Jim has some nickels, dimes and quarters worth $5.30. There are five times as many dimes as quarters, and six more nickels than dimes. How many of each kind of coin does he have?
5D = Q
D + 6 = N
.05N + .10D + .25Q = 5.30
Substitute terms.
.5(D+6) + .10D + .25(5D) = 5.30
Solve for D, then N and Q.
To solve this problem, let's assign variables to each kind of coin.
Let q represent the number of quarters.
Since there are five times as many dimes as quarters, let d represent the number of dimes, which would be 5q.
And since there are six more nickels than dimes, let n represent the number of nickels, which would be 5q + 6.
Now we can set up an equation to represent the total value of the coins:
0.25q + 0.10d + 0.05n = 5.30
Let's substitute the expressions for d and n into the equation:
0.25q + 0.10(5q) + 0.05(5q + 6) = 5.30
Simplifying the equation:
0.25q + 0.50q + 0.25q + 0.30 = 5.30
Combining like terms:
1.00q + 0.30 = 5.30
Subtracting 0.30 from both sides:
1.00q = 5.00
Dividing both sides by 1.00:
q = 5
Now that we know the number of quarters, we can find the number of dimes and nickels:
d = 5q = 5(5) = 25
n = 5q + 6 = 5(5) + 6 = 31
So, Jim has 5 quarters, 25 dimes, and 31 nickels.