Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Solution)
Line 13: Line 13:
  
 
==Solution==
 
==Solution==
 +
Let the price be <math>x</math>. Then, for option <math>1</math>, the discounted price is <math>(1-.15)(1-.15)x = .7225x</math>. For option <math>2</math>, the discounted price is <math>(1-.1)(1-.1)(1-.1)x = .729x</math>. Finally, for option <math>3</math>, the discounted price is <math>(1-.25)(1-.05) = .7125x</math>. Therefore, the discount must be greater than <math>\max(x - .7225x, x-.729x, x-.7125x)</math>. Thus, the discount must be greater than <math>.2875</math>. We multiply this by <math>100</math> to get the percent value, and then round up because <math>x</math> is the smallest integer that provides a greater discount than <math>28.75</math>, leaving us with the answer of <math>\boxed{\textbf{(C) } 29}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2014|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:15, 20 February 2014

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_11&oldid=60120"