AMC 8 · 2002 · #14
Grade 7 rate-ratioPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The original price is never given, so Tool #4 (Introduce a Variable) — let the original price be P — turns the abstract "percent of percent" question into clean arithmetic in P. Tool #7 (Identify Subproblems) splits the chain of discounts into two clear steps: first apply the 30% discount to get the sale price, then apply the 20% discount to that sale price. Comparing the final price to P at the end gives the total percent off.
Let the original price be P, so we can track what fraction of P is left at each step.
Naming the unknown original price is the Tool #4 move and matches the Grade 6 idea of writing an expression with a letter that stands for a number.
6.EE.A.2Use Matrix LogicA 30% discount leaves 70% of the price, so the sale price is 0.7P.
Treating a 30% discount as multiplying by 0.7 is the Grade 6 percent-of-a-quantity move.
6.RP.A.3Identify SubproblemsTaking 20% off keeps 80% of the sale price, so the final price is 0.7P × 0.8 = 0.56P.
Stacking a second percent discount on an already-discounted price is a Grade 7 multi-step percent-change problem.
7.RP.A.3Identify SubproblemsSince 0.56P is 56% of P, the customer paid 56% and saved 44% off the original.
Subtracting the "percent paid" from 100% to get the "percent off" is the standard Grade 7 percent-change wrap-up.
7.RP.A.3Identify SubproblemsStacked discounts multiply, not add — once you write the original price as P, this AMC 8 question is just Grade 7 percent reasoning.