In 1929, Farmer Giles took his animals to market. He had some cows, sheep and chickens to sell. Cows sold for ten pounds each. Sheep sold for one pound each. Chickens were sold for one pound for a brood of eight (they were only sold in whole broods). Farmer Giles had a good day and sold 100 animals for exactly £100. The day did have its down side; he had to sell Daisy, his favourite cow. How many sheep did Farmer Giles sell?

Question

Reuben Patrick

Mathematics

11 May, 00:52

In 1929, Farmer Giles took his animals to market. He had some cows, sheep and chickens to sell. Cows sold for ten pounds each. Sheep sold for one pound each. Chickens were sold for one pound for a brood of eight (they were only sold in whole broods). Farmer Giles had a good day and sold 100 animals for exactly £100. The day did have its down side; he had to sell Daisy, his favourite cow. How many sheep did Farmer Giles sell?

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Answers (2)

Know the Answer?

Trace Keith · Answer 1

90 sheep?

Step-by-step explanation:

1 sheep = 1 pound

1 cow = 10 pounds

total amount made = 100 pounds

100 : 1 (the cost for sheep) = 100 sheep

100 - 90 (the cost of cows) = 10

10 sheep

Zoey Herman · Answer 2

21 sheep

Step-by-step explanation:

100 = x + y + 8z

10x + y + z = 100

x = 1

y + 8z = 99

y + z = 90

x = 2

y + 8z = 98

y + z = 80

x = 3

y + 8z = 97

y + z = 70

x = 4

y + 8z = 96

y + z = 60

x = 5

y + 8z = 95

y + z = 50

x = 6

y + 8z = 94

y + z = 40

x = 7

y + 8z = 93

y + z = 30

7z = 63

z = 9

y = 30 - 9 = 21

x = 7 (7 cows)

y = 21 (21 sheep)

z = 9 (9*8 = 72 chicken)