Lesson 4.3: Mixtures, Averages, and Weighted Averages

Introduction

In this lesson, students, we will explore the concepts of mixtures, averages, and weighted averages, which are essential for solving quantitative reasoning problems on the GMAT. Understanding these concepts will help you translate word problems into mathematical expressions and find the solution systematically.

Learning Objectives

• Understand mixture and concentration problems.
• Differentiate between simple averages and weighted averages.
• Utilize the weighted-average relationship effectively.
• Solve mixture and weighted-average problems.
• Recognize when a weighted average shortcut is applicable.

Mixture and Concentration Problems

Mixture problems often involve blending two or more substances and determining the overall characteristics of the resulting mixture. These problems can appear in various real-world contexts, such as calculating concentrations in chemistry or mixing ingredients in cooking.

Example 1: Simple Mixture Problem

Suppose you have 5 liters of a 30% salt solution, and you want to mix it with 3 liters of a 70% salt solution to find the concentration of the resulting mixture.

1. Calculate the amount of salt in each solution:

• From the first solution: $5 \text{ liters} \times 0.30 = 1.5 \text{ liters of salt}$
• From the second solution: $3 \text{ liters} \times 0.70 = 2.1 \text{ liters of salt}$

1. Total amount of salt in the mixture:

$$ \text{Total salt} = 1.5 + 2.1 = 3.6 \text{ liters of salt} $$

1. Total volume of the mixture:

$$ \text{Total volume} = 5 + 3 = 8 \text{ liters} $$

1. Concentration of the resulting mixture:

$$ \text{Concentration} = \frac{\text{Total salt}}{\text{Total volume}} = \frac{3.6}{8} = 0.45 \text{ or } 45\% $$

Thus, the resulting mixture has a concentration of 45% salt.

Common Misconceptions

One common misconception is to directly average the percentages of the individual solutions without considering their volumes. This mistake can lead to inaccurate results. It's crucial to calculate the total amounts first and then find the average based on those totals.

Averages and Weighted Averages

Next, let’s distinguish between simple averages and weighted averages. A simple average is calculated by summing all values and dividing by the number of values. In contrast, a weighted average takes into account the varying significance of different values.

Simple Average Formula

The simple average can be computed using the formula:

$$\text{Simple Average} = \frac{x_1 + x_2 + ... + x_n}{n}$$

where $x_i$ are the individual values and $n$ is the number of values.

Example 2: Simple Average

Consider the following test scores: 70, 80, 90.

1. Calculate their average:

$$ \text{Simple Average} = \frac{70 + 80 + 90}{3} = \frac{240}{3} = 80 $$

Weighted Average Formula

The weighted average is computed using the formula:

$$\text{Weighted Average} = \frac{w_1x_1 + w_2x_2 + ... + w_nx_n}{w_1 + w_2 + ... + w_n}$$

where $w_i$ are the weights associated with each $x_i$.

Example 3: Weighted Average

Suppose you receive two grades: a test score of 90 that counts for 40% of your grade and a project score of 80 that counts for 60%.

1. Determine the weighted average:

$$ \text{Weighted Average} = \frac{(0.40 \times 90) + (0.60 \times 80)}{0.40 + 0.60} $$

$$ = \frac{36 + 48}{1.00} = 84 $$

Thus, your overall grade would be 84.

When to Use Weighted Averages

Not all situations require the calculation of a weighted average. You should apply it when the individual contributions to the average are not equal in significance. If all contributions are treated equally, a simple average suffices.

Solving Mixture and Weighted-Average Problems

Now, let's tackle problems that require a blend of various principles discussed.

Example 4: Mixture Problem with Multiple Ingredients

Imagine you have one solution that is 20% sugar and another that is 50% sugar. You want to create 10 liters of a solution that is 30% sugar. How many liters of each solution do you need?

1. Let $x$ be the volume of the 20% solution:

• Therefore, $10 - x$ will be the volume of the 50% solution.

1. Set up the equation for sugar content:

• Amount of sugar from the 20% solution: $0.20x$
• Amount of sugar from the 50% solution: $0.50(10 - x)$
• Total sugar in the final solution: $0.30 \times 10 = 3$ liters
• Thus, the equation is:

$$ 0.20x + 0.50(10 - x) = 3 $$

1. Solve the equation:

$$ 0.20x + 5 - 0.50x = 3 $$

$$ -0.30x + 5 = 3 $$

$$ -0.30x = -2 $$

$$ x = \frac{2}{0.30} \approx 6.67 \text{ liters} $$

1. Determine the volumes of each solution:

• Volume of the 20% solution: $6.67$ liters
• Volume of the 50% solution: $10 - 6.67 \approx 3.33$ liters

Conclusion

In this lesson, students, we have covered the fundamental concepts of mixtures, simple averages, and weighted averages. Understanding how to approach these problems step-by-step allows you to tackle a wide range of quantitative reasoning questions effectively. Remember, the key to solving these problems is setting up a clean mathematical representation and applying the appropriate formulas and relationships.

Study Notes

• Mixture problems involve combining substances and calculating resultant concentration.
• Use total amounts for proper concentrations rather than directly averaging percentages.
• Simple average formula: $ \frac{x_1 + x_2 + ... + x_n}{n} $
• Weighted average formula: $ \frac{w_1x_1 + w_2x_2 + ... + w_nx_n}{w_1 + w_2 + ... + w_n} $
• Weighted averages are used when contributions to the average vary in significance.
• A clear setup is crucial for success in mixture and weighted average problems.