Art of Problem Solving
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

2010 AMC 10B Problems/Problem 11: Difference between revisions

Newer edit →
Srumix (talk | contribs)
4 edits
Created page with 'Let the listed price be <math>(100 + p)</math>, where <math>p > 0</math> Coupon A saves us: <math>0.15(100+p) = (0.15p + 15)</math> Coupon B saves us: <math>30</math> Coupon …'
 
Gina (talk | contribs)
783 edits
No edit summary
Line 1: Line 1:
== Problem ==
A shopper plans to purchase an item that has a listed price greater than <math>\textdollar 100</math> and can use any one of the three coupons. Coupon A gives <math>15\%</math> off the listed price, Coupon B gives <math>\textdollar 30</math> off the listed price, and Coupon C gives <math>25\%</math> off the amount by which the listed price exceeds
<math>\textdollar 100</math>. <br/>
Let <math>x</math> and <math>y</math> be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is <math>y - x</math>?
<math>\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80  \qquad \textbf{(E)}\ 100</math>
==Solution==
Let the listed price be <math>(100 + p)</math>, where <math>p > 0</math>
Let the listed price be <math>(100 + p)</math>, where <math>p > 0</math>
   
   
Line 9: Line 17:
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 or C which give us the following inequalities:


<math>A \geq B \Leftrightarrow 0.15p + 15 \geq 30 \Leftrightarrow p \geq 100</math>
<math>A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100</math>
 
<math>A \geq C \Leftrightarrow 0.15p + 15 \geq 0.25p \Leftrightarrow p \leq 150</math>


We see here that the greatest possible value for p is <math> 150 = y </math> and the smallest is <math> 100 = x </math>
<math>A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150</math>


The difference between <math>y</math> and <math>x</math> is <math>y - x \Leftrightarrow 150 - 100 = 50</math>
We see here that the greatest possible value for <math>p</math> is <math> 150 = y </math> and the smallest is <math> 100 = x </math>


Our answer is:
The difference between <math>y</math> and <math>x</math> is <math>y - x = 150 - 100 = \boxed{\textbf{(A)}\ 50}</math>


<math> \boxed{\mathrm{(A)}= 50} </math>
==See Also==
{{AMC10 box|year=2010|ab=B|num-b=10|num-a=12}}

Revision as of 00:58, 26 November 2011

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 or 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 $150 = y$ and the smallest is $100 = x$

The difference between $y$ and $x$ is $y - x = 150 - 100 = \boxed{\textbf{(A)}\ 50}$

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
Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_11&oldid=43412"