Answer:
at least three are dimes
so we have 20 coins with a total of 3.50
if the rest are all quarters then there would be 4*3+2=14 but we need 20
so we are 6 short
we must have dimes in multiples of 5
12 q + 5 d=3.50 but only 17 coins
10q +10 dimes=3.50 with 20 coins
10 q +13 dimes total
algebraic solution
d+q=23
25q+10d=380
d=23-q
