In what ratio must tea worth of Rs. 10 per kg be mixed with tea worth Rs. 15 per kg, so that the resultant mixture costs Ra.12 per kg?
3 ∶ 2
This question asks about mixing two types of tea with different costs to achieve a specific target cost for the resulting mixture. We need to find the ratio in which the two types of tea should be mixed.
Understanding the Mixture Problem
We have two ingredients (tea types) with known costs per unit weight and a desired cost per unit weight for the final mixture. We need to determine the proportion or ratio of the weights of the two ingredients to be mixed.
Applying the Rule of Alligation
This type of problem can be efficiently solved using the Rule of Alligation. This rule helps in finding the ratio in which two ingredients at given prices should be mixed to produce a mixture of a desired price.
Let's define the terms:
- Cost of the cheaper ingredient (C)
- Cost of the dearer ingredient (D)
- Mean price of the mixture (M)
According to the Rule of Alligation, the ratio of the quantity of the cheaper ingredient to the quantity of the dearer ingredient is given by the difference between the dearer price and the mean price, and the difference between the mean price and the cheaper price.
Ratio of Quantity of Cheaper : Quantity of Dearer = (D − M) ∶ (M − C)
Solving the Tea Mixture Problem
In this problem:
- Cost of the cheaper tea (C) = Rs. 10 per kg
- Cost of the dearer tea (D) = Rs. 15 per kg
- Desired cost of the mixture (M) = Rs. 12 per kg
We can visualize this using the alligation diagram:
| Quantity | Price per kg | |
| Cheaper Tea | 10 | |
| Mixture | 12 | |
| Dearer Tea | 15 |
Now, calculate the differences diagonally:
- Difference between Dearer Price and Mean Price = D − M = 15 − 12 = 3
- Difference between Mean Price and Cheaper Price = M − C = 12 − 10 = 2
The ratio of the quantities of the cheaper tea to the dearer tea is the ratio of these differences, but in the opposite direction on the diagram.
Quantity of Cheaper Tea ∝ (D − M)
Quantity of Dearer Tea ∝ (M − C)
So, the ratio of Cheaper Tea : Dearer Tea = (D − M) ∶ (M − C) = (15 − 12) ∶ (12 − 10) = 3 ∶ 2.
| Cheaper Tea (10 Rs/kg) | Dearer Tea (15 Rs/kg) | |
| Mixture (12 Rs/kg) | ||
| Difference (15 − 12) = 3 | Difference (12 − 10) = 2 |
The quantities are mixed in the ratio 3 ∶ 2.
- Quantity of tea at Rs. 10 per kg is proportional to 3.
- Quantity of tea at Rs. 15 per kg is proportional to 2.
Thus, the two varieties of tea must be mixed in the ratio 3 ∶ 2.
Revision Table: Key Concepts
| Concept | Explanation |
|---|---|
| Mixture Problems | Questions involving combining two or more items with different properties (like price, concentration) to form a mixture with a desired property. |
| Rule of Alligation | A shortcut method to find the ratio in which two ingredients at given prices or concentrations must be mixed to obtain a mixture of a desired mean price or concentration. It's derived from the weighted average concept. |
| Ratio in Alligation | The ratio of the quantity of the cheaper item to the dearer item is in inverse proportion to the differences between their prices and the mean price. Specifically, QuantityCheaper ∶ QuantityDearer = (PriceDearer − Mean Price) ∶ (Mean Price − PriceCheaper). |
Additional Information: Weighted Average Approach
The rule of alligation is a specific application of the concept of weighted average. Let's say we mix quantity \(Q_1\) of tea with price \(P_1\) and quantity \(Q_2\) of tea with price \(P_2\). The total cost of the mixture will be \(Q_1 P_1 + Q_2 P_2\). The total quantity will be \(Q_1 + Q_2\). The mean price of the mixture (M) is the total cost divided by the total quantity:
\(M = \frac{Q_1 P_1 + Q_2 P_2}{Q_1 + Q_2}\)
If we assume \(P_1\) is the cheaper price (Rs. 10) and \(P_2\) is the dearer price (Rs. 15), and the mean price is Rs. 12, then:
\(12 = \frac{Q_1 \times 10 + Q_2 \times 15}{Q_1 + Q_2}\)
\(12(Q_1 + Q_2) = 10 Q_1 + 15 Q_2\)
\(12 Q_1 + 12 Q_2 = 10 Q_1 + 15 Q_2\)
\(12 Q_1 - 10 Q_1 = 15 Q_2 - 12 Q_2\)
\(2 Q_1 = 3 Q_2\)
To find the ratio \(Q_1 : Q_2\), we rearrange the equation:
\(\frac{Q_1}{Q_2} = \frac{3}{2}\)
So, \(Q_1 : Q_2 = 3 : 2\). This confirms the result obtained using the Rule of Alligation. The ratio is Quantity of tea at Rs. 10 : Quantity of tea at Rs. 15, which is 3 : 2.
Important Questions from Mixture Problems
A mixture of acid and water contains 20 percent acid. When 10 litres of water is added to the mixture, then the percentage of acid becomes 15 percent. What is the original quantity of mixture ?
A container contains 20 L mixture in which there is 10% sulphuric acid. Find the quantity of sulphuric acid to be added in it to make the solution to contain 25% sulphuric acid.
The ratio of milk to water in a 100 litres mixture is 2 ∶ 3. 10 litres of this mixture is withdrawn and replaced with milk. This process is repeated 2 more times, What is the percentage of milk in final mixture ?
- 80% and 90% pure acid solutions are mixed to obtain 20 litres of 87% pure acid solution. Find the quantity (in litres) of 80% pure acid solution taken to form the mixture.
Some fruits are bought at a rate of 11 for Rs. 100 and an equal number at a rate of 9 for Rs. 100. If all the fruits are sold at a rate of 10 for Rs. 100, then what is the gain or loss percent in the entire transaction?
