# Difference between revisions of "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 $\boxed{\textbf{(C)}\ \textdollar 2.50}$.

## See Also

 2005 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 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

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