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 "2005 AMC 8 Problems/Problem 2"
(→‎Solution)
Line 9: Line 9:
 
==Solution==
 
==Solution==
 
Karl paid <math>5 \cdot 2.50 = \textdollar 12.50</math>. <math>20 \%</math> of this cost that he saved is <math>12.50 \cdot .2 = \boxed{\textbf{(C)}\ \textdollar 2.50}</math>.
 
Karl paid <math>5 \cdot 2.50 = \textdollar 12.50</math>. <math>20 \%</math> of this cost that he saved is <math>12.50 \cdot .2 = \boxed{\textbf{(C)}\ \textdollar 2.50}</math>.
 +
 +
==Solution 2==
 +
Each folder can also be <math> 5/2</math> dollars, and <math>20\%</math> can be shown as <math>(1/5)</math>. We can multiply <math>(5/2) \cdot (1/5) = (1/2)</math>. <math>(1/2)</math>  is also <math>50</math> cents or the amount of money that is saved after the <math>20\%</math> discount. So each folder is <math>2.50-0.5 = \textdollar2</math>.Since Karl bought 5 folders all of the folders after the discount is <math>(5)(2) = 10</math>, and the money bought before the discount is <math>(5)(2.50) = \textdollar12.50</math>. To find the money Karl saves all we have to do is subtract <math>12.50 - 10 = 2.50</math>. Thus the answer is  <math>\boxed{\textbf{(C)}\ \textdollar 2.50}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=1|num-a=3}}
 
{{AMC8 box|year=2005|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:26, 4 November 2018

Problem

Karl bought five folders from Pay-A-Lot at a cost of $\textdollar 2.50$ each. Pay-A-Lot had a 20%-off sale the following day. How much could Karl have saved on the purchase by waiting a day?

$\textbf{(A)}\ \textdollar 1.00 \qquad\textbf{(B)}\ \textdollar 2.00 \qquad\textbf{(C)}\ \textdollar 2.50\qquad\textbf{(D)}\ \textdollar 2.75 \qquad\textbf{(E)}\ \textdollar 5.00$

Solution

Karl paid $5 \cdot 2.50 = \textdollar 12.50$. $20 \%$ of this cost that he saved is $12.50 \cdot .2 = \boxed{\textbf{(C)}\ \textdollar 2.50}$.

Solution 2

Each folder can also be $5/2$ dollars, and $20\%$ can be shown as $(1/5)$. We can multiply $(5/2) \cdot (1/5) = (1/2)$. $(1/2)$ is also $50$ cents or the amount of money that is saved after the $20\%$ discount. So each folder is $2.50-0.5 = \textdollar2$.Since Karl bought 5 folders all of the folders after the discount is $(5)(2) = 10$, and the money bought before the discount is $(5)(2.50) = \textdollar12.50$. To find the money Karl saves all we have to do is subtract $12.50 - 10 = 2.50$. Thus the answer is $\boxed{\textbf{(C)}\ \textdollar 2.50}$.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 AJHSME/AMC 8 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=2005_AMC_8_Problems/Problem_2&oldid=98504"