Difference between revisions of "2010 AMC 8 Problems/Problem 7"
Basketball8 (talk | contribs) |
Basketball8 (talk | contribs) |
||
| Line 7: | Line 7: | ||
<math>4+1+2+3 = \boxed{\textbf{(B) } 10}</math> | <math>4+1+2+3 = \boxed{\textbf{(B) } 10}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2010|num-b=4|num-a=6}} | ||
Revision as of 23:28, 3 November 2012
We need:
4 Pennies, 1 Nickel, 2 Dimes, and 3 Quarters
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
