If d is the number of dimes, then there are d-3 quarters, so 5.55=0.05n+0.1d+0.25(d-3), where n=number of nickels. So 5.55=0.05n+0.35d-0.75; and n+d+d-3=40. Therefore 0.05n+0.35d=6.30 and n+2d=43. Multiply the first equation by 100: 5n+35d=630; 5(43-2d)+35d=630 substituting for n; 215-10d+35d=630. 25d=415, so d=415/25=83/5=16.6, which is not a whole number. If we take d=16, then this is $1.60, and 13*0.25=$3.25, making $4.85 with 29 coins. This is a shortfall of $0.70 which is 14 nickels, making 43 coins altogether, not 40.

If the question should have read "3 times less quarters than dimes": 5.55=0.05n+0.1d+0.25d/3 and n+4d/3=40. So 16.65=0.15n+0.3d+0.25d and 3n+4d=120; 16.65=0.15n+0.55d; 1665=15n+55d=15(120-4d)/3+55d; 4995=1800-60d+165d; 105d=3195 and d=30.43, which is not much better than the first solution! There would be 30 dimes ($3), 10 quarters ($2.50) and 1 nickel ($0.05), a total of 41 coins.