In a collection of nickels, dimes and quarters, there are twice as many dimes as nickels, and 4 fewer quarters than dimes. The total value of the coins is $5.75.
How many dimes are there in the collection ?
See previous post.
89
To solve this problem, we'll use algebraic equations. Let's break down the information given:
1. There are twice as many dimes as nickels: Let's represent the number of nickels as "n" and the number of dimes as "2n" because there are twice as many dimes.
2. There are 4 fewer quarters than dimes: We can represent the number of quarters as "2n - 4" because there are 4 fewer quarters than dimes.
3. The total value of the coins is $5.75: We need to calculate the total value of each coin and set it equal to $5.75.
The value of a nickel is $0.05, the value of a dime is $0.10, and the value of a quarter is $0.25.
Now let's set up the equation:
0.05n + 0.10(2n) + 0.25(2n - 4) = 5.75
Simplifying the equation:
0.05n + 0.20n + 0.50n - 1.00 = 5.75
Combining like terms:
0.75n - 1.00 = 5.75
Adding 1.00 to both sides:
0.75n = 6.75
Dividing both sides by 0.75:
n = 9
Now that we have the value for "n" (number of nickels), we can find the number of dimes.
Number of Dimes = 2n = 2(9) = 18
Therefore, there are 18 dimes in the collection.