A box contains $7.80 in nickels, dimes, and quarters. There are 46 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?
5n+10d+25q = 780
n+d+q = 46
n+d = q-2
Now just solve.
56456
To solve this problem, let's assign variables to represent the number of nickels, dimes, and quarters in the box.
Let's say:
- N represents the number of nickels
- D represents the number of dimes
- Q represents the number of quarters
We are given the following information:
1) The total value of the coins is $7.80, and we know the values of each coin:
- Each nickel is worth $0.05
- Each dime is worth $0.10
- Each quarter is worth $0.25
We can use this information to write an equation for the value of the coins in terms of our variables:
0.05N + 0.10D + 0.25Q = 7.80
2) There are 46 coins in total, so the sum of the number of nickels, dimes, and quarters must be equal to 46:
N + D + Q = 46
3) The sum of the numbers of nickels and dimes is 2 less than the number of quarters:
N + D = Q - 2
Now we have a system of equations. Let's solve it to find the values of N, D, and Q.
We can use the substitution method to solve this system of equations. Starting with equation 3), we can rewrite it as:
N + D - Q = -2
Now we can rearrange equations 2) and 4) to solve for N and D in terms of Q:
N = 46 - D - Q
D = Q - N - 2
Substituting these expressions into equation 1) gives us:
0.05(46 - D - Q) + 0.10(Q - N - 2) + 0.25Q = 7.80
We can simplify and solve this equation to find the value of Q. Then we can substitute this value back into equations 2) and 4) to find the values of N and D.
I will now solve this equation to find the values of Q, N, and D.