total $3.25 worth of coins nickels, dimes, quarters 2morenickels than dimes 1 more quarter than nickels how many of each coin
5n+10d+25q = 325
n = d+2
q = n+1
Now just start substituting in to solve for the variables.
To solve this problem, we can set up a system of equations based on the given information.
Let's denote:
N = number of nickels
D = number of dimes
Q = number of quarters
From the given information, we know:
1) The total value is $3.25, which we can write as 0.05N + 0.10D + 0.25Q = 3.25.
2) There are 2 more nickels than dimes, so we can write N = D + 2.
3) There is 1 more quarter than nickels, so we can write Q = N + 1.
Now, let's substitute N = D + 2 and Q = N + 1 into the first equation to solve for the values of N, D, and Q.
0.05N + 0.10D + 0.25Q = 3.25
Substituting the equations for N and Q:
0.05(D + 2) + 0.10D + 0.25(D + 1) = 3.25
Simplifying and solving the equation:
0.05D + 0.10 + 0.05 + 0.10D + 0.25D + 0.25 = 3.25
0.40D + 0.40 = 3.25
0.40D = 2.85
D = 7.125
Since the number of coins must be a whole number, D = 7.
Now, substitute D = 7 back into the equation N = D + 2 to find N:
N = 7 + 2
N = 9
Finally, substitute N = 9 into the equation Q = N + 1 to find Q:
Q = 9 + 1
Q = 10
Therefore, there are 9 nickels, 7 dimes, and 10 quarters.