During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

1999 AMC 8 Problems/Problem 9

Revision as of 08:06, 17 June 2011 by Kingofmath101 (talk | contribs) (Solution to 1998 AMC 8, #9)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Okay. Given that the area common to A and C has 100 plants, and C has a total of 350 plants, there are 250 plants in the area exclusive to C. Area B has a total of 450 plants, but 50 of those are shared with Area A, so the number of plants exclusive to Area B is $450 - 50 = 400$. Lastly, in all of Area A, there are 500 plants, but 50 of those are shared with Area C, and 100 of those plants are shared with Area B. So the number of plants in the area exclusive to Area A is $500 - 100 - 50 = 400 - 50 = 350$. None of the plants are double-counted, so to find the total, the number of plants in all of the zones are added together: $250 + 100 + 350 + 50 + 400 = 350 + 400 + 400 = 750 + 400 = 1150$ plants total.

Invalid username
Login to AoPS