Question

In some solution of milk, 60% is pure milk white 40% is water. 125L of pure milk is added to the
solution, the strength of milk becomes 20% Find the original quantity of the solution

Answer 1

$\color{red} \bold{Given \: that: } \\ \\ \sf \bull\: Amount \: of \: pure \: milk \: i n\: the \: solution \: = \: 60 \: \% \\ \\ \sf \: \bull \: Amount \: of \: water \: = \: 40 \: \% \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: Amount \: of \: pure \: milk \: added \: = \: 125 \: L\: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: New \: strength \: of \: the \: milk \: = \: 20 \: \%\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\color{lime} \bold{To \: find : \: Original \: quantity \: of \: the \: solution} \\ \\ \color{deeppink} \bold{Solution : } \\ \\ \sf \: Let \: the \: origina l\: quantity \: of \: the \: solution \: be \: x \: . \: \: \: \: \: \: \: \: \\ \\ \sf \: So, \: quantity \: \: milk \: in \: the \: new \: solution \: = \: 60\% \: of \: x \: = \: \frac{60}{100} \: \times \: x \: = \: \frac{3x}{5} \\ \\ \sf \: Quantity \: of \: milk \: in \: the \: pure \: milk \: solution \: = \: 125 \: L \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: So \: quantity \: of \: milk \: in \: the \: new \: solution \: = \: \frac{3x}{5} \: + \: 125 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: Total \: quantity \: of \: new \: solution \: = \: original \: quantity \: + \: 125 \: L \: = \: x \: + \: 125 \\ \\ \sf \: Percentage \:o f \: milk \:i n \: the \: new \: solution \: = \: \frac{Quantity \: of \: milk}{Total \: quantity} \: \times \: 100 \: \% \: \: \: \: \\ \\ \sf \: \frac{ \frac{3x}{5} \: + \: 125 }{x \: + \: 125} \: \times \: 100 \: \%$

$\sf \color{skyblue} \:This \: is \: equal \: to \: 20 \: \% \: \: (given) \\ \\ \sf \: 20 \: \% \: = \:\frac{ \frac{3x}{5} \: + \: 125 }{x \: + \: 125} \: \times \: 100 \: \% \\ \\ \color{aqua}\sf \: { \underline{ Cancel \: \% \: from \: both \:the \: side\: and \: transpose \: 100 \: to \: the \: left : }} \\ \\ \sf \: \frac{20}{100} \: = \: \frac{ \frac{3x}{5} \: + \: 125 }{x \: + \: 125} \\ \\ \sf \: \frac{1}{5} \: = \: \frac{ \frac{3x}{5} \: + \: 125 }{x \: + \: 125} \\ \\ \sf \: Cross \: Multiplying, \\ \\ \sf \: x \: + \: 125 \: = \: \left( \frac{3x}{5} \: + \: 125 \: \right) \: \times \: 5 \\ \\ \sf \: Solve \: the \: brackets, \: \\ \\ \sf \: x \: + \: 125 \: = \: 3x \: + \: 625 \\ \\ \sf \: Now, \: transpose \: x \: to \: the \: right \: and \: 625 \: to \: the \: left, \\ \\ \sf \: 125 \: - \: 625 \: = \: 3x \: - \: x \\ \\ \sf \: - 500 \: = \: 2x \\ \\ \sf \: \implies \: 2x \: = \: - 500 \\ \\ \sf \: x \: = \: \frac{ - 500}{2} \\ \\ \sf \: x \: = \: - 250 \\ \\ \sf \color{gold}\: Hence, \: the \: original \: quantity \: of \: the \: solution \: was \: 250 \: L . \: Negative \: shows \: that \: the \\ \\ \color{gold} \sf \: 125 \: L \: milk \: was \: removed \: for m\: the \: solution. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$