Problem 60 - Algebra and Trigonometry

Question

Admission Fees The admission fee at an amusement park is \(\$ 1.50\) for children and \(\$ 4.00\) for adults. On a certain day, 2200 people entered the park, and the admission fees that were collected totaled \(\$ 5050 .\) How many children and how many adults were admitted?

Step-by-Step Solution

Answer

1500 children and 700 adults were admitted.

Step 1: Define Variables

Let \( c \) be the number of children and \( a \) be the number of adults admitted to the park. The problem provides us with two primary conditions to work with.

Step 2: Set Up Equations

Based on the problem statement, we establish the following two equations:1. The total number of people equation: \[ c + a = 2200 \]2. The total admission fee equation: \[ 1.5c + 4a = 5050 \]

Step 3: Solve One Equation for One Variable

Let's solve the first equation for \( c \):\[ c = 2200 - a \]

Step 4: Substitute in the Second Equation

Substitute \( c = 2200 - a \) into the second equation:\[ 1.5(2200 - a) + 4a = 5050 \]

Step 5: Simplify and Solve for \( a \)

Distribute and combine like terms:\[ 3300 - 1.5a + 4a = 5050 \]This simplifies to:\[ 3300 + 2.5a = 5050 \]Subtract 3300 from both sides:\[ 2.5a = 1750 \]Finally, divide by 2.5:\[ a = 700 \]

Step 6: Solve for \( c \)

Now that we know \( a = 700 \), substitute it back into the equation for \( c \):\[ c = 2200 - 700 \]Which gives:\[ c = 1500 \]

Step 7: Verify the Solution

Check both conditions with \( c = 1500 \) and \( a = 700 \):1. \( c + a = 1500 + 700 = 2200 \)2. \( 1.5 \times 1500 + 4 \times 700 = 2250 + 2800 = 5050 \)Both conditions are satisfied, confirming the solution.

Key Concepts

Algebra

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It's like a universal language for expressing mathematical ideas in a systematic way. In our amusement park example, we used algebra to form equations that represent the real-world situation—counting children and adults using the symbols \(c\) and \(a\).

Here are some highlights of algebraic techniques used in solving the problem:

• **Defining Variables:** This involves assigning letters to unknown quantities, like \(c\) for children and \(a\) for adults. It allows us to create concise mathematical expressions.
• **Equation Formulation:** Equations are built from word problems by translating the given conditions into mathematical statements. In our case, we translated the total number of people and fees into two equations.

These algebraic tools help us model and solve problems logically and are fundamental in fields ranging from basic mathematics to advanced scientific calculations.

Linear Equations

Linear equations are mathematical statements where the highest power of the variable is one, which makes them easy to work with. They often appear in the form \(ax + by = c\). In our example, linear equations helped us establish relationships between the number of people and their respective admission fees.

Let's dig into how linear equations were applied:

• **Two Equations Approach:** We began with two distinct linear equations: \(c + a = 2200\) and \(1.5c + 4a = 5050\). Each represents a different aspect of the problem—number count and total fees.
• **Substitution Method:** This method involves solving one of the equations for a single variable and substituting it into the other equation. It simplifies our two-equations system into one, making it easier to solve. Here, \(c\) was expressed in terms of \(a\) and substituted.

By breaking it down into manageable pieces, linear equations enable us to solve complex real-life financial and logistical problems efficiently.

Problem-Solving

Problem-solving in mathematics involves a series of strategic steps to find solutions to challenges, and it's as much about analytical thinking as it is about computation. The admission fees problem teaches us several problem-solving strategies:

• **Understand and Analyze the Problem:** Before jumping into equations, ensure you thoroughly understand the problem and identify what is being asked. In this case, the task was to determine the numbers of children and adults, given total admissions and fees.
• **Organize Information:** Break down the information offered in the problem into smaller parts, and think about how they connect to each other logically—like the total number of people and how they contribute to ticket pricing.
• **Verification of Solution:** Once a solution is found, always double-check by substituting numbers back into original conditions. We verified our solution to confirm both people count and fee collected match given values.

These strategies make solving complex equations approachable and open the door to tackling a multitude of real-world scenarios.