Question 1011499
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C+D+P=100
C=100-D-P

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$10D+$3P+$0.50C=$100
$10D+$3P+$0.50(100-D-P)=$100
$10D+$3P+$50-$0.50D-$0.50P=$100
$9.5D+$2.5P=$50
$19D+$5P=$100
$5P=$100-$19D
P=20-(19D/5)
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Since we know P must be a positive integer (or zero,
no partial pigs), we know 19D/5 must be a
positive integer (or zero), so 19D must be evenly divisible 
by 5. (i.e. , D must be a multiple of 5 - or zero)
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For D=0:
P=20-19(0)/5=20-0=20
We have 20 pigs and no ducks.
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For D=5:
P=20-(19(5)/5)=20-19=1
We have 1 pig and 5 ducks.
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For D=10:
P=20-(19(10)/5)=20-38=-18
Since P is negative,  this cannot be a solution
(no negative pigs, no anti-pigs)
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So we have either 20 pigs and no ducks or 1 pig and 5 ducks.
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For P=20, D=0:
C=100-D-P
C=100-0-20
C=80
ANSWER 1: We have 80 chicks, 20 pigs, and no ducks.
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For P=1, D=5:
C=100-D-P
C=100-1-5
C=94
ANSWER 2: We have 94 chicks, 1 pig, and 5 ducks.
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CHECK 1:
C=80; P=20; D=0
$0.50C+$3P+$10D=$100
$0.50(80)+$3(20)+$10(0)=$100
$40+$60=$100
$100=$100
Answer 1 checks.
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CHECK 2:
C=94; P=1; D=5
$0.50C+$3P+$10D=$100
$0.50(94)+$3(1)+$10(5)=$100
$47+$3+$50=$100
$100=$100
Answer 2 checks, so we have our two solutions.
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