Question

In a mixture of milk and water, there is only \(26 \%\) water. After replacing the mixture with 7 litres of pure milk, the percentage of milk in the mixture become \(76 \%\). The quantity of mixture is : (a) 65 litre (b) 91 litre (c) 38 litre (d) none of these

Short answer

Answer: (d) none of these (The initial quantity of the mixture is 84 litres)

Step-by-step solution

Set up the initial equation

Let x be the total quantity of the mixture. Since there is \(26 \%\) of water in the mixture, there is \(100\%-26\%=74\%\) of milk. So we have: \(0.26x\) - Quantity of water in the mixture \(0.74x\) - Quantity of milk in the mixture

Apply the replacement of 7 litres of pure milk

After 7 litres of pure milk is added, the total quantity of the mixture becomes \(x+7\). The proportion of milk in the new mixture is \(76\%\), which can be written as: \(0.76(x+7)\) Since the only change is the addition of 7 litres of pure milk, the quantity of water remains the same, \(0.26x\). Thus, we can write our second equation: \(0.74x + 7 = 0.76(x+7)\)

Solve the equation

To solve this equation, we first distribute the 0.76 on the right-hand side: \(0.74x + 7 = 0.76x + 5.32\) Now we can isolate x by moving variables to opposite sides of the equation: \(7 - 5.32 = 0.76x - 0.74x\) Simplifying, we get: \(1.68 = 0.02x\) Now, we can find the value of x by dividing both sides by 0.02: \(x = \frac{1.68}{0.02} = 84\) So, the original quantity of the mixture is 84 litres. The correct answer is not among the options provided, which means that the correct option is: (d) none of these

Key concepts

Percentage

Understanding percentages is crucial in solving mixture problems because it helps us determine the ratio of different components in a solution or mixture. In this exercise, we began by identifying that the milk and water mixture contained 26% water. Since percentages represent parts out of 100, it's easy to convert these into expressions. For example:

• 26% water can be expressed as \(0.26\).
• The remaining percentage for milk is given as \(100\% - 26\% = 74\%\), or \(0.74\).

When the percentage changes after adding or removing components (like adding pure milk), these percentages help us create equations to solve for unknown quantities.

Equation Solving

Once we know the percentages, we can set up equations to solve for the quantities involved. In this exercise, we first labeled the total quantity of the mixture as \(x\). Next, we found the initial amounts of milk and water in terms of \(x\):

• Water: \(0.26x\)
• Milk: \(0.74x\)

After the mixture is altered by adding 7 litres of pure milk, the percentage of milk in the mixture becomes 76%. This change lets us derive a new equation:

• New milk quantity: \(0.74x + 7\)
• Equation: \(0.74x + 7 = 0.76(x + 7)\)

To solve this equation, we first distribute \(0.76\) through the bracket, leading to \(0.76x + 5.32\). By moving terms across the equation, we isolate \(x\), leading us to \(x = 84\). This shows how logical operations are applied in an orderly manner to solve real-world problems.

Milk and Water Problems

This particular exercise highlights a common type of mixture problem involving milk and water percentages. One typical characteristic of such problems is the need to adjust for changes when ingredients are added or removed. Here, despite changing the component with the addition of pure milk, knowing the quantity of initial water (which remains unchanged) was fundamental in finding a solution. Milk and water problems usually involve:

• Determining initial ratios or percentages.
• Changing one or more components of the mixtures.
• Calculating new percentages based on the alterations.

In problems like ours, understanding the unchanged quantity (here, water) can significantly simplify solving the problem. By keeping track of what's added or stays the same, you can arrive more quickly at an accurate solution.