A farmer owns pigs, chickens, and ducks. When all her animals are together, she has 30 feathered animals, and the animals all together have a total of 120 legs and 90 eyes. (All the animals have the expected number of parts.) How many of each animal might she have?

She has 18 pigs, 13 chickens, and 17 ducks.

She has 10 pigs, 15 chickens, and 15 ducks.

She has 15 pigs, 12 chickens, and 18 ducks.

She has 12 pigs, 15 chickens, and 15 ducks.

To solve this problem, let us first assign some variables. Let us say that:

x = pigs

y = chickens

z = ducks

From the problem statement, we can formulate the following equations:

1. y + z = 30 - - - > only chicken and ducks have feathers

2. 4 x + 2 y + 2 z = 120 - - - > pig has 4 feet, while chicken and duck has 2 each

3. 2 x + 2 y + 2 z = 90 - - - > each animal has 2 eyes only

Rewriting equation 1 in terms of y:

y = 30 - z

Plugging this in equation 2:

4 x + 2 (30 - z) + 2 z = 120

4 x + 60 - 2z + 2z = 120

4 x = 120 - 60

4 x = 60

x = 15

From the given choices, only one choice has 15 pigs. Therefore the answers are:

She has 15 pigs, 12 chickens, and 18 ducks.