Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes was tripled and the number of nickels was decreased by 49, the value of the coins would be $14.05. How many nickels and dimes does he have?

Question

anonymous · Answer

Let:
n = number of nickles
d = number of dimes

we know from the problem statement that:
0.05n + 0.1d = 6.10
and
0.05(n-49) + 0.1*(3d) = 14.05

To solve, we have to solve one equation for one variable (doesn't matter which):
0.05n + 0.1d = 6.10
n = (6.10 - 0.1d)/0.05
n = 122 - 2d

then substitute that into the other equation and solve:
0.05(n-49) + 0.1*(3d) = 14.05
0.05([122-2d]-49) + 0.1*(3d) = 14.05
6.10 - 0.1d - 2.45 + 0.3d = 14.05
.2d = 10.4
d = 52

then plug that into any of the previous equations to find n:
n = 122 - 2d
n = 122 - 2(52)
n = 18