Jillian Is Selling Boxes Of Cookies To Raise Money For Her Basketball Team. The 10 Oz. Box Costs $ 3.50 \$3.50 $3.50 , While The 16 Oz. Box Costs $ 5.00 \$5.00 $5.00 . At The End Of One Week, She Collected $ 97.50 \$97.50 $97.50 , Selling A Total Of 24 Boxes. The

Jillian's Cookie Sales: A Math Problem

Jillian is a young basketball player who is selling boxes of cookies to raise money for her team. She has two types of boxes available: a 10 oz. box that costs $\$3.50$ and a 16 oz. box that costs $\$5.00$. At the end of one week, Jillian collected a total of $\$97.50$ by selling a total of 24 boxes. In this article, we will use mathematical concepts to analyze Jillian's cookie sales and determine how many of each type of box she sold.

Let's denote the number of 10 oz. boxes sold as $x$ and the number of 16 oz. boxes sold as $y$. We know that the total number of boxes sold is 24, so we can write the equation:

$$x + y = 24$$

We also know that the total amount of money collected is $\$97.50$. Since the 10 oz. box costs $\$3.50$ and the 16 oz. box costs $\$5.00$, we can write the equation:

$$3.50x + 5.00y = 97.50$$

To solve this system of equations, we can use the method of substitution or elimination. Let's use the elimination method. We can multiply the first equation by 3.50 to get:

$$3.50x + 3.50y = 84.00$$

Now, we can subtract this equation from the second equation to get:

$$1.50y = 13.50$$

Dividing both sides by 1.50, we get:

$$y = 9$$

Now that we know the value of $y$, we can substitute it into the first equation to get:

$$x + 9 = 24$$

Subtracting 9 from both sides, we get:

$$x = 15$$

Jillian sold 15 boxes of 10 oz. cookies and 9 boxes of 16 oz. cookies. This means that she sold a total of 24 boxes, which is consistent with the information given in the problem. We can also calculate the total amount of money collected by multiplying the number of each type of box by its price and adding the results:

$$15 \times 3.50 = 52.50$$

$$9 \times 5.00 = 45.00$$

$$52.50 + 45.00 = 97.50$$

This confirms that Jillian collected a total of $\$97.50$.

This problem has real-world applications in business and finance. For example, a company that sells two types of products may use a similar system of equations to determine how many of each product to produce and sell. This can help the company to maximize its profits and minimize its costs.

When solving a system of equations, it's often helpful to use the elimination method. This involves multiplying one or both of the equations by a constant to eliminate one of the variables. In this problem, we multiplied the first equation by 3.50 to eliminate the variable $x$.

Try solving the following system of equations:

$$x + y = 20$$

$$2x + 3y = 40$$

To solve this system of equations, we can use the elimination method. We can multiply the first equation by 2 to get:

$$2x + 2y = 40$$

Now, we can subtract this equation from the second equation to get:

$$y = 0$$

Now that we know the value of $y$, we can substitute it into the first equation to get:

$$x + 0 = 20$$

Subtracting 0 from both sides, we get:

$$x = 20$$

In this article, we used mathematical concepts to analyze Jillian's cookie sales and determine how many of each type of box she sold. We also discussed real-world applications of this problem and provided tips and tricks for solving systems of equations. Finally, we provided a practice problem for readers to try.
Jillian's Cookie Sales: A Math Problem Q&A

In our previous article, we analyzed Jillian's cookie sales and determined how many of each type of box she sold. We also discussed real-world applications of this problem and provided tips and tricks for solving systems of equations. In this article, we will answer some frequently asked questions about Jillian's cookie sales.

Q: How did Jillian determine how many of each type of box to sell?

A: Jillian used a system of equations to determine how many of each type of box to sell. She knew that the total number of boxes sold was 24 and that the total amount of money collected was $\$97.50$. She also knew the prices of each type of box and used this information to set up a system of equations.

Q: What is the elimination method?

A: The elimination method is a technique used to solve systems of equations. It involves multiplying one or both of the equations by a constant to eliminate one of the variables. In Jillian's case, we multiplied the first equation by 3.50 to eliminate the variable $x$.

Q: How can I use this problem in real-world applications?

A: This problem has many real-world applications in business and finance. For example, a company that sells two types of products may use a similar system of equations to determine how many of each product to produce and sell. This can help the company to maximize its profits and minimize its costs.

Q: What are some tips and tricks for solving systems of equations?

A: Some tips and tricks for solving systems of equations include:

• Using the elimination method to eliminate one of the variables
• Multiplying one or both of the equations by a constant to eliminate one of the variables
• Using substitution to solve for one of the variables
• Checking your work to make sure that the solution is correct

Q: Can I use this problem to practice solving systems of equations?

A: Yes, you can use this problem to practice solving systems of equations. Try solving the system of equations:

$$x + y = 20$$

$$2x + 3y = 40$$

To solve this system of equations, we can use the elimination method. We can multiply the first equation by 2 to get:

$$2x + 2y = 40$$

Now, we can subtract this equation from the second equation to get:

$$y = 0$$

Now that we know the value of $y$, we can substitute it into the first equation to get:

$$x + 0 = 20$$

Subtracting 0 from both sides, we get:

$$x = 20$$

In this article, we answered some frequently asked questions about Jillian's cookie sales. We also provided tips and tricks for solving systems of equations and a practice problem for readers to try. We hope that this article has been helpful in understanding how to solve systems of equations and how to apply this concept to real-world problems.

If you are interested in learning more about systems of equations, we recommend checking out the following resources:

• Khan Academy: Systems of Equations
• Mathway: Systems of Equations
• Wolfram Alpha: Systems of Equations

These resources provide a wealth of information on systems of equations, including tutorials, examples, and practice problems.

Try solving the following system of equations:

$$x + y = 25$$

$$3x + 2y = 60$$

To solve this system of equations, we can use the elimination method. We can multiply the first equation by 3 to get:

$$3x + 3y = 75$$

Now, we can subtract this equation from the second equation to get:

$$-y = -15$$

Dividing both sides by -1, we get:

$$y = 15$$

Now that we know the value of $y$, we can substitute it into the first equation to get:

$$x + 15 = 25$$

Subtracting 15 from both sides, we get:

$$x = 10$$

In this article, we provided additional resources for learning about systems of equations and a practice problem for readers to try. We hope that this article has been helpful in understanding how to solve systems of equations and how to apply this concept to real-world problems.