2010 AMC 8 Problems/Problem 7

Revision as of 13:41, 15 August 2021 by Raina0708 (talk | contribs) (Solution)

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?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$

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 (ProblemsAnswer KeyResources)
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

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