Calvin Has 13 Coins, All Of Which Are Quarters Or Nickels. The Coins Are Worth $\$2.45$. How Many Of Each Coin Does Calvin Have?

Introduction

Calvin has 13 coins, a mix of quarters and nickels, totaling $\$2.45$. The task at hand is to determine the number of each type of coin Calvin possesses. This problem requires a combination of algebraic thinking and logical reasoning. In this article, we will delve into the world of mathematics and explore the solution to Calvin's coin conundrum.

The Problem

Let's denote the number of quarters as $q$ and the number of nickels as $n$. We know that Calvin has a total of 13 coins, so we can write the equation:

$$q + n = 13$$

We also know that the total value of the coins is $\$2.45$. Since quarters are worth $\$0.25$ and nickels are worth $\$0.05$, we can write the equation:

$$0.25q + 0.05n = 2.45$$

Solving the System of Equations

To solve this system of equations, we can use the method of substitution or elimination. Let's use the elimination method to eliminate one of the variables.

First, we can multiply the first equation by 0.05 to make the coefficients of $n$ in both equations equal:

$$0.05q + 0.05n = 0.65$$

Now, we can subtract this equation from the second equation to eliminate $n$:

$$(0.25q + 0.05n) - (0.05q + 0.05n) = 2.45 - 0.65$$

Simplifying the equation, we get:

$$0.20q = 1.80$$

Dividing both sides by 0.20, we get:

$$q = 9$$

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

$$q + n = 13$$

Substituting $q = 9$, we get:

$$9 + n = 13$$

Subtracting 9 from both sides, we get:

$$n = 4$$

Conclusion

Calvin has 9 quarters and 4 nickels. This solution satisfies both equations and meets the given conditions. The elimination method was used to solve the system of equations, and the values of $q$ and $n$ were found to be 9 and 4, respectively.

The Importance of Algebraic Thinking

This problem requires algebraic thinking and logical reasoning. The use of variables and equations allows us to represent the problem mathematically and solve it systematically. The elimination method is a powerful tool for solving systems of equations, and it is essential to understand how to apply it in various contexts.

Real-World Applications

This problem has real-world applications in finance, economics, and other fields where mathematical modeling is used to solve problems. The ability to solve systems of equations is a fundamental skill that is essential in many areas of mathematics and science.

Tips and Tricks

When solving systems of equations, it is essential to:

• Use the elimination method to eliminate one of the variables.
• Multiply both equations by the same value to make the coefficients of one of the variables equal.
• Subtract one equation from the other to eliminate the variable.
• Check the solution by substituting the values back into the original equations.

By following these tips and tricks, you can solve systems of equations efficiently and effectively.

Conclusion

Introduction

In our previous article, we solved the problem of Calvin's coin conundrum, where he had 13 coins, a mix of quarters and nickels, totaling $\$2.45$. We used the elimination method to solve the system of equations and found that Calvin had 9 quarters and 4 nickels. In this article, we will answer some frequently asked questions related to this problem.

Q: What is the total value of the quarters and nickels?

A: The total value of the quarters is $9 \times \$0.25 = \$2.25$. The total value of the nickels is $4 \times \$0.05 = \$0.20$. Therefore, the total value of the coins is $\$2.25 + \$0.20 = \$2.45$.

Q: How many more quarters does Calvin have than nickels?

A: Calvin has 9 quarters and 4 nickels. Therefore, he has $9 - 4 = 5$ more quarters than nickels.

Q: What is the ratio of quarters to nickels?

A: The ratio of quarters to nickels is $9:4$. This can be simplified to $9/4$ or $2.25:1$.

Q: How many more nickels would Calvin need to have a total of $\$3.00$?

A: To find the number of nickels needed to reach a total of $\$3.00$, we can subtract the total value of the quarters from $\$3.00$. This gives us $\$3.00 - \$2.25 = \$0.75$. Since each nickel is worth $\$0.05$, we can divide $\$0.75$ by $\$0.05$ to find the number of nickels needed: $\$0.75 / \$0.05 = 15$. Therefore, Calvin would need 15 more nickels to have a total of $\$3.00$.

Q: What if Calvin had 15 nickels instead of 4? How many quarters would he have?

A: If Calvin had 15 nickels, the total value of the nickels would be $15 \times \$0.05 = \$0.75$. Since the total value of the coins is still $\$2.45$, we can subtract the total value of the nickels from $\$2.45$ to find the total value of the quarters: $\$2.45 - \$0.75 = \$1.70$. Since each quarter is worth $\$0.25$, we can divide $\$1.70$ by $\$0.25$ to find the number of quarters: $\$1.70 / \$0.25 = 6.8$. Since we cannot have a fraction of a coin, we can round down to 6 quarters. Therefore, Calvin would have 6 quarters if he had 15 nickels.

Q: Can we use the elimination method to solve this problem?

A: Yes, we can use the elimination method to solve this problem. We can multiply the first equation by 0.05 to make the coefficients of $n$ in both equations equal:

$$0.05q + 0.05n = 0.65$$

Then, we can subtract this equation from the second equation to eliminate $n$:

$$(0.25q + 0.05n) - (0.05q + 0.05n) = 2.45 - 0.65$$

Simplifying the equation, we get:

$$0.20q = 1.80$$

Dividing both sides by 0.20, we get:

$$q = 9$$

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

$$q + n = 13$$

Substituting $q = 9$, we get:

$$9 + n = 13$$

Subtracting 9 from both sides, we get:

$$n = 4$$

Therefore, Calvin has 9 quarters and 4 nickels.

Conclusion

In this article, we answered some frequently asked questions related to Calvin's coin conundrum. We used the elimination method to solve the system of equations and found that Calvin had 9 quarters and 4 nickels. We also explored some alternative scenarios and used the elimination method to solve them. By understanding how to solve systems of equations, you can develop essential skills that are applicable in many areas of mathematics and science.