Difference between revisions of "2014 AMC 10B Problems/Problem 11"

Revision as of 22:24, 10 July 2015

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 smallest possible positive integer value of $n$ ?

$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33$

Solution

Let the original 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, $n$ must be greater than $\max(x - .7225x, x-.729x, x-.7125x)$ . It follows $n$ must be greater than $.2875$ . We multiply this by $100$ to get the percent value, and then round up because $n$ is the smallest integer that provides a greater discount than $28.75$ , leaving us with the answer of $\boxed{\textbf{(C) } 29}$

2014 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 What is the smallest possible positive integer value of <math>n</math>?
-<math> \textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}}\ 31\qquad\textbf{(E)}\ 33 </math>
+<math> \textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33 </math>
 ==Solution==

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Difference between revisions of "2014 AMC 10B Problems/Problem 11"

Revision as of 22:24, 10 July 2015

Problem

Solution

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