Difference between revisions of "2021 JMPSC Problems/Problem 2"
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== Problem == | == Problem == | ||
− | Brady has an unlimited supply of quarters (<math>0.25), dimes (< | + | Brady has an unlimited supply of quarters (<math>\$0.25</math>), dimes (<math>\$0.10</math>), nickels (<math>\$0.05</math>), and pennies (<math>\$0.01</math>). What is the least number (quantity, not type) of coins Brady can use to pay off <math>\$2.78</math>? |
== Solution == | == Solution == | ||
To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth <math>\$2.75</math>, which is the most quarters we can use to get a value less than or equal to <math>\$2.78</math>. Finally, we can add 3 pennies to get a total of <math>\$2.87</math>, so the answer is <math>11+3=14</math>. | To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth <math>\$2.75</math>, which is the most quarters we can use to get a value less than or equal to <math>\$2.78</math>. Finally, we can add 3 pennies to get a total of <math>\$2.87</math>, so the answer is <math>11+3=14</math>. |
Latest revision as of 18:30, 10 July 2021
Problem
Brady has an unlimited supply of quarters (), dimes (), nickels (), and pennies (). What is the least number (quantity, not type) of coins Brady can use to pay off ?
Solution
To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth , which is the most quarters we can use to get a value less than or equal to . Finally, we can add 3 pennies to get a total of , so the answer is .