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?
Puzzle Copyright © Kevin Stone
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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.
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