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2010 AMC 10B Problem 11

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2010 AMC 10B Problem 11AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 3 of 5 · about 4 min · position P11-15

A shopper plans to purchase an item that has a listed price greater than $100\$100 and can use any one of the three coupons. Coupon A gives 15%15\% off the listed price, Coupon B gives $30\$30 off the listed price, and Coupon C gives 25%25\% off the amount by which the listed price exceeds $100\$100 .
Let xx and yy be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or Coupon C. What is y−xy - x ?

  1. A)

    5050

  2. B)

    6060

  3. C)

    7575

  4. D)

    8080

  5. E)

    100100

Answer

A

Key insight

Express each coupon's savings in terms of the price p; A beats B when p >= 200 and beats C when p <= 250.

Solution

Let the listed price be p>100p > 100. The savings are
A=0.15p,B=30,C=0.25(p−100).A = 0.15p,\qquad B = 30,\qquad C = 0.25(p - 100).

Coupon A saves at least as much as B when 0.15p≥300.15p \ge 30, i.e. p≥200p \ge 200.

Coupon A saves at least as much as C when
0.15p≥0.25p−25⟹25≥0.10p⟹p≤250.0.15p \ge 0.25p - 25 \quad\Longrightarrow\quad 25 \ge 0.10p \quad\Longrightarrow\quad p \le 250.

Both conditions must hold, so the valid prices form the interval 200≤p≤250200 \le p \le 250. Thus x=200x = 200, y=250y = 250, and y−x=50y - x = 50.

The answer is (A) 50\boxed{\textbf{(A)}\ 50}.

Why this works

Each coupon is a linear function of the price, and "A is at least as good as the others" is a pair of linear inequalities. A percentage coupon beats a flat coupon for large prices, but a coupon that takes a larger percentage of the excess eventually overtakes it, so the winning region is a bounded interval; the answer is its length.

The trap

Reading 'B or C' as needing A to beat only one of them, or comparing final prices with inconsistent signs and flipping an inequality.

Common mistakes

  • Reading 'B or C' as needing A to beat only one of them, or comparing final prices with inconsistent signs and flipping an inequality.
  • Forgetting the −100-100 in coupon C and solving 0.15p≥0.25p0.15p \ge 0.25p, which has no solution above 100100.

Techniques

Set up the equation/formula and compute; no special trick needed