AMC 10 · 2010 · #11

Grade 7 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
percentagelinear-equations-one-varinterval-arithmetic convert-to-algebra ↑ Prerequisites: percentage
📏 Medium solution 💡 3 insights
Problem
A shopper buys one item whose listed price is more than 100 dollars, using exactly one coupon. Coupon A takes 15% off the listed price, Coupon B takes 30 dollars off, and Coupon C takes 25% off only the amount by which the listed price exceeds 100 dollars. Let x be the smallest and y the largest listed price at which Coupon A saves at least as many dollars as Coupon B and at least as many dollars as Coupon C. What is y - x?

Pick an answer.

(A)
50
(B)
60
(C)
75
(D)
80
(E)
100

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The price is not given, so Tool #4 (Introduce a Variable) names it P and turns each coupon's savings into an expression in P. Tool #13 (Convert to Algebra) then rewrites the words "A saves at least as much as B" and "A saves at least as much as C" as two inequalities. Solving them pins down a range of allowed prices, and Tool #14 (Extreme Principle) reads off the two endpoints of that range as the smallest price x and largest price y.

1STEP 1

Name the price, write each saving

Let the price be P dollars, P above 100: A saves 0.15P, B saves a flat 30, C saves 0.25(P-100).

A=0.15P, B=30, C=0.25(P-100)
2STEP 2

A at least as good as B

A must match B: 0.15P ≥ 30. Dividing by 0.15, the price must be at least 200 dollars.

0.15P ≥ 30 → P ≥ 30/0.15=200
3STEP 3

A at least as good as C

A must match C too: 0.15P ≥ 0.25P-25, so 25 ≥ 0.10P and the price is at most 250 dollars.

0.15P ≥ 0.25(P-100) → 25 ≥ 0.10P → P ≤ 250
4STEP 4

Read the endpoints and subtract

Both must hold, so 200 ≤ P ≤ 250: x=200, y=250, and y-x = 50, choice (A).

200 ≤ P ≤ 250 → x=200, y=250, y-x=50 → (A)
Answer
50
Check the endpoints. At P=200: A saves 0.15 × 200 = 30, tying B's 30 dollars, while C saves only 0.25 × 100 = 25, so A wins — valid. At P=250: A saves 0.15 × 250 = 37.5 and C saves 0.25 × 150 = 37.5, a tie, while B is only 30 dollars — valid. Just outside the band A loses: at P=199, A=29.85 falls short of B's 30; at P=251, A=37.65 falls short of C's 37.75. So x=200 and y=250 are genuine boundaries and y-x=50.
💡Key takeaway

Turn each coupon into a dollar formula, write "A beats the others" as two inequalities, and the price range you get has the smallest and largest prices sitting at its two ends.

  • Name the price, write each saving
  • A at least as good as B
  • A at least as good as C
  • Read the endpoints and subtract

More like this

Same archetype — closest grade level first.