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 15:26, 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 (ProblemsAnswer KeyResources)
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