AMC 10 · 2014 · #8
Grade 7 algebraPick an answer.
Let the listed price be P (Tool #4). One variable lets us write all three reductions as expressions instead of testing five prices by hand five times over. Then Tool #13 turns the phrase "Coupon 1 gives a greater reduction" into two inequalities — Coupon 1 vs Coupon 2, and Coupon 1 vs Coupon 3 — that together pin P to a band. Finally Tool #3 checks which of the five listed prices lands inside that band, since only a multiple-choice value can be the answer.
Write the three reductions
Each discount is one short formula.
Naming the price P once lets every coupon speak the same language, so they can be compared directly.
7.RP.A.3Introduce A VariableCoupon 1 must beat Coupon 2
The first comparison gives a floor.
A fixed 200.
A fixed discount only loses to a percentage discount once the price climbs past a certain point.
▸ Why?
A percent is a count out of a hundred of the price, so it grows as the price grows while a fixed amount does not.
▸ Why?
Because one grows and the other stays put, they cross exactly once and the order never reverses again.
Coupon 1 must beat Coupon 3
The second gives a ceiling.
Coupon 3 grows faster than Coupon 1 as the price rises, so Coupon 1 only wins while the price stays below $ 225.
7.EE.B.4Convert To AlgebraFind the price in the band
Only one listed price falls in that band, choice (C).
The two inequalities carve out one narrow window, and just one listed price sits inside it.
6.EE.B.5Eliminate PossibilitiesWrite each coupon's savings in terms of the price, then two inequalities squeeze the price into 200 ＜ P ＜ 225 — and only $ 219.95 fits.
- Write the three reductions
- Coupon 1 must beat Coupon 2
- Coupon 1 must beat Coupon 3
- Find the price in the band