In Farmer Stone's hay loft, there were several animals, in particular birds, mice, and cockroaches.
One day, feeling bored, I decided to count the animals. I 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?
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Answer
35 birds, 5 mice, and 10 cockroaches.
Reasoning
Cockroaches have 6 feet, mice have 4, and birds have 2. They each have one head.
We can write down expressions for the heads and the feet, calling birds B, mice M, and cockroaches C.
Counting the heads:
(1) B + M + C = 50
Counting the feet:
(2) 2B + 4M + 6C = 150
We are told that for every mouse there are two cockroaches so, C = 2M. Update (1) and (2) to give:
(3) B + 3M = 50
(4) 2B + 16M = 150
If we double (3) we get:
(5) 2B + 6M = 100
We can now do (4) – (5) to give:
10M = 50
M = 5
So, we have 5 mice (and 10 cockroaches).
We can use M = 5 in (3) to give:
B + 3 x 5 = 50
B + 15 = 50
B = 35
So, C = 10, M = 5 and B = 35.
Double-Checking
10 cockroaches have 10 heads and 60 feet.
5 mice have 5 heads and 20 feet.
35 birds have 35 heads and 70 feet.
The total number of heads = 10 + 5 + 35 = 50.
The total number of feet = 60 + 20 + 70 = 150.