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2014 AMC 10A Problem 11

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2014 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: October 28, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 3 min · position P11-15

A customer who intends to purchase an appliance has three coupons, only one of which may be used:

Coupon 1: $10\%$ off the listed price if the listed price is at least $\$50$

Coupon 2: $\$20$ off the listed price if the listed price is at least $\$100$

Coupon 3: $18\%$ off the amount by which the listed price exceeds $\$100$

For which of the following listed prices will coupon $1$ offer a greater price reduction than either coupon $2$ or coupon $3$ ?

A)$179.95
B)$199.95
C)$219.95
D)$239.95
E)$259.95

Answer

Key insight

Write each coupon's savings for a price P. Beating the 20-dollar coupon needs P above 200; beating the 18 percent coupon needs P below 225. Only 219.95 fits.

Solution

Call the listed price $P$ . Every choice is above $\$100$ , so all three coupons can be used. Write down how much each coupon saves.

coupon	savings
1	$10\%$ of $P$ , which is $0.1P$
2	$\$20$
3	$18\%$ of the part above $\$100$ , which is $0.18(P - 100)$

Coupon 1 has to beat both of the others. Check them one at a time.

Beat coupon 2. We need $0.1P > 20$ . Multiply both sides by $10$ : $P > 200$ .
Beat coupon 3. We need $0.1P > 0.18(P - 100) = 0.18P - 18$ . Add $18$ to both sides and subtract $0.1P$ : $18 > 0.08P$ . Divide by $0.08$ : $P < 225$ .
So the price must be between $\$200$ and $\$225$ . Of the choices, only $\$219.95$ is in that range.

Check it: coupon 1 saves $\$21.995$ , about $\$22.00$ ; coupon 2 saves $\$20$ ; coupon 3 saves $0.18 \cdot 119.95 \approx \$21.59$ . Coupon 1 wins.

The answer is $\boxed{\textbf{(C)}\ \$219.95}$ .

Why this works

A flat discount stays the same, while a percentage discount grows with the price. So coupon 1 overtakes coupon 2 once the price is high enough. Coupon 3 takes a bigger percentage (of a smaller amount), so for very high prices it catches up and passes coupon 1. Each comparison is a simple inequality, and together they give a window of prices where coupon 1 is best.

Alternative approach

Just test the choices, computing each coupon's savings.

price	coupon 1	coupon 2	coupon 3	coupon 1 best?
$179.95$	$\approx 18.00$	$20$	$\approx 14.39$	no
$199.95$	$\approx 20.00$ (just under)	$20$	$\approx 17.99$	no
$219.95$	$\approx 22.00$	$20$	$\approx 21.59$	yes
$239.95$	$\approx 24.00$	$20$	$\approx 25.19$	no
$259.95$	$\approx 26.00$	$20$	$\approx 28.79$	no

The trap

Checking coupon 1 only against coupon 2 and picking the first price above 200, without checking coupon 3.

Concise solution

Let the listed price be $P$ ; every choice exceeds $\$100$ , so all three coupons apply. The savings are
$\text{Coupon 1: } 0.1P,\qquad \text{Coupon 2: } 20,\qquad \text{Coupon 3: } 0.18(P - 100).$

Coupon 1 must beat coupon 2:
$0.1P > 20 \quad\Longrightarrow\quad P > 200.$

Coupon 1 must beat coupon 3:
$0.1P > 0.18P - 18 \quad\Longrightarrow\quad 18 > 0.08P \quad\Longrightarrow\quad P < 225.$

So the price must satisfy $200 < P < 225$ . Among the choices only $\$219.95$ lies in that range.

The answer is $\boxed{\textbf{(C)}\ \$219.95}$ .

Review in the AoPS books

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 (§7.1, §7.2, §7.3)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Set up the equation/formula and compute; no special trick needed