Difference between revisions of "2014 AMC 8 Problems/Problem 2"
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==Solution== | ==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 <math>7-2=\boxed{\textbf{(E)}~5}</math>. | + | 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 20:11, 24 January 2021
Contents
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?
Video Solution
https://youtu.be/OOdK-nOzaII?t=454
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 
.
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. 
