Problem 96 - Algebra for College Students

Question

Solve each problem. Traveling club. The members of a traveling club plan to share equally the cost of a \(\$ 150,000\) motorhome. If they can find 10 more people to join and share the cost, then the cost per person will decrease by \(\$ 1250 .\) How many members are there originally in the club?

Step-by-Step Solution

Answer

There are originally 34 members in the club.

Step 1: Define variables

Let the original number of members in the club be denoted as \( x \).

Step 2: Express the original cost per person

The cost per person originally is \( \frac{150,000}{x} \).

Step 3: Express the cost per person after 10 more people join

If 10 more people join, the total number of members becomes \( x + 10 \). The new cost per person is \( \frac{150,000}{x + 10} \).

Step 4: Set up the equation based on the cost decrease

According to the problem, the cost per person decreases by \$1250 when 10 more people join. Thus, we have the equation: \[ \frac{150,000}{x} - \frac{150,000}{x + 10} = 1250. \]

Step 5: Eliminate fractions by finding a common denominator

Multiply both sides of the equation by \( x(x + 10) \) to clear the denominators: \[ 150,000(x + 10) - 150,000x = 1250x(x + 10). \]

Step 6: Simplify the equation

Simplify the left-hand side and expand the right-hand side: \[ 150,000x + 1,500,000 - 150,000x = 1250x^2 + 1250x \] \[ 1,500,000 = 1250x^2 + 1250x. \]

Step 7: Solve the quadratic equation

Divide by 1250 to simplify: \[ 1200 = x^2 + x \] Rearrange to standard form: \[ x^2 + x - 1200 = 0. \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 1 \), and \( c = -1200 \): \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1200)}}{2(1)} \] \[ x = \frac{-1 \pm \sqrt{1 + 4800}}{2} \] \[ x = \frac{-1 \pm 69}{2}. \]

Step 8: Evaluate the solutions

Calculate the two potential solutions for \( x \): \[ x = \frac{68}{2} = 34 \] \[ x = \frac{-70}{2} = -35. \] Since the number of members cannot be negative, the valid solution is \( x = 34 \).

Key Concepts

quadratic equations

Quadratic equations are an essential part of algebra that involve finding the values of a variable that satisfy the equation. A quadratic equation is generally in the form of \[ ax^2 + bx + c = 0. \] In our exercise, we ended up with the quadratic equation \[ x^2 + x - 1200 = 0. \] To solve it, we used the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \) is the constant term. This formula helps us find the roots of the quadratic equation, which are the values of \( x \) that make the equation true. By substituting \( a = 1 \), \( b = 1 \), and \( c = -1200 \), we solved the quadratic equation and found that \( x = 34 \) is the number of original members in the traveling club.

cost-sharing problems

Cost-sharing problems often involve distributing a total cost among several participants. In our exercise, the members of a traveling club aimed to share the cost of a \(\textdollar 150,000\) motorhome equally. When tackling cost-sharing problems, it's helpful to:

• Define the total cost
• Identify the number of participants
• Formulate how the cost per person changes with the number of participants

The initial step was to assume an original number of members, denoted as \( x \). Each person initially paid \(\frac{150,000}{x}\). When 10 additional members joined, the number of participants became \( x + 10 \), and the new cost per person became \( \frac{150,000}{x + 10} \). We were given that this cost decreased by \(\textdollar 1250\), leading us to the equation: \[ \frac{150,000}{x} - \frac{150,000}{x + 10} = 1250. \] By solving this equation, we determined how the cost sharing impacts each member.

variable definition in equations

Defining variables is a critical first step in solving algebraic word problems. It simplifies complex scenarios into manageable equations. In our exercise, we defined \( x \) as the original number of members in the traveling club. Setting this variable allowed us to express the initial cost per person as \( \frac{150,000}{x} \) and the cost per person with new members as \( \frac{150,000}{x + 10} \). Clearly defining the variable gave us a foundation to set up our equation based on the decrease in cost per person. Always define your variables early. Ensure that each variable represents a specific element of the problem to avoid confusion as you work through the algebra.

common denominators in algebra

Finding a common denominator is a crucial technique in algebra for solving equations involving fractions. In our exercise, we had the equation: \[ \frac{150,000}{x} - \frac{150,000}{x + 10} = 1250. \] To solve this, we multiplied both sides by \( x(x + 10) \), the common denominator. This step eliminated the fractions and provided us with a clearer, more manageable equation: \[ 150,000(x + 10) - 150,000x = 1250x(x + 10). \] Simplifying this equation further led us to a quadratic equation. Remembering to find and use common denominators can significantly simplify solving equations involving fractions, making it easier to focus on solving for the variable.