Problem 30 - Algebra 2

Question

ACTINIIIES The band has 30 more members than the school chorale. If each group had 10 more members, the ratio of their membership would be \(3 : 2\) . How many members are in each group?

Step-by-Step Solution

Answer

Chorale: 50 members, Band: 80 members.

Step 1: Define Variables

Let \( x \) be the number of members in the school chorale. Therefore, the number of members in the band is \( x + 30 \) because the band has 30 more members than the chorale.

Step 2: Formulate the Equation for 10 More Members

If each group had 10 more members, the number of chorale members would be \( x + 10 \) and the number of band members would be \( x + 40 \). The problem states that this ratio would be \( 3 : 2 \), so we have the equation: \( \frac{x + 40}{x + 10} = \frac{3}{2} \).

Step 3: Cross-Multiply to Solve

Cross-multiply the equation from step 2 to eliminate the fraction:\[ 2(x + 40) = 3(x + 10) \] which simplifies to \[ 2x + 80 = 3x + 30 \].

Step 4: Isolate the Variable

Subtract \( 2x \) from both sides to get:\[ 80 = x + 30 \].Then, subtract 30 from both sides to solve for \( x \):\[ x = 50 \].

Step 5: Find the Number of Members in Each Group

Since \( x = 50 \), the chorale has 50 members. The band has \( x + 30 = 80 \) members.

Key Concepts

Ratio and Proportion

In algebra, ratios and proportions are a way to compare two quantities. In this problem, the concept of ratio is key to understanding how the band and school chorale groups compare to each other. A ratio of \(3:2\) means that for every 3 members in the band, there are 2 members in the chorale. Ratios simplify the relationship between quantities into simple terms that are easy to understand.

Proportion, on the other hand, takes this comparison a step further by setting two ratios equal to each other. This equality allows us to solve problems by using known quantities to find missing ones. In our problem, the stated future membership figures result in a ratio that is expressed as a proportion: \(\frac{x+40}{x+10} = \frac{3}{2}\). Using proportionate relationships gives us a reliable method to solve algebraic equations involving ratios.

Variables and Equations

The use of variables and equations is fundamental in solving algebra word problems. Variables act as placeholders for unknown values that we need to discover. In this exercise, we define \( x \) as the number of members in the chorale.

Once we establish a relationship with another group, like the band, equations become essential tools. We express the band members in terms of \( x + 30 \), since the band has 30 more members than the chorale. By formulating equations, we systematically lay out a plan to solve for the unknowns, relying on mathematical relationships and operations.

Cross-Multiplication

Cross-multiplication is a useful technique for solving equations involving fractions, typically found in problems involving proportions. By cross-multiplying, we eliminate the fractions and simplify the process of finding solutions.

In our example, the equation \(\frac{x + 40}{x + 10} = \frac{3}{2}\) can be resolved using cross-multiplication. This involves multiplying across the equals sign diagonally: \[2(x + 40) = 3(x + 10)\]. This action clears the fraction and creates a simpler equation that we can solve by applying basic algebraic principles to isolate the variable.

With the fraction gone, solving the equation becomes straightforward. Cross-multiplication transforms a complex problem into a manageable task, providing clarity and direction for finding the solution.