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 <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>. | 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>. | ||
| − | + | ==Solution 2== | |
| − | + | Quarter and Dime 2 coins All nickels 35/5 = 7 coins | |
| + | 7 coins - 2 coins = 5 coins | ||
| + | -JasonDaGoat | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
Latest revision as of 14:08, 25 August 2025
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?
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 
.
Solution 2
Quarter and Dime 2 coins All nickels 35/5 = 7 coins
7 coins - 2 coins = 5 coins
-JasonDaGoat
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. 
