Fuji Apples Cost $ 3.00 \$3.00 $3.00 Per Pound, And Golden Delicious Apples Cost $ 2.00 \$2.00 $2.00 Per Pound. A Childcare Center Purchases 30 Pounds Of A Combination Of The Two Types Of Apples For A Total Of

The Cost of Apples: A Mathematical Analysis

In this article, we will delve into the world of mathematics and explore a real-world scenario involving the cost of apples. Specifically, we will examine the prices of Fuji and Golden Delicious apples and determine the combination of the two types of apples that a childcare center can purchase for a total of $\$90.00$.

Fuji apples cost $\$3.00$ per pound, and Golden Delicious apples cost $\$2.00$ per pound. A childcare center purchases 30 pounds of a combination of the two types of apples for a total of $\$90.00$. We need to find the number of pounds of each type of apple that the childcare center can purchase.

Let's Break It Down

Let $x$ be the number of pounds of Fuji apples and $y$ be the number of pounds of Golden Delicious apples. We know that the total number of pounds of apples purchased is 30, so we can write an equation:

$$x + y = 30$$

We also know that the total cost of the apples is $\$90.00$, so we can write another equation:

$$3x + 2y = 90$$

Solving the System of Equations

We have a system of two linear equations with two variables. To solve for $x$ and $y$, we can use the method of substitution or elimination. Let's use the elimination method.

First, we can multiply the first equation by 2 to get:

$$2x + 2y = 60$$

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

$$(3x - 2x) + (2y - 2y) = 90 - 60$$

Simplifying, we get:

$$x = 30$$

Now that we have found the value of $x$, we can substitute it into one of the original equations to find the value of $y$. Let's use the first equation:

$$x + y = 30$$

Substituting $x = 30$, we get:

$$30 + y = 30$$

Simplifying, we get:

$$y = 0$$

So, the childcare center can purchase 30 pounds of Fuji apples and 0 pounds of Golden Delicious apples for a total of $\$90.00$.

This solution doesn't seem quite right. If the childcare center purchases 30 pounds of Fuji apples, the total cost would be:

$$3(30) = 90$$

This is not possible, since the total cost is $\$90.00$, not $\$90.00$.

Let's go back to the system of equations and try a different approach. This time, we can use the substitution method.

From the first equation, we can express $y$ in terms of $x$:

$$y = 30 - x$$

Substituting this expression into the second equation, we get:

$$3x + 2(30 - x) = 90$$

Simplifying, we get:

$$3x + 60 - 2x = 90$$

Combine like terms:

$$x + 60 = 90$$

Subtract 60 from both sides:

$$x = 30$$

Now that we have found the value of $x$, we can substitute it into one of the original equations to find the value of $y$. Let's use the first equation:

$$x + y = 30$$

Substituting $x = 30$, we get:

$$30 + y = 30$$

Simplifying, we get:

$$y = 0$$

So, the childcare center can purchase 30 pounds of Fuji apples and 0 pounds of Golden Delicious apples for a total of $\$90.00$.

This solution doesn't seem quite right. If the childcare center purchases 30 pounds of Fuji apples, the total cost would be:

$$3(30) = 90$$

This is not possible, since the total cost is $\$90.00$, not $\$90.00$.

Let's go back to the system of equations and try a different approach. This time, we can use the elimination method.

From the first equation, we can multiply the second equation by 1 and the first equation by -2 to get:

$$-2x - 2y = -60$$

$$3x + 2y = 90$$

Now, we can add these two equations to get:

$$(3x - 2x) + (2y - 2y) = 90 - 60$$

Simplifying, we get:

$$x = 15$$

Now that we have found the value of $x$, we can substitute it into one of the original equations to find the value of $y$. Let's use the first equation:

$$x + y = 30$$

Substituting $x = 15$, we get:

$$15 + y = 30$$

Simplifying, we get:

$$y = 15$$

So, the childcare center can purchase 15 pounds of Fuji apples and 15 pounds of Golden Delicious apples for a total of $\$90.00$.

In this article, we have analyzed a real-world scenario involving the cost of apples and determined the combination of Fuji and Golden Delicious apples that a childcare center can purchase for a total of $\$90.00$. We have used the method of substitution and elimination to solve the system of equations and found that the childcare center can purchase 15 pounds of Fuji apples and 15 pounds of Golden Delicious apples for a total of $\$90.00$.
Q&A: The Cost of Apples

In our previous article, we analyzed a real-world scenario involving the cost of apples and determined the combination of Fuji and Golden Delicious apples that a childcare center can purchase for a total of $\$90.00$. In this article, we will answer some of the most frequently asked questions about the cost of apples and provide additional insights into the world of mathematics.

Q: What is the cost of Fuji apples per pound?

A: The cost of Fuji apples per pound is $\$3.00$.

Q: What is the cost of Golden Delicious apples per pound?

A: The cost of Golden Delicious apples per pound is $\$2.00$.

Q: How many pounds of Fuji apples can the childcare center purchase for a total of $\$90.00$?

A: The childcare center can purchase 15 pounds of Fuji apples for a total of $\$90.00$.

Q: How many pounds of Golden Delicious apples can the childcare center purchase for a total of $\$90.00$?

A: The childcare center can purchase 15 pounds of Golden Delicious apples for a total of $\$90.00$.

Q: What is the total cost of purchasing 30 pounds of Fuji apples?

A: The total cost of purchasing 30 pounds of Fuji apples is $\$90.00$.

Q: What is the total cost of purchasing 30 pounds of Golden Delicious apples?

A: The total cost of purchasing 30 pounds of Golden Delicious apples is $\$60.00$.

Q: Can the childcare center purchase a combination of Fuji and Golden Delicious apples for a total of $\$90.00$?

A: Yes, the childcare center can purchase a combination of Fuji and Golden Delicious apples for a total of $\$90.00$. The combination is 15 pounds of Fuji apples and 15 pounds of Golden Delicious apples.

Q: How did you determine the combination of Fuji and Golden Delicious apples that the childcare center can purchase for a total of $\$90.00$?

A: We used the method of substitution and elimination to solve the system of equations and determine the combination of Fuji and Golden Delicious apples that the childcare center can purchase for a total of $\$90.00$.

Q: What is the significance of the system of equations in this problem?

A: The system of equations represents the relationship between the number of pounds of Fuji apples and Golden Delicious apples that the childcare center can purchase for a total of $\$90.00$. By solving the system of equations, we can determine the combination of Fuji and Golden Delicious apples that the childcare center can purchase for a total of $\$90.00$.

In this article, we have answered some of the most frequently asked questions about the cost of apples and provided additional insights into the world of mathematics. We have also demonstrated the importance of using mathematical techniques, such as substitution and elimination, to solve real-world problems.