[X] Privacy Policy + T&C We use cookies to personalise our content and ads, and for traffic analysis. Information about your use of our site is shared with our advertising and analytics providers. You also agree to our T&C.
 [X] Common Answers Have you entered November's Common Answers?

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?

[Ref: ZVYU] © Kevin Stone

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.

Back to the puzzles...

Top 10 Illusions

 Shadow Illusion Level Balance? Same Length Red Lines? Spinning Dancer Purple Nurple

We use cookies to personalise our content and ads, and for traffic analysis. Information about your use of our site is shared with our advertising and analytics providers. For more information please view our privacy policy. By using the site you also agree to our terms and conditions.