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2014 AMC 8 Problems/Problem 2

Revision as of 10:18, 2 July 2023 by Thestudyofeverything (talk | contribs) (→‎Video Solution)

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?

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

Video Solution (CREATIVE THINKING)

https://youtu.be/Gm7gDXHjnUU

~Education, the Study of Everything


Video Solution

https://youtu.be/OOdK-nOzaII?t=454

https://youtu.be/scOob5X-l6g

~savannahsolver

Solution

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

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
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

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