2010 AMC 8 Problems/Problem 7: Difference between revisions

Revision as of 17:52, 4 November 2012

We need:

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

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

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

Revision as of 23:28, 3 November 2012 view source Basketball8 (talk \| contribs) 67 edits No edit summary ← Older edit		Revision as of 17:52, 4 November 2012 view source Mathway (talk \| contribs) 254 edits →See Also Newer edit →
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	==See Also==		==See Also==
	{{AMC8 box\|year=2010\|num-b=4\|num-a=6}}		{{AMC8 box\|year=2010\|num-b=6\|num-a=8}}