1999 AMC 8 Problems/Problem 9: Difference between revisions

Latest revision as of 06:03, 29 May 2025

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

Video Solution 3

https://www.youtube.com/watch?v=oWkBtswXl1A by YippieMath

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

@@ Line 1: / Line 1: @@
-==problem==
+==Problem==
-Three fower beds overlap as shown. Bed A has
+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
-plants, bed B has 450 plants, and bed C has
-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==
+<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===
-(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 +
-= 1150 plants.
-OR
 Plants shared by two beds have been counted
-twice, so the total is 500 + 450 + 350 ¡ 50 ¡ 100 =
+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
+==Video Solution 3==
+https://www.youtube.com/watch?v=oWkBtswXl1A by YippieMath
 ==See Also==
 {{AMC8 box|year=1999|num-b=8|num-a=10}}
+{{MAA Notice}}

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