Problem 272 - Elementary Algebra

Question

As the treasurer of her daughter's Girl Scout troop, Laney collected money for some girls and adults to go to a three-day camp. Each girl paid \(\$ 75\) and each adult paid \(\$ 30\). The total amount of money collected for camp was \(\$ 765 .\) If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?

Step-by-Step Solution

Answer

There are 3 adults and 9 girls.

Step 1: Define Variables

Let the number of adults be denoted by the variable \(a\). The problem states that the number of girls is three times the number of adults, so let the number of girls be denoted by \(g\), where \(g = 3a\).

Step 2: Set Up the Equation Based on Total Money Collected

Each girl paid \(\$75\) and each adult paid \(\$30\). The total amount collected is \(\$765\). Therefore, the equation based on the total money is: \[ 75g + 30a = 765 \]

Step 3: Substitute the Expression for Girls

Substitute the expression \(g = 3a\) into the equation: \[ 75(3a) + 30a = 765 \]

Step 4: Simplify the Equation

Simplify the left side of the equation: \[ 225a + 30a = 765 \] Combine like terms: \[ 255a = 765 \]

Step 5: Solve for the Number of Adults

To find the number of adults, divide both sides of the equation by 255: \[ a = \frac{765}{255} \] Therefore, \[ a = 3 \]

Step 6: Solve for the Number of Girls

Substitute \(a = 3\) back into the expression \(g = 3a\): \[ g = 3(3) \] Therefore, \[ g = 9 \]

Key Concepts

algebraic equations

Algebraic equations are mathematical statements that show the equality between two expressions. In this word problem, we need to form an algebraic equation to represent the relationship between the number of girls and adults and the money collected. To set up the equation, we must first define the variables. Let’s denote the number of adults by the variable \(a\). According to the problem, each girl (represented by \(g\), with \(g = 3a\)) paid \(\$75\), and each adult paid \(\$30\). The total amount collected is \(\$765\). Thus, our algebraic equation, representing the sum of money collected, will be: \[ 75g + 30a = 765 \]

variable substitution

Variable substitution is a method used to simplify algebraic equations. This involves replacing one variable with an equivalent expression containing another variable. In our exercise, we know that the number of girls \(g\) is three times the number of adults \(a\). So, we express this as \[ g = 3a \]. Now, we can substitute \(g\) in our main equation: \[ 75(3a) + 30a = 765 \]. This substitution reduces the complexity of the equation by eliminating one of the variables and expressing everything in terms of a single variable.

solving linear equations

Solving linear equations involves finding the value of unknown variables that make the equation true. After substituting \(g = 3a\) into our equation, we get: \[ 225a + 30a = 765 \]. Simplifying it further, we combine like terms to obtain: \[ 255a = 765 \]. Next, we solve for \(a\) by dividing both sides of the equation by 255: \[ a = \frac{765}{255} \] Hence, \[ a = 3 \]. To find the number of girls \(g\), we substitute \(a = 3\) back into the expression \[ g = 3a \], yielding: \[ g = 3(3) = 9 \]

real-world applications in math

Algebra isn't just an abstract part of math. It has real-world applications that allow you to solve everyday problems. In this example, algebra helps us determine the number of girls and adults who paid for the Girl Scout camp. Through creating and solving algebraic equations, we resolved how the collected money, divided based on participants' payments, relates to the counts of attendees. This problem shows how mathematical concepts are tangible and useful in managing finances, organizing events, and solving practical issues. Using algebraic techniques, such as defining variables and setting up equations, we can tackle a broad range of real-world scenarios.