2005 AMC 8 Problems/Problem 2

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

2005 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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2005 AMC 8 Problems/Problem 2

Contents

Problem

Solution

Solution 2

Solution 3

See Also