Question

A container filled with milk-water mixture contains \[75\% \] milk. \[5{\text{ litre}}\] of this mixture is replaced by water. Next, \[10{\text{ litres}}\] of the mixture is replaced by water. If the final percentage of milk in the container is \[54\% \], find the initial quantity of mixture in the container.

Answer 1

Hint: Here we will find the quantity of milk left also with the help of the final percentage and the found quantity of milk we will find the initial quantity of mixture in the container.

Complete step-by-step answer:
Suppose a container contains \[x\] units of a liquid from which \[y\] units are taken out and replaced by water. After n operations,
\[\dfrac{{{\text{Quantity of the liquid left}}}}{{{\text{The quantity of the liquid originally present}}}} = {(1 - \dfrac{y}{x})^n}units.\]
It is given that a container filled with milk-water mixture contains \[75\% \] milk.
Also given that \[5{\text{ litre}}\] of this mixture is replaced by water. Next, \[10{\text{ litre}}\] of the mixture is replaced by water.
We have to find out the initial quantity of mixture in the container.
Let, the initial quantity of mixture in the container\[ = x{\text{ litre}}\]
Now let us find the initial quantity from the given percentage we get,
The initial quantity of milk in the container \[ = 75\% \times x{\text{ litre}}\]
\[ \Rightarrow \dfrac{{75}}{{100}} \times x\]
\[ \Rightarrow 0.75x\]
When \[5{\text{ litre}}\] of this mixture is replaced by water and next, \[10{\text{ litre}}\] of the mixture is replaced by water then by mixture formula,
\[\dfrac{{{\text{Quantity of the liquid left}}}}{{{\text{The quantity of the liquid originally present}}}} = (1 - \dfrac{5}{x})(1 - \dfrac{{10}}{x})\;{\text{units}}\]
Let us now find the quantity of the milk,
Quantity of the milk left = \[ = 0.75x(1 - \dfrac{5}{x})(1 - \dfrac{{10}}{x})\,{\text{units}}\]……(1)
Given that, the final percentage of milk in the container is \[54\% \]
Let us use percentage formula to convert percentage into units we get,
\[\dfrac{{54}}{{100}} \times x = 0.54x{\text{ units}}\]….…(2)
Let us equate the equations 1 and 2 we have,
\[0.75x(1 - \dfrac{5}{x})(1 - \dfrac{{10}}{x}) = .54x\]
We solve this to find the following quadratic equation,
\[0.75(\dfrac{{x - 5}}{x})(\dfrac{{x - 10}}{x}) = .54\]
On solving the above equation we get,
\[25(x - 5)(x - 10) = 18{x^2}\]
Let us multiply the terms in the above equation then we get,
\[25{x^2} - 250x - 125x + 1250 - 18{x^2} = 0\]
On further reduction we have,
\[7{x^2} - 375x + 1250 = 0\]
Let us solve the quadratic equation using the formula given in the note, we have,
\[x = \dfrac{{375 \pm \sqrt {{{375}^2} - 4 \times 7 \times 1250} }}{{2 \times 7}}\]
On solving the above equation we get,
\[x = \dfrac{{375 \pm \sqrt {140625 - 35000} }}{{14}}\]
\[x = \dfrac{{375 \pm 325}}{{14}} = \dfrac{{700}}{{14}},\dfrac{{50}}{{14}}\]
\[x = 50{\text{ or }}3.57\]
So, the initial quantity of mixture in the container is \[50{\text{ litre}}\].

Note: We know that, the percentage of \[x\] is denoted by \[x\% \] and defined by,
\[x\% = \dfrac{x}{{100}}\]
Sridharacharya’s formula to solve quadratic equation
\[a{x^2} + bx + c = 0\] is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here \[3.57\] is not taken into account since it is a very small quantity.