Competition · AMC preparation · step 4 of 4
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 savings45 / $150.
Price the first pair
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 SubproblemsPrice the second pair
Subproblem 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 SubproblemsPrice the third pair
Subproblem 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 prices
Add 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 SubproblemsSubtract to find the savings
Savings 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 SubproblemsTurn the savings into a percent
Divide the savings by the $150 regular total: = 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.
The percent of the 150-dollar regular price that Javier saved is how big his 45-dollar savings is compared with that 150 dollars — a comparison that reduces to 3 dollars saved for every 10 dollars of regular price.
▸ Why?
The word 'percent' means 'per hundred', so asking what percent of the 150-dollar regular price Javier saved is exactly asking how many dollars, out of every hundred dollars of that regular price, his 45-dollar savings makes up — that per-hundred count of saved dollars is what the phrase 'percent saved' means, which is why this step is a comparison of the 45 against the 150.
▸ Why?
Both amounts are built from equal 15-dollar blocks — 45 dollars is 3 blocks and 150 dollars is 10 blocks — so comparing 45 to 150 is the same as comparing 3 to 10.
▸ Why?
45 dollars is exactly 3 equal groups of 15 dollars, and 150 dollars is exactly 10 equal groups of 15 dollars, so each amount is just a count of 15-dollar blocks.
▸ Why?
Re-measuring both the savings and the total in 15-dollar blocks instead of single dollars scales the top and bottom of the comparison by the same 15 — that is multiplying the comparison by 15/15, which equals 1 — and multiplying by one leaves the value unchanged.
This 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.
- Price the first pair
- Price the second pair
- Price the third pair
- Add the three prices
- Subtract to find the savings
- Turn the savings into a percent
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