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Posted by **unknown** on Monday, October 24, 2011 at 9:52am.

- math -
**Steve**, Monday, October 24, 2011 at 11:12amSuppose there were p plates. If you count both the ships and the cracks, then you count chipped&cracked twice.

p - 2p/3 - p/2 + p/4 = 2

12p - 8p - 6p + 3p = 24

p = 24

So, 16 were chipped

12 were cracked

6 both

2 undamaged

- math -
**Anonymous**, Monday, November 5, 2012 at 1:32am36

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