student score $\approx84.78\%$, about $0.22$ percentage points below average. Problem 2 answer: Net change $-8\%$, an 8\% decrease. Problem 3 answer: $55\%$ play at least one sport, 330 students play at least one, and 270 play neither.