Question

# The owner of a milk store finds that he can sell $980$ litres of milk each week at $Rs14$ per litre and $1220$ litres of milk each week at $Rs16$ per litre. Assuming a linear relationship between selling price and demand, how many litres could you sell weekly at $Rs17$ per litre?

Hint: Assume one of the entities to be along x axis and the other to be along y axis. We have 2 points on the graph where each point has one x value and one y value, using these points form an equation of line and then put the value of the asked entity in the equation to obtain the answer.
Selling price be along the $x$â€“axis.
And demand for milk is along the $y$â€“axis.
Let, $y$ litre be sold at Rs.$x/litre$
Considering the given statement, the owner sells $980$ litres milk at Rs $14/litre$.
Let,
${x_1} = 14,{y_1} = 980$
And the next case is owner sells $1220$ litres of milk each week at $Rs16$ per litre,
Therefore,
${x_2} = 16,{y_2} = 1220$
Equation of line passing through $\left( {14,980} \right)$ and $\left( {16,1220} \right)$ .
So, on using the line formula, we have,
$\left( {y - 980} \right) = \dfrac{{1220 - 980}}{{16 - 14}}\left( {x - 14} \right)$
$\left( {y - 980} \right) = \dfrac{{240}}{2}\left( {x - 14} \right)$
$\left( {y - 980} \right) = 120\left( {x - 14} \right)$
$y = 120\left( {x - 14} \right) + 980$
Now, in the question they have asked us to find the milk that can be sold at the rate of $Rs17/litre$.
Therefore, on equating $x = 17$, we get,
$y = 120\left( {x - 14} \right) + 980$
$y = 120\left( {17 - 14} \right) + 980$
$y = 120 \times 3 + 980$
$y = 360 + 980$
$y = 1340$
Therefore, $1340$ litres of milk can be sold at the rate of $Rs17/litre$.
Note: In this question, we took selling price on x-axis and demand on y-axis and then took the values of both to plot the graph, in this graph we have 2 points using these points we formed equation of line and then finally equated x=17 in the equation to get the result.