Problem 205 - 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 3 -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 \( a \). Since the number of girls is three times the number of adults, let the number of girls be denoted by \( g = 3a \).

Step 2: Set Up the Equation for Total Money Collected

The total money collected from girls is given by \( 75g \) and from adults is given by \( 30a \). The sum of these amounts is the total money collected: \[ 75g + 30a = 765 \]

Step 3: Substitute the Expression for Girls

Since \( g = 3a \), substitute this into the equation: \[ 75(3a) + 30a = 765 \]

Step 4: Simplify the Equation

Simplify the terms to get a single equation in terms of \( a \): \[ 225a + 30a = 765 \]This simplifies to \[ 255a = 765 \]

Step 5: Solve for the Number of Adults

Divide both sides of the equation by 255 to solve for \( a \): \[ a = \frac{765}{255} \]So \[ a = 3 \]

Step 6: Calculate the Number of Girls

Using the relationship \( g = 3a \), substitute \( a \) back in to find the number of girls: \[ g = 3(3) = 9 \]

Key Concepts

linear equations

Linear equations are fundamental in algebra. They are equations of the first degree, meaning they involve variables raised to the power of one. Think of them as straight lines when plotted on a graph.
In our exercise, the problem involves summing payments to reach a total amount. We form a linear equation by identifying constant rates (75 dollars per girl and 30 dollars per adult) and combining them to match the collected total (765 dollars).
To write it simply, a linear equation looks like this: x + y = c,
where x and y are variables, and c is a constant. Here, our equation is: 75g + 30a = 765.
Understanding how to set up these equations is essential because they allow us to solve for unknown values by manipulating the equation through algebraic operations.

substitution method

The substitution method is a technique to solve systems of linear equations. You solve one equation for one variable, then substitute that solution into another equation.
In this exercise, we were given the relationship g = 3a (the number of girls is three times the number of adults).
Here's what we did:

• First, we expressed one variable (number of girls) in terms of the other (number of adults): g = 3a
• We then took this expression and substituted it into the total money equation: 75(3a) + 30a = 765.
  This substitution allows us to work with a single equation in one variable, simplifying the solving process. Finally, as shown in the steps, we simplified and solved for a
• , then used that value to find g.

The substitution method is effective because it reduces complex systems into simpler, more manageable parts.

problem-solving steps

Solving algebra problems systematically can help. Always follow clear steps:

• Step 1: Define Variables: Identify what each part of the problem represents. Here, a represents adults and g represents girls.
• Step 2: Form the Equations: Use information given to create equations. (Total payments: 75g + 30a = 765.)
• Step 3: Substitution: When one variable is defined in terms of another, substitute it into the equation ( g = 3a into the first equation).
• Step 4: Simplify & Solve: Simplify the new equation to isolate and solve for the remaining variable. (225a + 30a = 765 to find a.)
• Step 5: Back-Substitute: Replace the solved variable to find other quantities ( g = 3a).

Following these steps meticulously can help solve similar algebra problems.
It ensures every aspect of the problem is addressed logically and accurately.