A man has three times as many quarters as half-dollars, four times as many dimes as quarters, and two more nickels than dimes. If he has $9.25, how many of each coin has he?
50h + 25(3h) + 10(12h) + 5(12h + 2) = 925
50h + 75h + 120h + 60h + 10 = 925
305h = 915
To solve this problem, let's assign variables to represent the number of each type of coin.
Let:
- "Q" be the number of quarters,
- "H" be the number of half-dollars,
- "D" be the number of dimes,
- "N" be the number of nickels.
Now, let's set up the equations based on the given information:
1) "A man has three times as many quarters as half-dollars."
This can be written as: Q = 3H.
2) "Four times as many dimes as quarters."
This can be written as: D = 4Q.
3) "Two more nickels than dimes."
This can be written as: N = D + 2.
Finally, we know that the total value of the coins is $9.25.
The value of each coin, in cents, is:
- Quarter: 25 cents,
- Half-dollar: 50 cents,
- Dime: 10 cents,
- Nickel: 5 cents.
So, the equation based on the value is:
25Q + 50H + 10D + 5N = 925 cents.
Now, let's substitute the values from equations (1), (2), and (3) into the value equation and solve the system of equations.
Replace Q with 3H:
25(3H) + 50H + 10D + 5N = 925
Replace D with 4Q:
25(3H) + 50H + 10(4Q) + 5N = 925
Replace N with D + 2:
25(3H) + 50H + 10(4Q) + 5(D + 2) = 925
Simplify the equation:
75H + 50H + 40Q + 5D + 10 = 925
Combine like terms:
125H + 40Q + 5D + 10 = 925
Now we have a system of equations:
Q = 3H (Equation 1)
D = 4Q (Equation 2)
N = D + 2 (Equation 3)
125H + 40Q + 5D = 915 (Equation 4)
Now we can use substitution or elimination method to solve the system of equations.
We can start by solving Equation 2 for Q in terms of D:
D = 4Q
Q = D/4
Now substitute this value of Q into Equation 1 and Equation 3:
Q = 3H (Equation 1)
D/4 = 3H
Q = D/4 = 3H (Equation 5)
N = D + 2 (Equation 3)
To simplify Equation 4, let's substitute Q with D/4 from Equation 5:
125H + 40(D/4) + 5D = 915
Simplify Equation 4:
125H + 10D + 5D = 915
125H + 15D = 915
Now, let's solve Equation 5 and Equation 3 simultaneously to find the values of D, Q, and N:
D/4 = 3H (Equation 5)
N = D + 2 (Equation 3)
Based on Equation 5, we can write H in terms of D:
H = (D/4)/3
H = D/12
Now substitute this value of H into Equation 3:
N = D + 2
Now we can further simplify Equation 4:
125(D/12) + 15D = 915
Simplify the equation:
125D/12 + 15D = 915
Multiply the equation by the common denominator, 12:
125D + 180D = 915 * 12
Combine like terms:
305D = 10980
Solve for D:
D = 10980 / 305 ≈ 36
Now substitute the value of D back into Equations 3 and 5 to find N and H:
N = D + 2
N = 36 + 2 = 38
H = D/12
H = 36/12 = 3
Finally, substitute the values of H, D, and N into Equation 1 to find Q:
Q = 3H
Q = 3(3) = 9
Therefore, the man has 9 quarters, 3 half-dollars, 36 dimes, and 38 nickels.