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

Hint: To solve this question we need to have the knowledge for solving the simultaneous equations. We are given 3 unknown variables. The first step is to form the equation as per the conditions given in the question after writing the conditions which will be calculating for the unknowns.

Complete step-by-step solution:
The questions ask us to find the value of the number of rams, pigs and oxen if each cost $4$ pounds, $2$ pounds and $17$pounds respectively. The total money spent on buying $40$animals is $301$ pounds. The first step will be to consider the unknown variables. Let the number of rams, pigs and oxen be $x,y$ and $z$ respectively. Since the total number of animals bought in the question is given as $40$. So on writing it mathematically we get:
$\Rightarrow x+y+z=40$ ………… (i)
The second condition is that the total amount required to buy $40$animals is $301$. On writing this in equation form we get:
$\Rightarrow 4x+2y+17z=301$ …….(ii)
With the help of equation (i) we will write the equation (ii) in terms of $y$ and $z$.
 $\Rightarrow 4\left( 40-y-z \right)+2y+17z=301$
On calculating the above expression we get:
$\Rightarrow 160-4y-4z+2y+17z=301$
$\Rightarrow 13z-2y=141$……………. (iii)
Now the unknown variables should be positive as it represents the number of animals.
$\Rightarrow \dfrac{13z}{2}-y=\dfrac{141}{2}$
$\Rightarrow 6z+\dfrac{z}{2}-y=70+\dfrac{1}{2}$
$\Rightarrow 6z-y+\dfrac{z-1}{2}=70$
As $y$ and $z$are integers, consider $\dfrac{z-1}{2}=p$ . On rearranging we get:
$\Rightarrow z=2p+1$ ………… (iv)
On substituting the equation (iv) in equation (iii) we get:
$\Rightarrow 13\left( 2p+1 \right)-2y=141$
On calculating $y$in terms of $p$ , we get:
$\Rightarrow 2y=26p-128$
$\Rightarrow y=13p-64$……… (v)
In the similar manner we will find the value of $x$ in terms of $p$, by putting in the equation (i)
$\Rightarrow x+13p-64+2p+1=40$
$\Rightarrow x=103-15p$…….. (vi)
 We can see from equation (v) that $y<0$ for integer $p<5$ and from (vi) $x<0$ for integer $p>6$, which is not possible as $x$ and $y$ can only be positive integers.
So the value of $p$ can be $5,6$.
Now, on substituting the value of $p$as $5$ and $6$, in the expression for $x,y,z$ which are equation (iv), (v) and (vi), we get:
$\begin{align}
  & z=11,13 \\
 & y=1,14 \\
 & x=28,13 \\
\end{align}$
$\therefore $ He buys $28$ rams, $1$pig and $11$ oxen or $13$ rams, $14$pigs and $13$ oxen.

Note:To solve the question we will form equations on the basis of the conditions. The simultaneous equation can also be solved with the help of cross multiplication technique. We got two sets of answers because the total number unknown are three and the equations are two.