Question

From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

Answer 1

Assume the capacity of the container is ‘c’
and track down the milk which is retained.
The milk retained after taking first 9 litres from the container is \((c × c−\frac{9}{c })\)

The milk retained after taking next 9 litres from the container is \(( c × c−\frac{9}{c}× c−\frac{9}{c})\)
The volumes of milk and water in the container are now in the ratio of 16 : 9.
The milk in the container is \(\frac{16}{25}\)
\(\frac{c×\frac{c−9}{c}×\frac{c−9}{c}}{c} = \frac{16}{25}\)

\((c−\frac{9}{c} )^2 = \frac{16}{25}\)

\(c−\frac{9}{c}= \frac{45}{45}\)

c = 45

So, the answer is 45 liters.

Answer 2

Let initial volume be V, final be F for milk.
The formula is given by: \(F = V. (1-\frac{K}{V})^n\) n is the number of times the milk is drawn and replaced.
so \(F = V (1-\frac{K}{V})^2\)
Given, K =9
So, \(\frac{16}{25} V=V(1-\frac{9}{V})^2\) 

\(1-\frac{9}{V} = \frac{4}{5} \space Or -\frac{4}{5}\)

\(1- \frac{9}{V}=-\frac{4}{5}\)

V = 5, but 9 liters is drawn every time.
Hence, \(1-\frac{9}{V}=\frac{4}{5}\)
V=45 liters

So, the answer is 45 liters.