Answer:
25 quarters, 23 dimes, and 0 nickels.
Explanation:
Let n = number of nickels, q = number of quarters, and d = number of dimes
n + d = q - 2
n + d + q = 48
8.05 = .25q + .1d + .05n
so
(q-2) + q = 48
2q = 50
q = 25
Using that q value:
n + d = q - 2 --> n + d = 25 - 2 --> n + d = 23 --> n = 23 - d
Using that n value and q value:
8.05 = .25(23) + .1d + .05(23 - d)
8.05 = 5.75 + .1d + 1.15 - .05d
2.30 = 1.15 + .05d
1.15 = .05d
23 = d
Using that d value and q value:
n + d + q = 48 --> n + 23 + 25 = 48
n + 48 = 48
n = 0
Therefore there are 25 quarters, 23 dimes, and 0 nickels.
Checking that with:
8.05 = .25q + .1d + .05n
8.05 = .25(23) + .1(23) + .05(0)
8.05 = 5.75 + 2.30 + 0
8.05 = 8.05 ✓
