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

m (Removed superfluos "okay")
Line 1: Line 1:
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.
+
==problem==
 +
 
 +
Three fower beds overlap as shown. Bed A has
 +
500 plants, bed B has 450 plants, and bed C has
 +
350 plants. Beds A and B share 50 plants, while
 +
beds A and C share 100. The total number of
 +
plants is
 +
(A) 850 (B) 1000 (C) 1150 (D) 1300
 +
(E) 1450
 +
 
 +
==solution==
 +
 
 +
(C) 1150: Bed A has 350 plants it doesn't
 +
share with B or C. Bed B has 400 plants it doesn't
 +
share with A or C. And C has 250 it doesn't share
 +
with A or B. The total is 350 + 400 + 250 + 50 +
 +
100 = 1150 plants.
 +
OR
 +
Plants shared by two beds have been counted
 +
twice, so the total is 500 + 450 + 350 ¡ 50 ¡ 100 =
 +
1150 .
 +
 
 +
==See Also== 
 +
 
 +
{{AMC8 box|year=1999|num-b=8|num-a=10}}

Revision as of 14:02, 4 November 2012

problem

Three fower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is (A) 850 (B) 1000 (C) 1150 (D) 1300 (E) 1450

solution

(C) 1150: Bed A has 350 plants it doesn't share with B or C. Bed B has 400 plants it doesn't share with A or C. And C has 250 it doesn't share with A or B. The total is 350 + 400 + 250 + 50 + 100 = 1150 plants. OR Plants shared by two beds have been counted twice, so the total is 500 + 450 + 350 ¡ 50 ¡ 100 = 1150 .

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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