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

Latest revision as of 12:22, 27 February 2022

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

Video Solution 2

https://youtu.be/UGElt-v9n7A Soo, DRMS, NM

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 1: / Line 1: @@
-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 <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 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>
+<math>\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450</math>
+==Solution==
+===Solution 1===
+Plants shared by two beds have been counted
+twice, so the total is <math>500 + 450 + 350 - 50 - 100 = \boxed{\text{(C)}\ 1150}</math>.
+===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 +
+= \boxed{\text{(C)}\ 1150}</math> plants.
+==Video Solution==
+https://youtu.be/lajfUn8R6M4 ~DSA_Catachu
+==Video Solution 2==
+https://youtu.be/UGElt-v9n7A Soo, DRMS, NM
+==See Also==
+{{AMC8 box|year=1999|num-b=8|num-a=10}}
+{{MAA Notice}}

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