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Question

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)

The correct answer is

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.
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Important Questions from To Make a Mixture from Two Mixtures

  1. One cup has juice and water in the ratio 5 ∶ 2, while another cup of the same capacity has them in the ratio 7 ∶ 4, respectively. If contents of both the cups (when full) are poured in a vessel, then what will be the final ratio of water to juice in the vessel?

  2. A and B are solutions of acid and water. The ratios of water and acid in A and B are 4 : 5 and 1 : 2 respectively. If x liters of A is mixed with y liters of B, then the ratio of water and acid in the mixture becomes 8 : 13 What is x : y?

  3. A drink of chocolate and milk contains 8% pure chocolate by volume. If 10 litres of pure milk are added to 50 litres of this drink, the percentage of chocolate in the new drink is:

  4. Mixture A contains chocolate and milk in the ratio 4 ∶ 3 and mixture B contains chocolate and milk in the ratio 5 ∶ 2. A and B are taken in the ratio 5 ∶ 6 and mixed to form a new mixture. The percentage of chocolate in the new mixture is closest to:

  5. If 80 litres of milk solution has 60% milk in it, then how much milk should be added to make milk 80% in the solution?

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