2014 AMC 8 Problems/Problem 2

Problem

Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution

The fewest amount of coins that can be used is 2 (a quarter and a dime). The greatest amount is 7, if he uses all nickels. Therefore we have $7-2=5$ , or $\fbox{E}$ .

2014 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2014 AMC 8 Problems/Problem 2

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