QUESTION

# Ramesh buys animals consisting of rams at 4 ponds, pigs at 2 pounds and oxen at 17 pounds. If I spend 301 pounds, how many of each do Ramesh buys?

Hint: In this question, we first need to assume three variables to represent the respective animals. Then from the given conditions we get two equations but as there are three variables we need to find the possible integral solutions by assuming some other variable and writing all the other terms in that and on further simplification we get only 2 possible ways.

Let us assume the number of rams as x, number of pigs as y and number of oxen as z.
Now, from the given conditions in the question we get,
\begin{align} & x+y+z=40.....\left( 1 \right) \\ & 4x+2y+17z=301.....\left( 2 \right) \\ \end{align}
Now, from the equation (1) we get,
$x=40-y-z$
Let us now substitute this value of x in the equation (2)
$\Rightarrow 4\left( 40-y-z \right)+2y+17z=301$
Now, on further simplifying this we get,
$\Rightarrow 160-4y-4z+2y+17z=301$
Let us now rearrange the terms accordingly.
$\Rightarrow 13z-2y=141.....\left( 3 \right)$
Here, we need to find the positive integral solutions as the number of animals are positive.
Now, on dividing the above equation (3) with 2 on both sides we get,
$\Rightarrow \dfrac{13z}{2}-y=\dfrac{141}{2}$
Now, this can be further written as
$\Rightarrow 6z+\dfrac{z}{2}-y=70+\dfrac{1}{2}$
Now, on rearranging the terms further we get,
$\Rightarrow 6z-y+\dfrac{z-1}{2}=70$
As we already know that y and z are integers it means
$\Rightarrow \dfrac{z-1}{2}=\text{integer}$
Let us assume that this integer as p.
\begin{align} & \Rightarrow \dfrac{z-1}{2}=p \\ & \Rightarrow z=2p+1 \\ \end{align}
Now, by substituting this value of z in equation (3) we get,
$\Rightarrow 13\left( 2p+1 \right)-2y=141$
Now, on rearranging the terms and further simplifying we get,
\begin{align} & \Rightarrow 2y=26p-128 \\ & \Rightarrow y=13p-64 \\ \end{align}
Now, by substituting these values of y and z we get the value of x from equation (1)
\begin{align} & \Rightarrow x+13p-64+2p+1=40 \\ & \Rightarrow x=103-15p \\ \end{align}
Now, as we know that x and y should be positive we get,
\begin{align} & \Rightarrow x>0 \\ & \Rightarrow 103-15p>0 \\ & \Rightarrow 103>15p \\ \end{align}
$\therefore p<6$
\begin{align} & \Rightarrow y>0 \\ & \Rightarrow 13p-64>0 \\ & \Rightarrow 13p>64 \\ & \therefore p>5 \\ \end{align}
As both the conditions should be satisfied so we have only 2 possible values of p.
$\therefore p=5,6$
Now, by substituting these values of p in the equations of x, y, z we get,
For $p=5$
\begin{align} & \Rightarrow x=103-15\times 5 \\ & \therefore x=28 \\ \end{align}
\begin{align} & \Rightarrow y=13\times 5-64 \\ & \therefore y=1 \\ \end{align}
\begin{align} & \Rightarrow z=2\times 5+1 \\ & \therefore z=11 \\ \end{align}
For $p=6$
\begin{align} & \Rightarrow x=103-15\times 6 \\ & \therefore x=13 \\ \end{align}
\begin{align} & \Rightarrow y=13\times 6-64 \\ & \therefore y=14 \\ \end{align}
\begin{align} & \Rightarrow z=2\times 6+1 \\ & \therefore z=13 \\ \end{align}

Note: It is important to note that as there are 3 unknowns and only 2 equations we cannot get the actual solution. So, we need to check the possibilities which can be more than one by considering that the number of animals can only be positive.
While assuming an integer and then writing all the variables in terms of it we should not neglect any of the terms because neglecting the condition so obtained changes accordingly so that we cannot get a possible solution.