Problem 62 - Algebra for College Students

Question

Set up an equation and solve each problem. A group of students agreed that each would chip in the same amount to pay for a party that would cost \(\$ 100\). Then they found 5 more students interested in the party and in sharing the expenses. This decreased the amount each had to pay by \(\$ 1\). How many students were involved in the party and how much did each student have to pay?

Step-by-Step Solution

Answer

Initially, there were 20 students each paying $5. With 5 more, each paid $4.

Step 1: Define the Variables

Let \( x \) be the initial number of students who agreed to chip in for the party, and \( y \) be the original amount each student was supposed to pay. We also know that the total cost of the party is \( \$100 \). So, the equation for the original setup is \( x \times y = 100 \).

Step 2: Set Up the Modified Equation

With 5 additional students, the total number of students becomes \( x + 5 \), and the amount each has to pay decreases by \( \$1 \). Therefore, the new amount each student has to pay is \( y - 1 \). The modified equation becomes \( (x + 5) \times (y - 1) = 100 \).

Step 3: Solve the System of Equations

We have two equations: \( x \, y = 100 \) and \((x + 5)(y - 1) = 100 \). Substituting \( y = \frac{100}{x} \) from the first equation into the second gives us \[ (x + 5) \left( \frac{100}{x} - 1 \right) = 100. \]Simplifying this equation gives:\[ (100 + 500/x - x - 5) = 100. \]This simplifies to \( 500/x - x = 5 \).

Step 4: Change to a Quadratic Equation

Multiply the entire equation by \( x \) to eliminate the fraction:\[ 500 - x^2 = 5x. \] Rearrange the terms to form a quadratic equation:\[ x^2 + 5x - 500 = 0. \]

Step 5: Solve the Quadratic Equation

We can solve \( x^2 + 5x - 500 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 1, b = 5, \) and \( c = -500 \).This gives:\[ x = \frac{-5 \pm \sqrt{25 + 2000}}{2}. \] Thus, \[ x = \frac{-5 \pm 45}{2}. \] This yields solutions \( x = 20 \) or \( x = -25 \). Since \( x \) represents the number of students, we take \( x = 20 \).

Step 6: Determine the Payment Amounts

Using \( x = 20 \) to find \( y \), substitute into the equation \( y = \frac{100}{x} \), which gives: \[ y = \frac{100}{20} = 5. \] So, initially each student was supposed to pay \( \\(5 \). With 5 more students, the new number of students is 25, making the new payment \[ y - 1 = 4. \] Therefore, the payment amount with extra students is \( \\)4 \).

Key Concepts

Quadratic Equations

Quadratic equations are very common in many mathematical problems. These equations have the general form of a polynomial equation:

• \(ax^2 + bx + c = 0\)

where \(a\), \(b\), and \(c\) are constants. A quadratic equation can have two solutions, which can be found using the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In the given problem, the equation \(x^2 + 5x - 500 = 0\) is a quadratic equation. To solve it, we use the quadratic formula:

• \(a = 1\), \(b = 5\), and \(c = -500\)
• Substitute these values into the formula
• Calculate the discriminant: \(b^2 - 4ac\)
• Find the roots using the plus-minus sign \(\pm\)

These steps help determine the possible values of \(x\). Here, \(x = 20\) is the logical solution, as negative students do not make sense.

System of Equations

A system of equations is a set of two or more equations that have common variables. The aim is to find the values of the variables that satisfy all the equations at the same time. In our word problem, we have two equations:

• \(xy = 100\)
• \((x + 5)(y - 1) = 100\)

The first equation is formed based on the initial setup where students share a fixed total cost. The second equation adjusts for additional students joining the group and sharing costs differently. Solving this system involves substituting one variable's expression derived from one equation into another equation, simplifying it, and solving further. Effectively, this method allows us to pair relationships and constraints into easier-to-manage terms.

Solving Equations

Solving equations involves finding the specific value(s) of the variable(s) that make an equation true. In this case, after forming a system of equations from the word problem, the next step is to solve them. By isolating one variable and expressing it in terms of the other variable, we can simplify equations easily. Substitution is key here:

• Start with \(xy = 100\), giving \(y = \frac{100}{x}\)
• Substitute \(y\) into the second equation: \((x + 5)(\frac{100}{x} - 1) = 100\)
• Simplify and multiply to clear fractions, resulting in a quadratic equation

Once a simplified quadratic or linear equation is achieved, techniques like factoring, quadratic formula, or simplification can be utilized to find the solutions for variables \(x\) and \(y\).

Variable Definition

Variable definition is a critical part of translating a word problem into mathematical equations. This step involves identifying and clearly naming the unknown quantities represented by variables. In this exercise:

• \(x\) is defined as the initial number of students contributing
• \(y\) is the original amount each student agreed to pay

Clear definition prevents confusion and aids in establishing relationships. For example, knowing that \(xy = 100\) ties the cost per student to the number of students, makes setting up equations straightforward. It also helps ensure that all aspects of the problem are represented in a way that leads to finding an effective solution. This clarity is pivotal as any complexity in understanding or translating the word problem can cause errors in solution development.