Question

I buy $40$ animals consisting of rams at $4$ pounds, pigs at $2$ pounds, and oxen at $17$ pounds. if I spend $301$ pounds, how many of each do I buy?

Answer 1

Let number of rams be $x$, pigs be $y$ and oxen be $z$

$x+y+z=40$ .......(i)

$4x+2y+17z=301$

Using $x$ from (i), we have

$4(40-y-z)+2y+17z=301$

$\Rightarrow 160-4y-4z+2y+17z=301$

$\Rightarrow 13z-2y=141$ ..........(ii)

We will approach the question by finding the positive integral solutions of the equations as the number of animals can olny be some positive integer.

$\frac{13z}{2}-y=\frac{141}{2}$

$\Rightarrow 6z+\frac{z}{2}-y=70+\frac{1}{2}$

$\Rightarrow 6z-y+\frac{z-1}{2}=70$

As $y$ and $z$ are integers

$\Rightarrow \frac{z-1}{2}=$ integer

Let the integer be $p$

$\frac{z-1}{2}=p$

$\Rightarrow z=2p+1$ ..........(iii)

Substituting $z$ in (ii)

$13(2p+1)-2y=141$

$\Rightarrow 2y=26p-128$

$\Rightarrow y=13p-64$ .........(iv)

Substituting $y$ and $z$ in (i)

$x+13p-64+2p+1=40$

$\Rightarrow x=103-15p$ ........(v)

We can see from (iv) that $y<0$ for integer $p<5$ and from (v) $x<0$ for integer $p>6$ , which is not possible as $x$ and $y$ can only be positive integers.

So, $p$ can take values $5,6$

Substituting $p$ in (iii),(iv) and (v) , we get

$\{\begin{array}{c}z=11,13\\ y=1,14\\ x=28,13\end{array}$

So, he can buy $28$ rams, $1$ pig and $11$ oxen or $13$ rams, $14$ pig and $13$ oxen.