AMC 8 · 1999 · #16
Grade 6 rate-ratioPick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
One big question ("how many more?") splits cleanly into smaller, easier questions — tool #9. The three categories are independent: each contributes its own count of correct answers. We solve three Grade 6 "percent of a number" sub-problems, organize the results in a short list (tool #2), add them up, then compare with the passing total of 60% of 75. The final answer is one subtraction.
Find each category's correct count — take that percent of that category's total.
Solving the test "piece by piece" is much easier than handling all 75 at once. Each piece is just one decimal multiplication.
6.RP.A.3Solve An Easier Related ProblemAdd the three category counts to get the total she answered correctly: 40.
Three numbers, one sum — keeping them in a short list makes sure no category is forgotten.
4.NBT.B.4Make A Systematic ListThe passing mark is 60% of all 75 problems: 45.
Same "percent of a quantity" idea as the category sub-problems, just applied to the full test.
6.RP.A.3Solve An Easier Related ProblemSubtract actual from passing: the number of extra correct answers she needed is 5 → (B).
"How many more" is a comparison subtraction: target minus actual.
4.NBT.B.4Solve An Easier Related ProblemBreak the test into its three categories, compute each "percent of a quantity" separately, add, and compare with the passing total — a Grade 6 percent problem dressed up as an AMC 8 word problem.