Solving Coin Problems with Algebra: A Case Study with Dimes and Nickels
Have you ever come across a mathematical problem involving coins with different denominations? In this article, we will walk through a detailed step-by-step solution to a classic coin problem, using algebra and systems of equations. We will use the example of James, who has 157 coins in his jar, consisting of 5-cent and 10-cent coins, with a total value of 1000 cents.
The Problem at Hand
James has a jar filled with 5-cent and 10-cent coins, with a total of 157 coins. The total value of the coins is 1000 cents. Let's denote the number of 5-cent coins as (x) and the number of 10-cent coins as (y).
Setting Up the Equations
We can create a system of equations based on the given information:
(x y 157) (5x 10y 1000)Notice that the second equation can be simplified by dividing everything by 5:
(x 2y 200)
Solving the System of Equations
Next, we can solve the system of equations. We start by subtracting the first equation from the second equation:
(x 2y - (x y) 200 - 157) (y 43)With (y 43), we can use the first equation to find (x):
(x 43 157) (x 157 - 43 114)
So, James has 114 5-cent coins and 43 10-cent coins.
Alternative Solution Method
Another way to solve this problem is to calculate the number of 10-cent coins if there were all 10-cent coins:
(157 times 0.10 15.70)
Since the total value is 1000 cents, there's a difference of 5.70 cents (because 15.70 - 10 5.70). This difference is due to the replacement of 10-cent coins with 5-cent coins. Therefore, the number of 5-cent coins needed is:
(5.70 / 0.05 114)
Subtracting the number of 5-cent coins from the total number of coins gives:
(157 - 114 43)
This confirms that James has 43 10-cent coins and 114 5-cent coins.
Verification
To verify the solution, we can check both the total number of coins and the total value:
Total number of coins: (114 43 157) Total value: (114 times 0.05 43 times 0.10 5.70 4.30 10.00) (1000 cents)Conclusion
This problem demonstrates how algebraic equations can be used to solve real-world problems involving money and coins. By setting up and solving a system of equations, or using alternative methods like consistent calculations, we can find the exact number of different types of coins in a given situation.