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

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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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==Solution==
 
==Solution==
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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 <math>7-2=\boxed{\textbf{(E)}~5}</math>.
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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>.
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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==
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https://youtu.be/OOdK-nOzaII?t=454
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https://youtu.be/scOob5X-l6g
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~savannahsolver
  
 
==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}}

Latest revision as of 23:33, 11 December 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?

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


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 $7-2=\boxed{\textbf{(E)}~5}$.


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

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

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

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