2014 AMC 10B Problem 11

Use a price of $\$100$, so every percent is just a number of dollars.

1. Two $15\%$ discounts. $15\%$ off $\$100$ leaves $\$85$. Another $15\%$ off $\$85$ takes off $\$12.75$, leaving $\$72.25$. Total discount: $\$27.75$, or $27.75\%$.
2. Three $10\%$ discounts. $\$100 \to \$90 \to \$81 \to \$72.90$. Total discount: $27.1\%$.
3. $25\%$ then $5\%$. $\$100 \to \$75$, then $5\%$ of $\$75$ is $\$3.75$, leaving $\$71.25$. Total discount: $28.75\%$.
4. The single discount of $n\%$ must be more advantageous than every one of these, so it must take off more than the biggest of them: $n > 28.75$.
5. The smallest whole number bigger than $28.75$ is $29$.

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

The second discount is taken from a price that is already lower, so two $15\%$ discounts take off less than $30\%$. To compare deals, track how much of the price is left after each one; multiplying the "left over" parts ($0.85 \times 0.85$, and so on) does this quickly. Also read the inequality carefully: "more advantageous than" is strict, so $n$ has to beat $28.75$, not tie it.

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 $\boxed{\textbf{(C)}\ 29}$.