Question

A 20 litre mixture of milk and water contains milk and water in the ratio 3:2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of milk and water in the resultant mixture?

Answer 1

Hint:
First we will find the amount of water and milk in the original mixture with the help of the ratio that is given. Then, we will find out what amount of milk and water will be removed if 10 litres of the mixture is removed. We will find the quantity of water and milk left in the new mixture by subtracting the amount removed from the original amount. We will repeat this process once again as the replacement is done twice. We will find the ratio of milk and water in the resultant mixture by dividing the amount of milk and water in the final mixture.

Complete step by step solution:
If the ratio of 2 things is $$a:b$$and the total amount of the mixture is $$T$$, then the amount of 1st thing is given by $${\displaystyle \frac{a}{a+b}}\times T$$ and the amount of the second thing is given by $${\displaystyle \frac{b}{a+b}}\times T$$.
The total amount of the mixture is 20 litres. Let’s find the amount of milk in the original mixture and call it $$Q_m$$ . We will substitute 3 for $$a$$, 2 for $$b$$ and 20 for $$T$$ in the formula.
$$\begin{array}{l}{Q}_m={\displaystyle \frac{3}{3+2}}\times 20\\ \Rightarrow Q_m={\displaystyle \frac{3}{5}}\times 20\\ \Rightarrow Q_m=12\end{array}$$
The original mixture has 12 litres of milk.
Let’s find the amount of water in the original mixture and call it $$Q_w$$ . We will substitute 3 for $$a$$, 2 for $$b$$ and 20 for $$T$$ in the formula.
$$\begin{array}{l}{Q}_w={\displaystyle \frac{2}{2+3}}\times 20\\ \Rightarrow Q_w={\displaystyle \frac{2}{5}}\times 20\\ \Rightarrow Q_w=8\end{array}$$
The original mixture has 8 litres of water.
10 litres of mixture is removed and replaced by 10 litres of pure milk. Milk and water will be removed in the ratio 3:2. Let’s find the amount of milk removed and call it $${Q_m}^{{}^{\prime }}$$. We will substitute 3 for $$a$$, 2 for $$b$$ and 10 for $$T$$ in the formula.

$$\begin{array}{l}{Q_m}^{{}^{\prime }}={\displaystyle \frac{3}{3+2}}\times 10\\ \Rightarrow {Q_m}^{{}^{\prime }}={\displaystyle \frac{3}{5}}\times 10\\ \Rightarrow {Q_m}^{{}^{\prime }}=6\end{array}$$
6 litres of milk is removed.
Let’s find the amount of water removed and call it $${Q_w}^{{}^{\prime }}$$. We will substitute 3 for $$a$$, 2 for $$b$$ and 10 for $$T$$ in the formula.
$$\begin{array}{l}{Q_w}^{{}^{\prime }}={\displaystyle \frac{2}{3+2}}\times 10\\ \Rightarrow {Q_w}^{{}^{\prime }}={\displaystyle \frac{2}{5}}\times 10\\ \Rightarrow {Q_w}^{{}^{\prime }}=4\end{array}$$
4 litres of water is removed.
The removed mixture is replaced with 10 litres of pure milk.
Let’s find the remaining amount of milk in the original mixture. We will subtract the amount of milk removed from the amount of milk present in the mixture and add the amount of milk added to the mixture.
$$12-6+10=16$$
16 litres of milk is left.
Let’s find the remaining amount of water in the original mixture. We will subtract the amount of water removed from the amount of water present in the mixture.
$$8-4=4$$
4 litres of water is left.
Let’s find the ratio of milk and water in the new mixture. We will divide the amount of milk by the amount of water and cancel common factors.
$$\begin{array}{l}{\displaystyle \frac{16}{4}}={\displaystyle \frac{4\times 4}{4}}\\ \\ \Rightarrow {\displaystyle \frac{4}{1}}\\ \\ \Rightarrow 4:1\end{array}$$
The process is repeated again. 10 litres of the mixture will be removed again.
Milk and water will be removed in the ratio 4:1. Let’s find the amount of milk removed and call it $${Q_m}^{{}^{″}}$$. We will substitute 4 for $$a$$, 1 for $$b$$ and 10 for $$T$$ in the formula.
$$\begin{array}{l}{Q}_m^{{}^{″}}={\displaystyle \frac{4}{4+1}}\times 10\\ \Rightarrow Q_m^{{}^{″}}={\displaystyle \frac{4}{5}}\times 10\\ \Rightarrow Q_m^{{}^{″}}=8\end{array}$$
8 litres of milk is removed.
Let’s find the amount of water removed and call it $$Q_w^{{}^{″}}$$. We will substitute 4 for $$a$$, 1 for $$b$$ and 10 for $$T$$ in the formula.
$$\begin{array}{l}{Q}_w^{{}^{″}}={\displaystyle \frac{1}{4+1}}\times 10\\ \Rightarrow Q_w^{{}^{″}}={\displaystyle \frac{1}{5}}\times 10\\ \Rightarrow Q_w^{{}^{″}}=2\end{array}$$
2 litres of water is removed.
The removed mixture is replaced with 10 litres of pure milk.
 Let’s find the remaining amount of milk in the original mixture. We will subtract the amount of milk removed from the amount of milk present in the mixture and add the amount of milk added to the mixture.
$$\begin{array}{l}16-8+10\\ \Rightarrow 18\end{array}$$
18 litres of milk is left.
Let’s find the remaining amount of water in the original mixture. We will subtract the amount of water removed from the amount of water present in the mixture
$$\begin{array}{l}4-2\\ \Rightarrow 2\end{array}$$
2 litres of water is left.
Let’s find the ratio of milk and water in the new mixture. We will divide the remaining amount of water by the remaining amount of milk and cancel common factors.
$$\begin{array}{l}{\displaystyle \frac{18}{2}}\\ \Rightarrow {\displaystyle \frac{9\times 2}{2}}\\ \\ \Rightarrow {\displaystyle \frac{9}{1}}\\ \Rightarrow 9:1\end{array}$$

$$\therefore$$ The ratio of milk and water in the resultant mixture is 9:1.

Note:
We need to take care that the order in which the ratio is given should be followed. If the ratio of milk and water is given as 3:2, we have to take milk as 3 out of 5 parts and water as 2 out of 5 parts and not vice-versa. We should also be careful that when some amount of mixture (say 10 litres) is removed, water and milk will be removed in the same ratio in which they are present in the mixture. We cannot assume that only 10 litres of water or 10 litres of milk will be removed.