Solving Coin Problems: Susan's Dimes and Nickels

Many real-world scenarios can be modeled and solved with algebra, especially when dealing with coins. This article will walk you through a classic problem involving dimes and nickels, providing a step-by-step solution and exploring the underlying mathematical concepts.

Introduction to Coin Problems

Questions involving mixed coins, such as dimes and nickels, are common in algebra classes. These problems often require setting up and solving systems of equations. This article will demonstrate how to find the number of dimes and nickels Susan has, given certain constraints.

The Problem

Let's start with the problem: Susan has $4.50 in dimes and nickels in her purse. She has 15 more dimes than nickels. How many of each type does she have?

Defining Variables

To solve this, let's define:

n as the number of nickels. The number of dimes will be n 15.

Setting Up the Equations

We know that one nickel is worth 5 cents and one dime is worth 10 cents. Therefore, the total value of the nickels is 5n cents, and the total value of the dimes is 10(n 15) cents. The sum of these values is $4.50, which is 450 cents. Hence, we can set up the equation:

5n   10(n   15)  450

Solving the Equation

Let's simplify this equation step-by-step:

Multiply out the terms:

5n   10n   150  450

Combine like terms:

15n   150  450

Subtract 150 from both sides:

15n  300

Divide by 15:

n  20

Now that we have the number of nickels, we can find the number of dimes:

Number of dimes n 15 20 15 35

Thus, Susan has:

20 nickels

35 dimes

Additional Considerations

For those curious about the value of the coins, let's consider the basic worth of each type. A dime is 10 cents, and a nickel is 5 cents. Let's break down the problem in an alternative method:

If we subtract the value of 15 dimes (1.50 dollars) from the total, we have 3.00 dollars left. If there were the same number of dimes as nickels, we could set up the following equation:

0.10d   0.05n  3.00

Given that there are 15 more dimes, we can denote the number of dimes as d n 15. Substituting d into the equation, we get:

0.10(n   15)   0.05n  3.00

Simplifying this equation:

0.10n 1.50 0.05n 3.00

0.15n 1.50 3.00

0.15n 1.50

n 10

d n 15 25

The alternative method leads to a different result, indicating a need for careful variable assignment and equation setup.

Conclusion

Understanding coin problems and solving them step-by-step with algebraic methods can be a valuable skill in both academic and real-world contexts. The key is to define variables, set up equations based on the given conditions, and solve systematically.

Additional Information

For those familiar with non-American currency systems, it's worth noting that different regions, such as Australia, use different coin denominations. In Australia, despite using a decimal currency system, they do not use the terms 'dime' and 'nickel'. Instead, they use '5 cent piece' and '10 cent piece'. The concepts remain applicable regardless of the regional differences in coinage terminology and denominations.