Question

From a container, 6 litres of milk was drawn out and was replaced by water. Again 6 litres of mixture was drawn out and replaced by the water. Thus the quantity of milk and water in the container after these two operations is 9:16. The quantity of mixture is
A.15
B.16
C.25
D.31

Answer 1

Hint: Consider the original quantity of a liquid as ‘x’ and a quantity of ‘y’ units is taken out from ‘x’. So the remaining original liquid will be \[x - y\] . Now consider \[x - y\] as ‘m’ and a quantity of ‘y’ units is taken from ‘m’ where the original quantity of the liquid becomes \[m - \dfrac{{m\left( y \right)}}{x}\] which is \[\left( {x - y} \right) - \dfrac{{\left( {\left( {x - y} \right)y} \right)}}{x}\] as m=x-y. As per the given question, the original quantity of the milk changes when the water is added to it. The sum of the quantities of the milk and the water gives the total quantity of the mixture. Consider \[\left( {x - y} \right) - \dfrac{{\left( {\left( {x - y} \right)y} \right)}}{x}\] as the quantity of the milk , ‘x’ as the quantity of the mixture and y=6 as 6 litres of water is drawn out.

Complete step-by-step answer:
We are given that a container contains milk and 6 litres of milk is replaced with water. Again 6 litres of this mixture is replaced with water. And after these two actions, the quantities of milk and water are 9, 16 respectively. We need to find the whole quantity of the mixture.
As per the question given, the initial quantity of the milk= Quantity of the container.
Let us say quantity of the container and the milk is ‘x’. $ \to eq(1) $
6 litres of milk is replaced with water.
Quantity of the milk will become \[x - 6\] .
And again 6 litres of this mixture is replaced with water.
Quantity of the milk will become \[\left( {x - 6} \right) - \dfrac{{\left( {\left( {x - 6} \right)6} \right)}}{x}\] $ \to eq(2) $
And given that the ratio of the quantities of the milk and water is \[9:16\]
The sum of the quantities of the milk and the water gives the total quantity of the mixture.
So, the ratio of the quantity of the milk and the mixture is \[9:9 + 16 = 9:25\]
As the quantity of drawn out and replaced liquids is the same the original quantity of the container will not change which means the quantity of the mixture is still ‘x’ from eq(1).
And the quantity of milk after drawing and replacing it is \[\left( {x - 6} \right) - \dfrac{{\left( {\left( {x - 6} \right)6} \right)}}{x}\] from eq(2)
Finally, 9:25=Quantity of milk: Quantity of mixture
 \[\dfrac{9}{{25}} = \dfrac{{\left( {x - 6} \right) - \dfrac{{\left( {\left( {x - 6} \right)6} \right)}}{x}}}{x}\]
Taking out $ x - 6 $ common from RHS
 \[
  \dfrac{9}{{25}} = \dfrac{{\left( {x - 6} \right)\left( {1 - \dfrac{6}{x}} \right)}}{x} \\
  \dfrac{9}{{25}} = \dfrac{{x\left( {1 - \dfrac{6}{x}} \right)\left( {1 - \dfrac{6}{x}} \right)}}{x} \\
  \dfrac{9}{{25}} = \left( {1 - \dfrac{6}{x}} \right)\left( {1 - \dfrac{6}{x}} \right) \\
  \dfrac{9}{{25}} = {\left( {1 - \dfrac{6}{x}} \right)^2} \\
  {\left( {\dfrac{3}{5}} \right)^2} = {\left( {1 - \dfrac{6}{x}} \right)^2} \\
  \dfrac{3}{5} = \left( {1 - \dfrac{6}{x}} \right) \\
 \]
Send $ \dfrac{3}{5} $ to RHS and $ \dfrac{6}{x} $ to LHS
 $
  \dfrac{6}{x} = 1 - \dfrac{3}{5} \\
  \dfrac{6}{x} = \dfrac{{5 - 3}}{5} \\
  \dfrac{6}{x} = \dfrac{2}{5} \\
  x = \dfrac{{6 \times 5}}{2} \\
  x = \dfrac{{30}}{2} \\
  x = 15 \\
  $
Therefore, the quantity of the mixture is 15 litres.
So, the correct answer is “Option A”.

Note: Finally we can derive a formula for the quantity of milk remained which is
 \[x{\left( {1 - \dfrac{y}{x}} \right)^n}\] , where ‘y’ is the quantity of the milk drawn out and ‘n’ is the no. of times the milk is drawn out. The formula is valid when the quantity of milk drawn out is equal throughout ‘n’ actions.