# Difference between revisions of "1999 AMC 8 Problems/Problem 9"

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− | + | 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 <math>450 - 50 = 400</math>. 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 <math>500 - 100 - 50 = 400 - 50 = 350</math>. None of the plants are double-counted, so to find the total, the number of plants in all of the zones are added together: <math>250 + 100 + 350 + 50 + 400 = 350 + 400 + 400 = 750 + 400 = 1150</math> plants total. |

## Revision as of 22:12, 23 October 2012

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 . 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 . None of the plants are double-counted, so to find the total, the number of plants in all of the zones are added together: plants total.