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Difference between revisions of "2005 AMC 8 Problems/Problem 2"

(Created page with "==Problem== Karl bought five folders from Pay-A-Lot at a cost of <math> \textdollar 2.50 </math> each. Pay-A-Lot had a 20%-off sale the following day. How much could Karl have s...")
 
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==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>.
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==Solution 2==
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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>.
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==Solution 3==
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Having a <math>20\%</math> off sale is equivalent to a "Buy 4 get 1 free" sale. Thus, he would have saved the price of 1 folder, which is <math>\boxed{\textbf{(C)}\ \textdollar 2.50}</math>.
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-JeffersonJ
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=1|num-a=3}}
 
{{AMC8 box|year=2005|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 14:26, 19 June 2022

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 $\boxed{\textbf{(C)}\ \textdollar 2.50}$.

Solution 3

Having a $20\%$ off sale is equivalent to a "Buy 4 get 1 free" sale. Thus, he would have saved the price of 1 folder, which is $\boxed{\textbf{(C)}\ \textdollar 2.50}$.

-JeffersonJ

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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