REPARTO PROPORCIONAL SIMPLE *INVERSO*

Understanding Proportional Distribution

Introduction to Proportional and Inverse Distribution

• The session begins with an introduction to solving problems related to proportional and inverse distribution, emphasizing the increasing difficulty of problems as the lesson progresses.

• The first example involves distributing 300 in inverse proportion to the values 15, 12, and 10. The goal is to find the smallest part among three parts labeled a, b, and c.

Converting Inverse to Direct Proportion

• To solve the problem, the instructor explains that they will convert inverse proportions into direct proportions by inverting the indices: 1/15, 1/12, 1/10 .

• The next step involves multiplying these fractions by their least common multiple (LCM), which needs to be determined for effective calculation.

Finding Least Common Multiple (LCM)

• A traditional method is used to find the LCM of 15, 12, and 10. The instructor demonstrates this through factorization.

• After calculating, it is established that the LCM is 60. This value will be used for further calculations involving parts a, b, and c.

Solving for Parts

• By substituting back into equations derived from direct proportions (a = 4k, b = 5k, c = 6k), they sum up to equal 300.

• Ultimately, it’s calculated that part a equals 80 when k is found to be 20; thus confirming it as the smallest part.

Example Two: Distributing Different Fractions

Setting Up Example Two

• In this example, participants are tasked with distributing 1260 inversely proportional to fractions 1/4, 1/7, 1/10 . They need to identify which part is largest.

Conversion Process

• Similar steps are taken where inverses are converted directly: 4, 7, and 10. No LCM calculation is necessary since there are no fractions involved here.

Calculating Parts Again

• Each part can now be expressed as multiples of k: a = 4k, b = 7k, c = 10k. Summing these gives us an equation equating them with total distribution (1260).

Finding Largest Part

• Upon solving for k (which equals 60), it's concluded that part c represents the largest portion at a value of 600.

Example Three: Roots in Distribution

Introduction of Cubic Roots

• The third example introduces cubic roots where participants must distribute a total of 620 inversely proportional to cube roots of numbers like 16, 54, and 250 while finding out which part is smallest.

Establishing Relationships

• As none of these numbers have exact cube roots readily available for simple calculations, factors need extraction from each number before proceeding with conversions.

Factor Extraction Methodology

• Factors such as extracting cube rootable components from each number are discussed—e.g., breaking down numbers like using factors such as 8, 27, or 125.

Simplifying Expressions

• All expressions can then be simplified by dividing or multiplying through by common terms like cube root values ensuring all parts remain consistent throughout calculations.

Understanding Proportional Relationships in Problem Solving

Simplifying Proportions

• The discussion begins with simplifying values 2, 3, and 5 to establish a direct proportional relationship. The speaker emphasizes the need to invert these indices for proper calculation.

• The minimum common multiple (LCM) of the numbers is identified as 30, which serves as a basis for further calculations. This step is crucial for ensuring accurate proportional relationships.

• A traditional method is suggested for finding the LCM, demonstrating how to simplify fractions derived from the original values (1/2, 1/3, and 1/5).

Establishing Constants

• After simplification, constants are introduced: a = 15, b = 10k, and c = 6k. These constants will be used in subsequent calculations to find specific values.

• The speaker calculates that when substituting back into the equation, one finds that the smallest part corresponds to a value of 20.

Example Problem: Distributing Money Among Children

• A new example involves distributing $480 among three children based on their academic performance (number of failures). It raises questions about whether distribution should be directly or inversely proportional.

• The conclusion drawn is that more failures result in less money received; thus, this scenario represents an inverse relationship between failures and monetary distribution.

Calculating Shares Based on Failures

• Each child's share is calculated based on their number of failures (2, 3, and 6). To solve this problem effectively, it’s necessary to convert these ratios into direct proportions by inverting them.

• The LCM of the new set of fractions (1/2, 1/3, and 1/6) is determined as 6. This helps facilitate further calculations regarding each child’s share.

Final Calculation Steps

• Using established constants and sums leads to determining that 6k = 480, allowing us to find k = 80.