2014 AMC 10B Problems/Problem 11

Revision as of 17:15, 20 February 2014 by Crastybow (talk | contribs) (Solution)

Problem

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 possible positive integer value of $n$?

$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}}\ 31\qquad\textbf{(E)}\ 33$ (Error compiling LaTeX. Unknown error_msg)

Solution

Let the price be $x$. Then, for option $1$, the discounted price is $(1-.15)(1-.15)x = .7225x$. For option $2$, the discounted price is $(1-.1)(1-.1)(1-.1)x = .729x$. Finally, for option $3$, the discounted price is $(1-.25)(1-.05) = .7125x$. Therefore, the discount must be greater than $\max(x - .7225x, x-.729x, x-.7125x)$. Thus, the discount must be greater than $.2875$. We multiply this by $100$ to get the percent value, and then round up because $x$ is the smallest integer that provides a greater discount than $28.75$, leaving us with the answer of $\boxed{\textbf{(C) } 29}$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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