2010 AMC 8 Problems/Problem 7

Revision as of 20:24, 20 September 2015 by Jimin0124 (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

To have less then a dollar, the most it could be is 99 cents. You could use 3 of the quarters, then 2 dimes, and 4 pennies to make 99 cents with least coins. So the total of coins is 6 coins.

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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