Final answer:
To solve this problem, we can set up a system of equations. Let's assume that the number of nickels is N, the number of dimes is D, and the number of quarters is Q. We can substitute the value of Q from equation 3 into equations 1 and 2 to eliminate Q. By solving the resulting equations, we find that there are 51 nickels, 40 dimes, and 61 quarters.
Step-by-step explanation:
To solve this problem, we can set up a system of equations. Let's assume that the number of nickels is N, the number of dimes is D, and the number of quarters is Q. We can write the following equations based on the information given:
1. N + D + Q = 131 (the total number of coins)
2. 0.05N + 0.10D + 0.25Q = 21.55 (the total value of the coins in dollars)
3. Q = D + 10 (there are 10 more quarters than dimes)
We can substitute the value of Q from equation 3 into equations 1 and 2 to eliminate Q:
N + D + (D + 10) = 131
0.05N + 0.10D + 0.25(D + 10) = 21.55
Simplifying these equations, we get:
2D + 10 = 131 - N
0.05N + 0.10D + 0.25D + 2.50 = 21.55
Multiplying the second equation by 100 to eliminate decimals and further simplifying, we have:
5N + 10D + 25D + 250 = 2155
Combining like terms, we get:
5N + 35D = 1905
Now, we can substitute the value of N from the first equation into the simplified second equation:
5(131 - D - 10) + 35D = 1905
Expanding and simplifying, we get:
655 - 5D - 50 + 35D = 1905
Combining like terms, we have:
30D = 1200
Dividing both sides by 30, we find that D = 40. Substituting this value of D back into the first equation, we can find the value of N:
N + 40 + (40 + 10) = 131
N + 80 = 131
N = 51
Therefore, there are 51 nickels, 40 dimes, and 51 + 10 = 61 quarters.