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Chapter 4: Problem 40

A charity fund-raiser consisted of a spaghetti supper where a total of 387 people were fed. They charged 6.80 dollars for adults and half-price for children. If they took in 2444.60 dollars, find how many adults and how many children attended the supper.

Short Answer

Expert verified
332 adults and 55 children attended the supper.

Step by step solution

01

Define Variables

Let's denote the number of adults who attended the supper as \( x \) and the number of children who attended as \( y \).
02

Set Up Equations

First, set up the equation for the total number of people: \( x + y = 387 \). This equation represents the total number of attendees. Next, set up the equation for the total revenue: \( 6.80x + 3.40y = 2444.60 \). This equation represents the total amount collected from adults and children, where children are charged half of the adult price.
03

Solve the System of Equations

We now have the system of equations:1. \( x + y = 387 \)2. \( 6.80x + 3.40y = 2444.60 \)Solve the first equation for \( y \):\[ y = 387 - x \]Substitute this into the second equation:\[ 6.80x + 3.40(387 - x) = 2444.60 \]
04

Simplify the Equation

Simplify the equation:\[ 6.80x + 1315.80 - 3.40x = 2444.60 \]Combine terms:\[ 3.40x + 1315.80 = 2444.60 \]
05

Solve for x

Subtract 1315.80 from both sides:\[ 3.40x = 1128.80 \]Divide both sides by 3.40 to solve for \( x \):\[ x = \frac{1128.80}{3.40} = 332 \]
06

Solve for y

Substitute \( x = 332 \) back into \( y = 387 - x \):\[ y = 387 - 332 = 55 \]
07

Check the Solution

Verify that the values of \( x = 332 \) and \( y = 55 \) satisfy both original equations:- Total people: \( 332 + 55 = 387 \)- Total revenue: \( 6.80 \times 332 + 3.40 \times 55 = 2257.60 + 187 = 2444.60 \)Both equations are satisfied, confirming the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebra
Algebra is a branch of mathematics that uses symbols and letters to represent numbers in equations and formulas. It is a powerful tool for solving problems as it allows us to understand relationships between different quantities.

In this exercise, algebra is used to solve a problem involving two variables: the number of adults (\( x \)) and the number of children (\( y \)) who attended a charity fundraiser. Solving such problems involves setting up and solving equations that represent the relationships described in the problem.

We start by creating two equations. One that captures the total number of attendees, (\( x + y = 387 \)), and another that represents the total revenue, (\( 6.80x + 3.40y = 2444.60 \)). This approach helps in structuring the problem so we can find a solution efficiently.
Word Problems
Word problems often involve real-life situations where we use language to describe mathematical concepts. These problems require us to carefully interpret the words to set up mathematical equations.

In the provided scenario, the key phrases are "total of 387 people," "charged 6.80 dollars for adults," and "half-price for children." Understanding these phrases enables us to transform them into mathematical language, crafting specific equations that represent the situation.

It's essential to read the problem a few times and identify important details before setting up the equations. These problems test our ability to translate everyday scenarios into mathematical terms, showcasing the practical applications of algebra.
Revenue Calculation
Revenue calculation is an important concept in business and finance where we determine the total income generated from selling goods or services. In algebraic word problems like this one, revenue calculation helps to find the relationships between different components involved, such as the number of items sold and the price per item.

The fundraiser problem utilizes revenue calculation by multiplying the number of adult tickets sold by their price (\( 6.80 \)) and adding it to the number of children's tickets sold multiplied by their half-price (\( 3.40 \)). The equation (\( 6.80x + 3.40y = 2444.60 \)) represents this total revenue.

Understanding how to calculate revenue is essential for solving business-related problems and making informed decisions. In practice, setting up these types of equations allows us to predict income or to analyze the effectiveness of pricing strategies.

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Most popular questions from this chapter

Wayne Osby blends coffee for a local coffee caf茅. He needs to prepare 200 pounds of blended coffee beans selling for 3.95 dollars per pound. He intends to do this by blending together a high-quality bean costing 4.95 dollars per pound and a cheaper bean costing 2.65 dollars per pound. To the nearest pound, find how much high-quality coffee bean and how much cheaper coffee bean he should blend.

Joan Gundersen rented a car from Hertz, which rents its cars for a daily fee plus an additional charge per mile driven. Joan recalls that a car rented for 5 days and driven for 300 miles cost her 178 dollars, while a car rented for 4 days and driven for 500 miles cost 197 dollars. Find the daily fee, and find the mileage charge.

Solve each system by substitution. When necessary, round answers to the nearest hundredth. \(\left\\{\begin{array}{l}{x+y=-15.2} \\ {-2 x+5 y=-19.3}\end{array}\right.\)

Twice last month, Judy Carter rented a car from Enterprise in Fresno, California, and traveled around the Southwest on business. Enterprise rents its cars for a daily fee, plus an additional charge per mile driven. Judy recalls that her first trip lasted 4 days, she drove 450 miles, and the rental cost her 240.50 dollars. On her second business trip she drove 200 miles in 3 days, and paid 146.00 dollars for the rental. Find the daily fee and the mileage charge.

Write a system of three linear equations in three variables that has \((2,1,5)\) as a solution. (There are many possibilities, Explain the process you used to write an equation.
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