TWIN QUESTION: https://gmatclub.com/forum/10-litres-ar ... 14164.html
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Initial quantity of milk= 80*(4/7)=320/7
Quantity after three successive removal=\((320/7)*(1-10/80)^3\)
=245/8
A:)

satya2029
How does this formula work?

If the 10 litres is removed and replaced with milk then?

Quantity after three successive removal=\((320/7)*(1+10/80)^3\) ??

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?