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 "2010 AMC 10B Problems/Problem 11"
(→‎Solution)
Line 15: Line 15:
 
Coupon C saves us: <math>0.25p</math>
 
Coupon C saves us: <math>0.25p</math>
  
−
Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities:
+
Now, the condition is that A has to be greater than or equal to either B and C which give us the following inequalities:
  
 
<math>A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100</math>
 
<math>A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100</math>

Revision as of 12:31, 26 June 2014

Problem

A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 100$.
Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x$?

$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80 \qquad \textbf{(E)}\ 100$

Solution

Let the listed price be $(100 + p)$, where $p > 0$

Coupon A saves us: $0.15(100+p) = (0.15p + 15)$

Coupon B saves us: $30$

Coupon C saves us: $0.25p$

Now, the condition is that A has to be greater than or equal to either B and C which give us the following inequalities:

$A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100$

$A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150$

We see here that the greatest possible value for $p$ is $150$, thus $y = 100 + 150 = 250$ and the smallest value for p is $100$ so $x = 100 + 100 = 200$.

The difference between $y$ and $x$ is $y - x = 250 - 200 = \boxed{\textbf{(A)}\ 50}$

See Also

2010 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=2010_AMC_10B_Problems/Problem_11&oldid=62334"