Difference between revisions of "1999 AMC 8 Problems/Problem 9"
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| − | == | + | ==Problem== |
| − | Three | + | Three flower 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 |
| − | 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 | ||
| − | |||
| − | |||
| − | == | + | <asy> |
| + | draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); | ||
| + | draw(circle((.3,-.1),.7)); | ||
| + | draw(circle((2.8,-.2),.8)); | ||
| + | label("A",(1.3,.5),N); | ||
| + | label("B",(3.1,-.2),S); | ||
| + | label("C",(.6,-.2),S); | ||
| + | </asy> | ||
| + | |||
| + | <math>\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450</math> | ||
| + | |||
| + | ==Solution== | ||
| + | ===Solution 1=== | ||
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Plants shared by two beds have been counted | Plants shared by two beds have been counted | ||
| − | twice, so the total is 500 + 450 + 350 | + | twice, so the total is <math>500 + 450 + 350 - 50 - 100 = \boxed{\text{(C)}\ 1150}</math>. |
| − | 1150 . | + | |
| + | ===Solution 2=== | ||
| + | Bed A has <math>350</math> plants it doesn't | ||
| + | share with B or C. Bed B has <math>400</math> plants it doesn't | ||
| + | share with A or C. And C has <math>250</math> it doesn't share | ||
| + | with A or B. The total is <math>350 + 400 + 250 + 50 + | ||
| + | 100 = \boxed{\text{(C)}\ 1150}</math> plants. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|num-b=8|num-a=10}} | {{AMC8 box|year=1999|num-b=8|num-a=10}} | ||
Revision as of 12:35, 23 December 2012
Problem
Three flower 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
![[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy]](http://latex.artofproblemsolving.com/e/c/e/ece582021fe59973b5c82e6a43d3522518c4c99c.png)
Solution
Solution 1
Plants shared by two beds have been counted
twice, so the total is 
.
Solution 2
Bed A has 
plants it doesn't
share with B or C. Bed B has 
plants it doesn't
share with A or C. And C has 
it doesn't share
with A or B. The total is 
plants.
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
