Question

There are multiple chickens and rabbits in a cage.

There are $72$ heads and $200$ feet inside the cage.

How many chickens and rabbits are in there?

Answer 1

Step -1: Model the given situation as a pair of linear equations:

Let there be $x$ chickens and $y$ rabbits in the cage.

Since each chicken has one head, $x$ chickens have $x$ heads.

Similarly, since each rabbit has one head, $y$ rabbits have $y$ heads.

Thus the total number of heads in the cage will be $x+y$.

It is given that the number of heads in the cage is $72$.

Thus, $x+y=72$.

Since each chicken has $2$ feet, $x$ chickens have $2x$ feet.

Similarly, since each rabbit has $4$ feet, $y$ rabbits have $4y$ feet.

Thus the total number of feet in the cage will be $2x+4y$.

It is given that the number of feet in the cage is $200$.

Thus, $2x+4y=200$.

Step- 2: Prepare to solve the pair of linear equations:

The equations $x+y=72$, and $2x+4y=200$ form a pair of linear equations in two variables.

Solve these by simultaneously using the substitution method.

Isolate $y$ from the equation $x+y=72$:

$\begin{array}{rcl}x+y& =& 72\\ & \Rightarrow & y=72-x\end{array}$

Step 3. Determine the number of chickens:

Substitute $y=72-x$ in $2x+4y=200$ and then solve for $x$:

$\begin{array}{rcl}2x+4y& =& 200\\ \Rightarrow 2x+4\left(72-x\right)& =& 200\\ & \Rightarrow & 2x+4\times 72-4x=200\\ & \Rightarrow & 2x-4x=200-4\times 72\\ & \Rightarrow & -2x=200-4\times 72\\ & \Rightarrow & x=\frac{200-4\times 72}{-2}\\ & \Rightarrow & x=\frac{200}{-2}-\frac{4\times 72}{-2}\\ & \Rightarrow & x=-100+2\times 72\\ & \Rightarrow & x=-100+144\\ & \Rightarrow & x=44\end{array}$

Thus, the number of chickens in the cage is $x=44$.

Step- 4: Determine the number of rabbits.

Substitute, $x=44$ in the equation $y=72-x$ and solve for $y$:

$\begin{array}{rcl}y& =& 72-x\\ & =& 72-44\\ & =& 28\end{array}$

Thus, the number of rabbits in the cage is $y=28$.

Hence, there are $\mathbf{44}$ chickens and $\mathbf{28}$ rabbits in the cage.