Total quantity of mixture = 80 L
Ratio of Milk & Water => \(\frac{M}{W}\) = \(\frac{4}{3}\)
4x+3x = 80
7x = 80
x = \(\frac{80}{7}\)
Quantity of Milk = 4x = \(\frac{320}{7}\)
Quantity of Water = 3x = \(\frac{240}{7}\)
10 Liters of mixture is replaced with water and 10 liters of mixture would contain:
Milk = \(\frac{4}{7}*10\) = \(\frac{40}{7}\)
Water = \(\frac{3}{7}*10\) = \(\frac{30}{7}\)
Milk and water remaining after replacement:
Milk = \(\frac{320}{7}-\frac{40}{7}\) = \(\frac{280}{7}\) = 40 L
Water = \(\frac{240}{7}-\frac{30}{7}+10\) = \(\frac{280}{7}\) = 40 L
New ratio of Milk and Water => \(\frac{M}{W}\) = \(\frac{1}{1}\)
Again, 10 Liters of mixture is replaced with water and 10 liters of mixture would contain:
Milk = \(\frac{1}{2}*10\) = \(5\)
Water = \(\frac{1}{2}*10\) = \(5\)
Milk and water remaining after replacement:
Milk = \(40-5\) = 35 L
Water = \(40-5+10\) = 45 L
New ratio of Milk and Water => \(\frac{M}{W}\) = \(\frac{7}{9}\)
And again, 10 Liters of mixture is replaced with water and 10 liters of mixture would contain:
Milk = \(\frac{7}{16}*10\) = \(\frac{35}{8}\)
Milk remaining after replacement:
Milk = \(35-\frac{35}{8}\) = \(\frac{245}{8}\) (A)
Bunuel chetan2u is there a quicker way to solve this question?