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

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==See Also==
 
==See Also==
{{AMC8 box|year=2010|num-b=4|num-a=6}}
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{{AMC8 box|year=2010|num-b=6|num-a=8}}

Revision as of 17:52, 4 November 2012

We need:

4 Pennies, 1 Nickel, 2 Dimes, and 3 Quarters

$4+1+2+3 = \boxed{\textbf{(B) } 10}$

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