Difference between revisions of "2010 AMC 8 Problems/Problem 7"
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<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99</math> | <math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99</math> | ||
==Solution== | ==Solution== | ||
− | You need | + | You need 2 dimes, 1 nickel, and 4 pennies for the first 25 cents. From 26 cents to 50 cents, you only need to add 1 quarter. From 51 cents to 75 cents, you also only need to add 1 quarter. The same for 76 cents to 99 cents. Notice that instead of 100, it is 99 We are left with 3 quarters, 1 nickel, 2 dimes, and 4 pennies. so the correct answer is 3+2+1+4=10 |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=6|num-a=8}} | {{AMC8 box|year=2010|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:41, 15 August 2021
Problem
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
Solution
You need 2 dimes, 1 nickel, and 4 pennies for the first 25 cents. From 26 cents to 50 cents, you only need to add 1 quarter. From 51 cents to 75 cents, you also only need to add 1 quarter. The same for 76 cents to 99 cents. Notice that instead of 100, it is 99 We are left with 3 quarters, 1 nickel, 2 dimes, and 4 pennies. so the correct answer is 3+2+1+4=10
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.