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Difference between revisions of "2014 AMC 8 Problems/Problem 2"
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==Problem==
 
Paul owes Paula <math>35</math> cents and has a pocket full of <math>5</math>-cent coins, <math>10</math>-cent coins, and <math>25</math>-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?
 
Paul owes Paula <math>35</math> cents and has a pocket full of <math>5</math>-cent coins, <math>10</math>-cent coins, and <math>25</math>-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?
  
 
<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==
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https://youtu.be/OOdK-nOzaII?t=454
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==Solution==
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The fewest amount of coins that can be used is <math>2</math> (a quarter and a dime). The greatest amount is <math>7</math>, if he only uses nickels. Therefore we have <math>7-2=\boxed{\textbf{(E)}~5}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=1|num-a=3}}
 
{{AMC8 box|year=2014|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:11, 24 January 2021

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

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

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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