Difference between revisions of "2014 AMC 8 Problems/Problem 2"

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(Video Solution)
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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>
 
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/Gm7gDXHjnUU
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~Education, the Study of Everything
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==Video Solution==
 
==Video Solution==

Revision as of 10:18, 2 July 2023

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 (ProblemsAnswer KeyResources)
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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