A newspaper carrier has $5.30 in change. He has ten more quarters than dimes but two times as many nickles as quarters. How many coins of each type does he has?
number of dimes --- x
number of quarters -- x+10
number of nickels = 2(x+10)
now for the "value":
10x + 25(x+10) + 5(2(x+10)) = 530
solve for x, then plug into my definitions.
To solve this problem, we can use algebraic equations to represent the given information.
Let's assume the number of dimes the newspaper carrier has is 'x'. Since he has ten more quarters than dimes, the number of quarters would be 'x + 10'. And since he has two times as many nickels as quarters, the number of nickels would be '2(x + 10)'.
Now let's assign values to the denominations:
The value of each dime is $0.10, so the total value of dimes is 0.10x.
The value of each quarter is $0.25, so the total value of quarters is 0.25(x + 10).
The value of each nickel is $0.05, so the total value of nickels is 0.05 * 2(x + 10).
According to the problem, the total change the newspaper carrier has is $5.30.
Therefore, we can set up the equation:
0.10x + 0.25(x + 10) + 0.05 * 2(x + 10) = 5.30
Now, let's solve this equation step by step:
0.10x + 0.25x + 2.50 + 0.10x + 1.00 = 5.30
0.45x + 3.50 = 5.30
0.45x = 1.80
x = 1.80 / 0.45
x = 4
So, the number of dimes the newspaper carrier has is 4. Now we can substitute this value back into our equations to find the number of quarters and nickels.
Number of quarters = x + 10 = 4 + 10 = 14
Number of nickels = 2(x + 10) = 2(4 + 10) = 28
Therefore, the newspaper carrier has 4 dimes, 14 quarters, and 28 nickels.