A23L
B27L
C30L
D48L
Answer:
A. 23L
Read Explanation:
Understanding Mixture Problems in Competitive Exams
Mixture and Alligation is a very common topic in quantitative aptitude sections of competitive exams, testing a candidate's ability to work with ratios, percentages, and basic algebra.
The core idea is to combine different substances (like milk and water, or different solutions) and determine the properties of the resulting mixture.
Step-by-Step Solution Breakdown
Initial State: Jar A initially contains only pure milk. Let its volume be X Litres. Since it's pure milk, the water content is 0 Litres.
Analyzing the Added Mixture: A 27 Litre mixture of milk and water is added. The ratio of milk to water in this mixture is 4:5.
To find the quantity of milk in this 27 L mixture: Milk = (Ratio of Milk / Sum of Ratios) * Total Volume = (4 / (4 + 5)) * 27 = (4/9) * 27 = 12 Litres.
To find the quantity of water in this 27 L mixture: Water = (Ratio of Water / Sum of Ratios) * Total Volume = (5 / (4 + 5)) * 27 = (5/9) * 27 = 15 Litres.
Calculating the Components of the New Mixture in Jar A:
Total Milk: The original milk in Jar A (X Litres) + milk from the added mixture (12 Litres) = (X + 12) Litres.
Total Water: The original water in Jar A (0 Litres) + water from the added mixture (15 Litres) = 15 Litres.
Total Volume of New Mixture: Sum of total milk and total water = (X + 12) + 15 = (X + 27) Litres.
Formulating the Equation Based on Final Percentage: It is given that the new mixture contains 70% milk.
The percentage of milk is calculated as (Total Milk / Total Volume of Mixture) * 100%.
So, (X + 12) / (X + 27) = 70 / 100.
Cross-multiplying: 100 * (X + 12) = 70 * (X + 27).
Expanding the equation: 100X + 1200 = 70X + 1890.
Rearranging to solve for X: 100X - 70X = 1890 - 1200.
30X = 690.
X = 690 / 30 = 23 Litres.