AMC 8 · 2013 · #12
Grade 6 arithmeticrate-ratioPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The three pairs have three different discount rules, so the cleanest path is Tool #7 (Identify Subproblems): compute the price of each pair on its own, then add them up, then compare to 150. No algebra is needed — each subproblem is a one-line decimal or fraction calculation. Tool #3 (Eliminate Possibilities) is the natural verification: once we get 30%, we cross-check against the five multiple-choice values and confirm only (B) matches the savings\frac{45}{150}$.
Subproblem 1 — pair 1 carries no discount, so it stays at the full $50.
Naming the first subproblem and stating its answer is the Tool #7 move — break a multi-part question into one-line pieces.
4.OA.A.3Identify SubproblemsSubproblem 2 — a 40% discount means paying only 60%, so pair 2 costs $30.
Flipping "40% off" into "60% of the price" is the standard shortcut for percent discounts.
5.NBT.B.7Identify SubproblemsSubproblem 3 — "half price" is a 50% discount, so pair 3 costs $25.
Taking half of a quantity is a Grade 4 fraction-of-a-whole calculation.
4.NF.B.4Identify SubproblemsAdd the three pair prices to get what Javier actually paid: $105.
After solving each subproblem separately, addition glues the pieces back into the full answer.
4.OA.A.3Identify SubproblemsSavings is the regular total minus what he paid: $45.
Savings = (what it would have cost) - (what it did cost). A subtraction, by definition.
4.OA.A.3Identify SubproblemsDivide the savings by the \frac{45}{150}$ = 30%, choice (B).
Percent saved is a part-of-whole ratio expressed per 100 — the core Grade 6 percent skill. Then Tool #3 confirms only choice (B) matches.
6.RP.A.3Eliminate PossibilitiesThis AMC 8 problem really is just "price of each pair, add them up, then turn the savings into a percent" — the Grade 6 percent step is the only new idea, and the rest is patient subproblem work.