Question

In a milk shoppe there are three varieties of milk, 'Pure', 'Cure' and 'Lure'. The 'Pure' milk has \(100 \%\) concentration of milk. The ratio of milk is to water in the 'Cure' is \(2: 5\) and in the Lure it is \(3: 8\) respectively. Sonali purchased 14 litres of Cure and 22 litres of Lure milk and mixed them. If she wanted to make the concentration of milk in the mixture of purchased milk to \(50 \%\). How many litres of 'Pure' milk she is needed? (a) 6 litres (b) 8 litres (c) 16 litres (d) 18 litres

Short answer

Answer: Sonali needs to add approximately 16 litres of 'Pure' milk.

Step-by-step solution

Find out concentration of milk in 'Cure' and 'Lure' mixtures.

The ratio of milk to water in 'Cure' is 2:5, i.e., there are 2 litres of milk for every 5 litres of water. So, the concentration of milk in 'Cure' is: (2 / (2+5)) x 100 = (2/7) x 100 The ratio of milk to water in 'Lure' is 3:8, i.e., there are 3 litres of milk for every 8 litres of water. So, the concentration of milk in 'Lure' is: (3 / (3+8)) x 100 = (3/11) x 100

Find out the total concentration of milk in the mixture of 'Cure' and 'Lure' milk.

Sonali purchased 14 litres of 'Cure' milk and 22 litres of 'Lure' milk. Therefore, the total milk and water in the 'Cure' and 'Lure' mixtures are: Milk in 'Cure': 14 * (2/7) litres Water in 'Cure': 14 * (5/7) litres Milk in 'Lure': 22 * (3/11) litres Water in 'Lure': 22 * (8/11) litres Now to find the total milk and water in the mixture, we add the milk from both mixtures and do the same for the water: Total milk = (14 * (2/7)) + (22 * (3/11)) Total water = (14 * (5/7)) + (22 * (8/11))

Find out how many litres of 'Pure' milk Sonali needs to add.

Let x litres of 'Pure' milk be added to the mixture. Since 'Pure' milk has a concentration of 100%, x litres of milk will be added and no water. To achieve a 50% concentration of milk in the mixture, we have: ((Total milk + x) / (Total milk + Total water + x)) x 100 = 50 Solve the equation for x: x = ((Total milk + Total water) / 2) - Total milk Now, substituting the values of Total milk and Total water that we found in Step 2, we get: x = ((((14 * (2/7)) + (22 * (3/11))) + ((14 * (5/7)) + (22 * (8/11)))) / 2) - ((14 * (2/7)) + (22 * (3/11))) x ≈ 16 litres Sonali needs to add approximately 16 litres of 'Pure' milk to get the desired 50% concentration of milk in the mixture. So the correct answer is (c) 16 litres.

Key concepts

Understanding Ratios and Proportions

Ratios and proportions are foundational concepts used to describe the relationship between quantities. In mixture problems, we often come across situations where these concepts are vital. For example, the 'Cure' milk in our exercise contains a ratio of milk to water as 2:5, meaning for every 2 litres of milk, there are 5 litres of water. Understanding this ratio allows us to calculate the concentration of milk in the mixture.

A proportion, on the other hand, is an equation that states two ratios are equal. If we were to compare the concentration of milk in two different mixtures, we might say the proportion of milk in mixture one is equal to the proportion of milk in mixture two. Mastery of these ideas is key for solving problems involving mixtures by enabling us to represent relationships precisely and compute the desired concentrations.

Percentage Calculation in Mixtures

Percentage calculations are integral when dealing with mixture problems, as they provide a way to express concentrations. In the shopkeeper's problem, determining the concentration of milk in 'Cure' and 'Lure' involves converting the ratios into percentage forms. For instance, calculating the concentration of milk in 'Cure' to be \(\frac{2}{7} \times 100\%\) evaluates to approximately 28.57%. This tells us that 'Cure' contains 28.57% milk.

Similarly, when the goal is to reach a 50% concentration in the final mixture, we use percentage calculation to ascertain what amount of 'Pure' milk must be added. It's essential to be comfortable with converting ratios to percentages and vice-versa to effectively deal with these types of problems.

Using Algebraic Equations to Solve Mixture Problems

Algebraic equations form the backbone of solving mixture problems quantitatively. They allow us to set up relationships between different parts of a mixture and find unknown quantities. In the given problem, we are required to determine the amount of 'Pure' milk needed to achieve a specific concentration. We express this goal through an algebraic equation: \[\frac{{\text{Total milk} + x}}{{\text{Total milk} + \text{Total water} + x}} \times 100 = 50\%\]

By representing the variables and their relationships in a clear equation, we create a solvable expression that leads us to the answer after substitution and simplification. Learning to craft these equations accurately is a vital skill for students tackling mixture problems, as it is the key to unlocking the quantities they seek.