A certain quantity of water is mixed with milk price at Rs. 48 per litre. The price of mixture is Rs. 30 per litre. The ratio of water and milk in the new mixture: (water is available free of cost)
3 : 5
Finding the Ratio of Water and Milk in a Mixture
This problem involves finding the ratio of two ingredients, water and milk, in a mixture given their individual prices and the price of the final mixture. Water is stated to be free of cost, meaning its price is Rs. 0 per litre. Milk has a price of Rs. 48 per litre. The resulting mixture is priced at Rs. 30 per litre.
Understanding the Problem Components
- Price of Water: Rs. 0 per litre
- Price of Milk: Rs. 48 per litre
- Price of Mixture: Rs. 30 per litre
- Goal: Find the ratio of Water : Milk in the mixture.
Method 1: Using the Rule of Alligation
The rule of alligation is a shortcut method used to find the ratio in which two ingredients at given prices should be mixed to produce a mixture of a desired price.
We place the price of the cheaper ingredient (Water) on the left and the price of the dearer ingredient (Milk) on the right. The mean price (Mixture) is placed in the centre. We then find the difference between the mean price and each ingredient's price diagonally.
| Price of Ingredients | Mean Price | ||
| Water (Cheaper) | Milk (Dearer) | Mixture | |
| Rs. 0 | Rs. 48 | Rs. 30 | |
| \( (48 - 30) \) | \( (30 - 0) \) | ||
| \( 18 \) | \( 30 \) | ||
The difference \( (48 - 30) = 18 \) gives the relative quantity of the cheaper ingredient (Water).
The difference \( (30 - 0) = 30 \) gives the relative quantity of the dearer ingredient (Milk).
The ratio of the quantity of the cheaper ingredient (Water) to the quantity of the dearer ingredient (Milk) is the ratio of these differences taken diagonally:
Ratio of Water : Milk = \( (48 - 30) : (30 - 0) \)
Ratio of Water : Milk = \( 18 : 30 \)
To simplify the ratio, we can divide both numbers by their greatest common divisor, which is 6.
\( 18 \div 6 = 3 \)
\( 30 \div 6 = 5 \)
So, the ratio of Water : Milk is \( 3 : 5 \).
Method 2: Algebraic Method
Let \( w \) be the quantity of water in litres and \( m \) be the quantity of milk in litres.
- Cost of \( w \) litres of water = \( w \times \text{Price of water} = w \times 0 = 0 \)
- Cost of \( m \) litres of milk = \( m \times \text{Price of milk} = m \times 48 = 48m \)
The total quantity of the mixture is \( (w + m) \) litres.
The total cost of the mixture is \( 0 + 48m = 48m \).
The price of the mixture per litre is given as Rs. 30.
The total cost of the mixture can also be expressed as the total quantity multiplied by the price per litre:
Total Cost = \( (w + m) \times 30 \)
Equating the two expressions for the total cost:
\[ 48m = 30(w + m) \]Divide both sides by 6:
\[ 8m = 5(w + m) \]Distribute the 5 on the right side:
\[ 8m = 5w + 5m \]Subtract \( 5m \) from both sides to isolate the terms with \( w \) and \( m \):
\[ 8m - 5m = 5w \] \[ 3m = 5w \]We want to find the ratio of water and milk, which is \( w : m \) or \( w/m \).
Rearrange the equation to find the ratio \( w/m \):
\[ \frac{w}{m} = \frac{3}{5} \]So, the ratio of Water : Milk is \( 3 : 5 \).
Conclusion
Both the Rule of Alligation and the algebraic method show that the ratio of water to milk in the mixture is \( 3 : 5 \).
The final answer is the ratio of water and milk, which is \( 3:5 \).
Mixture Problem Revision Table
| Component | Price per Litre (Rs.) | Quantity Ratio (from Calculation) |
|---|---|---|
| Water | 0 | 3 parts |
| Milk | 48 | 5 parts |
| Mixture | 30 | Total (3+5=8 parts) |
Additional Information on Mixture Problems
Mixture problems often involve combining two or more substances with different properties (like price, concentration, etc.) to form a mixture with a desired property. These problems can be solved using algebra or graphical methods like the Rule of Alligation.
- Rule of Alligation: This method is particularly useful when mixing two ingredients. It helps determine the ratio in which the ingredients should be mixed to achieve a specific mean value. The differences obtained diagonally represent the quantities of the components in the inverse ratio of the price differences from the mean.
- Concentration Problems: Similar to price-based problems, mixture problems can involve mixing solutions of different concentrations to get a mixture of a desired concentration. The same principles of weighted averages apply.
- Average Problems: Mixture problems are fundamentally applications of the concept of weighted averages. The price of the mixture is the weighted average of the prices of the ingredients, where the weights are the quantities of the ingredients.
Important Questions from To Make a Mixture from Two Mixtures
In a mixture of 75 liters, the ratio of milk to water is 3 : 2. If the ratio is to be 1 : 2, how much of water should be added?
A mixture contains acid and alcohol in the ratio of 3 : 2. On adding 10 litres of alcohol in mixture, the ratio of acid to alcohol becomes 3 : 5. The quantity of acid (in litres) in the original mixture was:
There are two containers Xand Y. Xcontains 100 ml of milk and Ycontains 100 ml of water. 20 ml of milk from Xis transferred to Y. After mixing well, 20 ml of the mixture in Yis transferred back to X. If mdenotes the proportion of milk in Xand ndenotes the proportion of water in Y, then which one of the following is correct?
Two vessels P and Q contain liquid A and liquid B in the ratio $4 : 3$ and $5 : 4$ respectively. In what ratio must the mixtures from vessel P and vessel Q be combined to obtain a new mixture in vessel R containing liquid A and liquid B in the ratio $11 : 8$?
In a vessel, a mixture of milk and water is in ratio $9 : 5$, while in another vessel mixture of milk and water is in ratio $3 : 8$. In what ratio mixture of both the vessels should be mixed together so that in the resultant mixture ratio of milk and water becomes $13 : 19$?
