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2014 AMC 10B Problem 11

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2014 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

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

For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:

(1) two successive $15\%$ discounts

(2) three successive $10\%$ discounts

(3) a $25\%$ discount followed by a $5\%$ discount

What is the smallest possible positive integer value of $n$ ?

A)
$27$
B)
$28$
C)
$29$
D)
$31$
E)
$33$

Answer

Key insight

Successive discounts multiply: the three options leave 72.25%, 72.9%, and 71.25% of the price, so n must beat 28.75%.

Solution

Compute the fraction of the price that remains after each scheme.

(1) $0.85 \cdot 0.85 = 0.7225$ , an effective discount of $27.75\%$ .

(2) $0.9^3 = 0.729$ , an effective discount of $27.1\%$ .

(3) $0.75 \cdot 0.95 = 0.7125$ , an effective discount of $28.75\%$ .

The single discount must be strictly better than all three, so $n > 28.75$ . The smallest integer is $n = 29$ .

The answer is $\boxed{\textbf{(C)}\ 29}$ .

Why this works

Percent discounts compound multiplicatively on the shrinking price, so two $15\%$ discounts remove less than $30\%$ . Convert each scheme to "fraction remaining," compare with $1 - n/100$ , and remember that "more advantageous than" is a strict inequality, so the threshold value itself is excluded.

The trap

Adding successive discounts (15 + 15 = 30) instead of multiplying the remaining fractions, or picking 28 when the bar is strictly above 28.75.

Common mistakes

Adding successive discounts (15 + 15 = 30) instead of multiplying the remaining fractions, or picking 28 when the bar is strictly above 28.75.
Comparing against only the first scheme (27.75%) and answering 28.

Techniques

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