Difference between revisions of "2010 AMC 8 Problems/Problem 7"

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==Solution==
 
==Solution==
It would have a total of 10 coins: 4 pennies, 1 nickel, 2 dimes, and 3 quarters, so the answer is <math>B</math>
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It would have a total of 10 coins: 4 pennies, 1 nickel, 2 dimes, and 3 quarters, so the answer is <math> \boxed{\textbf{(B)}\ 10}</math>
  
 
==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 11:41, 13 November 2016

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

It would have a total of 10 coins: 4 pennies, 1 nickel, 2 dimes, and 3 quarters, so the answer is $\boxed{\textbf{(B)}\ 10}$

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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