10 litres are removed from 80 litres of solution of milk and water hav

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?

Thank you for this simple solution

I have tried to solve this question using this formulae. Kindly check . KarishmaB

Since  after taking out 10 litres of solution , water is added , we will use the formulae from the milk's persepctive.

FInal amount of the milk = Initial amount  of the milk ( amount left after taking out 10 litres / total amount of the solution ) ^ 3

= 80* 4/ 7 * (( 80 - 10) / 80)^ 3

= 320 / 7 * (7/8)^3
=320/7 * 343 /512
= 49*5 / 8
= 245/8
KarishmaB Kindly check.­

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