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

Revision as of 11:26, 6 July 2020

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

$[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy]$

$\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

Solution

Solution 1

Plants shared by two beds have been counted twice, so the total is $500 + 450 + 350 - 50 - 100 = \boxed{\text{(C)}\ 1150}$ .

Solution 2

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 $350 + 400 + 250 + 50 + 100 = \boxed{\text{(C)}\ 1150}$ plants.

Video Solution

https://youtu.be/lajfUn8R6M4 ~DSA_Catachu

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 26: / Line 26: @@
 with A or B. The total is <math>350 + 400 + 250 + 50 +
 = \boxed{\text{(C)}\ 1150}</math> plants.
+==Video Solution==
+https://youtu.be/lajfUn8R6M4 ~DSA_Catachu
 ==See Also==

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