Chicken Rabbit Cage Problem: Formula and 4 Methods

The chicken-rabbit cage problem, also called Ji tu tong long or chickens and rabbits in the same cage, gives two facts:

• the total number of heads
• the total number of legs

The goal is to find how many chickens and how many rabbits are in the cage.

The most useful version is:

A cage has 10 heads and 28 legs. How many chickens and rabbits are there?

Direct Formula

If every animal is either a chicken or a rabbit:

rabbits = (legs - 2 x heads) / 2
chickens = heads - rabbits

For 10 heads and 28 legs:

rabbits = (28 - 2 x 10) / 2
rabbits = (28 - 20) / 2
rabbits = 4

chickens = 10 - 4 = 6

So the answer is 4 rabbits and 6 chickens.

Try the interactive version here: Chicken-Rabbit Cage Solver.

Method 1: Assumption Method

This is usually the cleanest method for elementary Olympiad problems.

Assume every head is a chicken. Since chickens have 2 legs, 10 heads would make:

10 x 2 = 20 legs

But the problem says there are 28 legs, so there are:

28 - 20 = 8 extra legs

Every rabbit has 2 extra legs compared with a chicken:

4 - 2 = 2

So the number of rabbits is:

8 / 2 = 4 rabbits

The remaining animals are chickens:

10 - 4 = 6 chickens

This method works because the head count stays fixed. You are only measuring how many 2-leg upgrades are needed.

Method 2: Equation Method

Let:

• c = chickens
• r = rabbits

The head count gives:

c + r = 10

The leg count gives:

2c + 4r = 28

From the first equation:

c = 10 - r

Substitute into the leg equation:

2(10 - r) + 4r = 28
20 - 2r + 4r = 28
20 + 2r = 28
2r = 8
r = 4

Then:

c = 10 - 4 = 6

The equation method is the same reasoning as the assumption method, just written with symbols.

Method 3: Table Method

A table helps students who are not ready for equations.

Each row swaps one chicken into one rabbit. The head count stays at 10, and the leg count increases by 2 each time.

The row with 28 legs is 6 chickens and 4 rabbits.

Method 4: Visual Swapping

Draw 10 animals as chickens first. That gives 20 legs.

Then swap one chicken into a rabbit:

20 -> 22 -> 24 -> 26 -> 28

It takes four swaps to reach 28 legs, so there are four rabbits. The other six animals stay chickens.

This is the method behind the interactive chicken-rabbit cage model.

Common Variations

More legs than heads allow

If the legs are too high, start from all chickens and add rabbits until the leg total matches.

Fewer legs than possible

If legs are less than 2 x heads, the problem has no solution because even all chickens would have too many legs.

Odd number of extra legs

If legs - 2 x heads is odd, the problem has no whole-number solution. Each rabbit adds exactly 2 extra legs.

Different animals

The same method works for any two types with different counts:

number of high-count animals =
(total count - low count x total items) / (high count - low count)

Chicken-rabbit is just the most famous version because the gap is easy:

4 legs - 2 legs = 2 extra legs

What to Practice Next

If this problem felt new, practice these related models:

• Chicken-Rabbit Cage Solver for visual swapping
• Sum-Difference-Multiple Bar Model for hidden quantities
• Groups Sharing Lab for equal-sharing structure
• Grade 4 multi-step word problems for Common Core practice

The goal is not to memorize one puzzle. The transferable move is: fix one quantity, compare the difference, then divide by the per-swap change.