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

(Solution 2)
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==problem==
+
==Problem==
  
Three fower beds overlap as shown. Bed A has
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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
 
(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 +
 
100 = 1150 plants.
 
OR
 
 
Plants shared by two beds have been counted
 
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>.
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.
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/lajfUn8R6M4 ~DSA_Catachu
  
 
==See Also==   
 
==See Also==   
  
 
{{AMC8 box|year=1999|num-b=8|num-a=10}}
 
{{AMC8 box|year=1999|num-b=8|num-a=10}}
 +
{{MAA Notice}}

Revision as of 12:26, 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

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

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