A jar contains $34.75 in quarters and dollar coins. If the number quarters is three more than four times the number of dollars, determine the number of each type of coin.
let's see what they say:
.25q + 1.00d = 34.75
q = 3+4d
Now just plug in q and you have
.25(3+4d) + d = 34.75
.75 + 2d = 34.75
d = 17
so, q = 3+4d = 71
To solve this problem, we can create a system of equations based on the given information:
Let's represent the number of quarters as "q" and the number of dollar coins as "d".
1. We know that the value of the quarters is $0.25 each. So, the total value of the quarters is 0.25q.
2. We also know that the value of the dollar coins is $1 each. So, the total value of the dollar coins is 1d.
3. According to the problem, the total value of the coins in the jar is $34.75. Therefore, we can form the equation:
0.25q + 1d = 34.75
4. Additionally, the problem states that the number of quarters is three more than four times the number of dollars. Mathematically, this can be represented as:
q = 4d + 3
Now, we have a system of two equations:
Equation 1: 0.25q + 1d = 34.75
Equation 2: q = 4d + 3
We can solve this system of equations to find the values of "q" and "d".
Using substitution or elimination method, let's solve the system of equations:
From Equation 2, we can express "q" in terms of "d":
q = 4d + 3
Substituting this expression for "q" into Equation 1, we have:
0.25(4d + 3) + 1d = 34.75
Now, we can solve for "d":
1 + d + 0.75d = 34.75
1.75d = 34.75 -1
1.75d = 33.75
Dividing both sides by 1.75, we get:
d = 33.75 / 1.75
d ≈ 19.2857
Since the number of coins cannot be fractional, we can approximate "d" to the nearest whole number:
d ≈ 19
Now that we know the value of "d", we can substitute it back into Equation 2 to find the value of "q":
q = 4d + 3
q = 4(19) + 3
q = 76 + 3
q = 79
Therefore, there are approximately 19 dollar coins and 79 quarters in the jar.