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Puzzle Details

In farmer Brown's hay loft there are a number of animals, in particular crows, mice and cockroaches.

Being bored one day, I decided to count the animals and found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice.

How many of each animal were there?

Answer: 10 cockroaches, 5 mice and 35 birds.

Cockroaches have 6 feet, mice have 4 and birds have 2.

For every mouse there are two cockroaches, so every mouse is worth 3 heads and 16 feet (4 + 6 + 6).

We can now write down an expression for the heads and the feet (calling birds B and mice M):

B +  3M = 50       (1)
Feet gives us:
2B + 16M = 150      (2)

If we double (1) we get:

2B +  6M = 100      (3)

We can now do (2) - (3) to give:

10M = 50
M = 5
So we have 5 mice (and 10 cockroaches). We can use M = 5 in (1) to give:

B + 3 x 5 = 50
B + 15 = 50
B = 35

So, C = 10, M = 5 and B = 35. Checking that 10 + 5 + 35 = 50, and 10 x 6 + 5 x 4 + 35 x 2 = 150. QED.

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