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Difficulty: Tricky

Farmer Doolittle has a farm with just cows and chickens. The farmer counts up all the animals and discovers that there are 43 heads and 120 legs on the farm (not counting the farmers).

How many cows live on farmer Doolittle’s farm?

Need a hint?

With these kinds of word problems, some of the key information isn’t said outright. For example, the key difference between cows and chickens is that each cow has 4 legs while each chicken just has 2. Both have just 1 head.

Now ask yourself, how many total animals are on the farm? How many legs would there be if all the animals were cows? Or if all were chickens?

Brainteaser answer

There are 17 cows and 26 chickens on farmer Doolittle’s farm.

First, we need to note the key difference between cows and chickens: each cow has 4 legs while each chicken only has 2 legs. This will help us figure out how many of each animal there are.

We also know that both kinds of animals have 1 head each. This means that there are 43 total animals.

To get a handle on this question, let’s start with an extreme. If all the animals were cows, there would be 4 x 43 = 172 legs, which is way too many legs.

If all the animals were chickens, there would be only 2 x 43 = 86 legs, which is not enough. But it might be a good place to start. We could try some guess and check, but there’s also a cleverer way to look at the question.

Imagine Farmer Doolittle sold one chicken and bought a cow instead. The number of heads stays the same, but the number of legs goes up by 2.

If we start with all chickens, we’re still short by 120 – 86 = 34 legs. We can make up the difference by selling chickens and buying cows! Selling 17 chickens and buying 17 cows will increase the total number of legs by 2 x 17 = 34, the number we’re hoping for.

To check, we have 17 cows and 43 – 17 = 26 chickens.

The total number of legs is:

17 x 4 + 26 x 2 = 68 + 52

= 120 legs

One response

  1. Steve Sugden Avatar
    Steve Sugden

    The cows and chickens brainteaser is interesting and your solution is clear and elegant while cleverly avoiding algebra. But that’s my point. I would like to see an algebraic solution also published side-by-side to that readers can compare the two methods. Who knows – maybe a compact algebraic solution may help convince us that algebra is a compact and powerful tool, used routinely many STEM professionals as part of their scientific toolkit. Keep up the good work!

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