2010 AMC 10B Problem 11

P11-15~4 minPrint

Linear Equations & Word Problems Absolute Value & Inequalities Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed

A shopper plans to purchase an item that has a listed price greater than $\$100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\$30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\$100$ .
Let $x$ and $y$ 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 $y - x$ ?

Show solution without answering

Key insight

For price p, Coupon A saves 0.15p, B saves 30, C saves 0.25(p - 100). A ties B at p = 200 and ties C at p = 250.

Solution

Start with a few real prices to see how the coupons compare. Savings in dollars:

Price	A: $15\%$ off	B: $\$30$ off	C: $25\%$ of the amount over $\$100$
$\$150$	$22.50$	$30$	$12.50$
$\$200$	$30$	$30$	$25$
$\$250$	$37.50$	$30$	$37.50$
$\$300$	$45$	$30$	$50$

At low prices A loses to B, and at high prices A loses to C. A is best somewhere in the middle. Now find the exact edges.

1. Let the price be $p$ dollars. The savings are
$\text{A: } 0.15p, \qquad \text{B: } 30, \qquad \text{C: } 0.25(p - 100) = 0.25p - 25.$
2. A against B. We need $0.15p \ge 30$ . Divide by $0.15$ : $p \ge 200$ .
3. A against C. We need $0.15p \ge 0.25p - 25$ . Move the $p$ terms to one side: $25 \ge 0.10p$ , so $p \le 250$ .
4. A must do at least as well as both coupons, so $p$ is from $200$ to $250$ . The smallest price is $x = 200$ and the largest is $y = 250$ .
5. So $y - x = 250 - 200 = 50$ .

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

Why this works

Each coupon's savings is a simple straight-line formula in the price. A percent-off coupon grows with the price, so it passes a flat $\$30$ coupon once the price is high enough. But Coupon C takes a bigger percent (of the part above $\$100$ ), so it eventually catches up with A. The prices where A wins form one stretch between those two crossing points, and the question asks for its length.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

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

Coupon A saves at least as much as C when
$0.15p \ge 0.25p - 25 \quad\Longrightarrow\quad 25 \ge 0.10p \quad\Longrightarrow\quad p \le 250.$

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

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

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 7: Proportion (§7.4)