2014 AMC 8 Problems/Problem 2: Difference between revisions
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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> | ||
==Solution== | |||
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>. | |||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
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~savannahsolver | ~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}} | ||
Revision as of 23:33, 11 December 2023
Problem
Paul owes Paula 
cents and has a pocket full of 
-cent coins, 
-cent coins, and 
-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?
Solution
The fewest amount of coins that can be used is 
(a quarter and a dime). The greatest amount is 
, if he only uses nickels. Therefore we have 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/OOdK-nOzaII?t=454
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
